Chapter 7 Geometrical Optics Introduction:
Blue lakes, ochre deserts, green forests, and multicolored rainbows can be enjoyed by anyone who has eyes with which to see them. But by studying the branch of physics called optics, which deals with the behavior of light and other electromagnetic waves, we can reach a deeper appreciation of the visible world.
A knowledge of the properties of light allows us to understand the blue color of the sky and the design of optical devices such as telescopes, microscopes, cameras, eyeglasses, and the human eye.
The same basic principles of optics also lie at the heart of modern developments such as the laser, optical fibers, holograms, optical computers, and new techniques in medical imaging.
1. Condition For Rectilinear Propagation Of Light
Some parts of the optics can be understood if we assume that light travels in a straight line and bends abruptly when it suffers reflection or refraction.
The assumption that the light travels in a straight line is correct if
- The medium is isotropic, i.e. its behavior is the same in all directions and
- The obstacle past which the light moves or the opening through which the light moves is not very small.
Geometrical Optics NEET Physics Class 12 Study Notes
Consider a slit of width ‘a’ through which monochromatic light rays pass and strike a screen, placed at a distance D as shown.
It is found that the light strikes in a band of width ‘b’ more than ‘a’. This bending is called diffraction.
Light bends by (b-a)/2 on each side of the central line. It can be shown by the wave theory of light that \(\sin \theta=\frac{\lambda}{\mathrm{a}} \ldots \ldots. (\mathrm{A})\).
where θ is shown in the figure.
This formula indicates that the bending is considerable only when a ≅ λ. Diffraction is more pronounced in sound because its wavelength is much more than that of light and it is of the order of the size of obstacles or apertures.
Formula (A) gives \(\frac{\mathrm{b}-\mathrm{a}}{2 \mathrm{D}} \approx \frac{\lambda}{\mathrm{a}}\)
It is clear that the bending is negligible if \(\frac{D \lambda}{a} \ll a \text { or } a \gg \sqrt{D \lambda}\).
If this condition is fulfilled, light is said to move rectilinearly. In most of the situations including geometrical optics, the conditions are such that we can safely assume that light moves in a straight line and bends only when it gets reflected or refracted.
∴ \(\frac{\mathrm{b}-\mathrm{a}}{2 \mathrm{D}} \approx \frac{\lambda}{\mathrm{a}}\)
Thus geometrical optics is an approximate treatment in which the light waves can be represented by straight lines which are called rays.
A ray of light is the straight line path of transfer of light energy. The arrow represents the direction of propagation of light.
The figure shows a ray which indicates light is moving from A to B.
2. Properties Of Light
The speed of light in a vacuum, denoted by c, is equal to 3 x 108 m/s approximately.
Light is an electromagnetic wave (proposed by Maxwell). It consists of varying electric fields and magnetic fields.
Light carries energy and momentum.
The formula v = fλ is applicable to light.
When the light gets reflected in the same medium, it suffers no change in frequency, speed, and wavelength.
The frequency of light remains unchanged when it gets reflected or refracted.
3. Reflection Of Light
When light rays strike the boundary of two media such as air and glass, a part of light is turned back into the same medium. This is called the Reflection of Light.
Regular Reflection:
When the reflection takes place from a perfect plane surface it is called Regular Reflection. In this case, the reflected light has a large intensity in one direction and a negligibly small intensity in other directions.
Diffused Reflection
When the surface is rough, we do not get a regular behavior of light.
Although at each point light rays get reflected irrespective of the overall nature of the surface, the difference is observed because even in a narrow beam of light many rays are reflected from different points of the surface and it is quite possible that these rays may move in different directions due to irregularity of the surface. This process enables us to see an object from any position.
Such a reflection is called a diffused reflection.
For example reflection from a wall, from a newspaper, etc. This is why you can not see your face in the newspaper and in the wall.
3.1 Laws Of Reflection
The incident ray, the reflected ray, and the normal at the point of incidence lie in the same plane. This plane is called the plane of incidence (or plane of reflection).
This condition can be expressed mathematically as \(R.\overrightarrow{\mathrm{R}} \cdot(\overrightarrow{\mathrm{I}} \times \overrightarrow{\mathrm{N}})=\overrightarrow{\mathrm{N}} \cdot(\overrightarrow{\mathrm{I}} \times \overrightarrow{\mathrm{R}})=\overrightarrow{\mathrm{I}} \cdot(\overrightarrow{\mathrm{N}} \times \overrightarrow{\mathrm{R}})=0 \text { where } \overrightarrow{\mathrm{I}}, \overrightarrow{\mathrm{N}} \text { and } \overrightarrow{\mathrm{R}}\) ≠ vectors of any magnitude along incident ray, the normal and the reflected ray respectively.
The angle of incidence (the angle between the normal and the incident ray) and the angle of reflection (the angle between the reflected ray and the normal) are equal, i.e.
Special Cases:
Normal Incidence: In case the light is incident normally, i = r = 0
Note: We say that the ray has retraced its path.
Grazing Incidence: In case light strikes the reflecting surface tangentially, i = r = 90 8 = 0° or 360°
Note: In the case of reflection, the speed (magnitude of velocity) of light remains unchanged but in Grazing incidence velocity remains unchanged.
Example: Show that for a light ray incident at an angle ‘i’ on getting reflected the angle of deviation is δ = π – 2i or π + 2i.
Solution:
From Figure (2) It is clear that light rays bend either by δ1 anticlockwise or by δ2 (= 2π – δ1) clockwise.
From Figure (1) δ1 = π – 2i.
∴ δ1 = π + 2i .
Class 12 NEET Geometrical Optics Notes
3.2 Object And Image
(1)Object (O): Object is defined as a point of intersection of incident rays.
Let us call the side in which incident rays are present the incident side and the side in which reflected (refracted) rays are present, the reflected (refracted) side.
Note: An object is called real if it lies on the incident side otherwise it is called virtual.
(2) Image (I): Image is defined as a point of intersection of reflected rays (in case of reflection) or refracted rays (in case of refraction).
Note: An image is called real if it lies on the reflected or refracted side otherwise it is called virtual.
4. Plane Mirror
A plane mirror is formed by polishing one surface of a plane thin glass plate. It is also said to be silvered on one side.
A beam of parallel rays of light, incident on a plane mirror will get reflected as a beam of parallel reflected rays.
Example: For a fixed incident light ray, if the mirror is rotated through an angle θ (about an axis which lies in the plane of mirror and perpendicular to the plane of incidence), show that the reflected ray turns through an angle 2θ in the same sense.
Solution:
M1, N1, and R1 indicate the initial position of the mirror, the initial normal, and the initial direction of the reflected light rays respectively. M2, N2, and R2 indicate the final position of the mirror, the final normal, and the final direction of reflected light rays respectively. From the figure, it is clear that ABC = 2Φ + δ = 2(Φ + θ) or δ = 2θ.
Note: Keeping the mirror fixed if the incident ray is rotated by an angle θ then the reflected ray rotates by the same angle in the opposite direction of rotation.
4.1 Point object: Characteristics Of Image Due To Reflection By A Plane Mirror
- Distance of object from mirror = Distance of image from the mirror.
- All the incident rays from a point object will meet at a single point after reflection from a plane mirror which is called an image.
- The line joining a point object and its image is normal to the reflecting surface.
- For a real object, the image is virtual and for a virtual object the image is real
- The region in which the observer’s eye must be present in order to view the image is called the field of view.
Example: Shows a point object A and a plane mirror MN. Find the position of the image of object A, in mirror MN, by drawing a ray diagram. Indicate the region in which the observer’s eye must be present in order to view the image. (This region is called the field of view).
Solution:
Consider any two rays emanating from the object. N1 and N2 are normals; i1 = r1 and i2 = r2
The meeting point of reflected rays R1 and R2 is image A’. Though only two rays are considered it must be understood that all rays from A reflect from the mirror such that their meeting point is A’.
To obtain the region in which reflected rays are present, join A’ with the ends of the mirror and extend. The following figure shows this region as shaded.
In the figure there are no reflected rays beyond the rays 1 and 2, therefore the observers P and Q cannot see the image because they do not receive any reflected rays.
4.2 Extended Object:
An extended object like AB shown in the figure is a combination of an infinite number of point objects from A to B. Image of every point object will be formed individually and thus infinite images will be formed.
A’ will be an image of A, C’ will be an image of C, B’ will be an image of B, etc. All point images together form an extended image. Thus, an extended image is formed of an extended object.
Properties Of Image Of An Extended Object, Formed By A Plane Mirror:
1. Size of extended object = size of the extended image.
2. The image is erect if the extended object is placed parallel to the mirror.
3. The image is inverted if the extended object lies perpendicular to the plane mirror.
If an extended horizontal object is placed in front of a mirror inclined 45° with the horizontal, the image formed will be vertical.
Example: Show that the minimum size of a plane mirror, required to see the full image of an observer is half the size of that observer.
Solution:
It is self-explanatory if you consider lengths ‘x’ and ‘y’ as shown in the figure.
Aliter: ΔE M1, M2, and ΔE H’F’ are similar
∴ \(\frac{M_1 M_2}{H^{\prime} F^{\prime}}=\frac{z}{2 z} \text { or } \quad M_1 M_2=H^{\prime} F^{\prime} / 2=H F / 2\)
4.3 Relation Between Velocity Of Object And Image: From mirror property: Xim = – Xom, yim = yom and zim = zom
Here xim means the ‘x’ coordinate of the image with respect to the mirror. Similarly, others have meaning.
Differentiating w.r.t time, we get \(\mathrm{v}_{(\mathrm{im}) \mathrm{x}}=-\mathrm{v}_{(\mathrm{om}) \mathrm{x}} ; \quad \mathrm{v}_{(\mathrm{im}) \mathrm{y}}=\mathrm{v}_{(\mathrm{om}) \mathrm{y}} ; \quad \mathrm{v}_{(\mathrm{im}) \mathrm{z}}=\mathrm{v}_{(\mathrm{om}) \mathrm{z}} \text {, }\)
⇒ for \(x \text { axis } \quad v_{i G}-v_{m G}=-\left(v_{o G}-v_{m G}\right)\)
but for y-axis and z-axis \(v_{i G}-v_{m G}=\left(v_{O G}-v_{m G}\right) \text { or } \quad v_{i G}=v_{O G}\)
here: ViG = Velocity of the image with respect to ground.
4.3 Relation Between Velocity Of Object And Image Solved Examples
Example 1. An object moves with 5 m/s towards the right while the mirror moves with 1m/s towards the left as shown. Find the velocity of the image.
Solution: Take → as + direction.
vi – vm = vm – v0
vi – (-1) = (-1) – 5
vi = – 7m/s.
7 m/s and direction towards the left.
Example 2. There is a point object and a plane mirror. If the mirror is moved by 10 cm away from the object find the distance at which the image will move.
Solution:
We know that \(x_{i m}=-x_{o m}\) or \(x_i-x_m=x_m-x_0\) or \(\Delta x_1-\Delta x_m=\Delta x_m-\Delta x_0\).
In this Q. \(\Delta x_o=0 ; \Delta x_m=10 \mathrm{~cm}\).
Therefore \(\Delta x_1=2 \Delta x_m-\Delta x_o=20 \mathrm{~cm}\).
NEET Physics Chapter 7 Geometrical Optics Summary
4.4 Images Formed By Two Plane Mirrors:
If rays after getting reflected from one mirror strike the second mirror, the image formed by first mirror will function as an object for the second mirror, and this process will continue for every successive reflection.
4.4 Images Formed By Two Plane Mirrors Solved Examples
Example 1. The figure shows a point object placed between two parallel mirrors. Its distance from M1 is 2 cm and that from M2 is 8 cm. Find the distance of images from the two mirrors considering reflection on mirror M1 first.
Solution:
To understand how images are formed see the following figure and table. You will be required to know what symbols like I121 stand for. See the following diagram.
Similarly, images will be formed by the rays striking mirror M2 first. Total number of images = ∞
Example 2. Consider two perpendicular mirrors. M1 and M2 and a point object O. Taking the origin at the point of intersection of the mirrors and the coordinate of the object as (x, y), find the position and number of images.
Solution:
Rays ‘a’ and ‘b’ strike mirror M, only and these rays will form image I1 at (x1 -y), such that O and I1 are equidistant from mirror M1. These rays do not form further images because they do not strike any mirror again.
Similarly, rays ‘d’ and ‘e’ strike mirror M2 only and these rays will form image l2 at (-x, y), such that O and l2 are equidistant from mirror M2.
Now consider those rays which strike mirror M2 first and then the mirror M1
For incident ray 1,2 the object is O, and reflected rays 3, and 4 form image l2.
Now rays 3, and 4 incident on M1 (object is l2) which reflect as rays 5, and 6 and form image I21. Rays 5, and 6 do not strike any mirror, so image formation stops.
I2 and I21, are equidistant from M1. To summarize see the following figure
For rays reflecting first from M1 and then from M2, the first image I1 (at (x1 -y)) will be formed and this will function as an object for mirror M2 and then its image I12 (at (-x, -y)) will be formed.
I12 and I21 coincide.
∴ A total of three images are formed
4.5 Locating All The Images Formed By Two Plane Mirrors
Consider two plane mirrors M1 and M2 inclined at an angle θ = α+β as shown in the figure.
Point P is an object kept such that it makes angle α with mirror M1 and angle β with mirror M2. The image of object P formed by M1, denoted by I1, will be inclined by angle α on the other side of mirror M1. This angle is written in brackets in the figure beside I1.
Similarly, the image of object P formed by M2, denoted by I2, will be inclined by angle β on the other side of mirror M2. This angle is written in brackets in the figure beside I2.
Now I2 will act as an object for M1 which is at an angle (α+2β) from M1. Its image will be formed at an angle (α+2β) on the opposite side of M1. This image will be denoted as I21, and so on. Think when this process will stop.
Hint: The virtual image formed by a plane mirror must not be in front of the mirror or its extension.
Number Of Images Formed By Two Inclined Mirrors
(1) If \(\frac{360^{\circ}}{\theta}=\) even number; number of image \(=\frac{360^{\circ}}{\theta}-1\)
(2) If \(\frac{360^{\circ}}{\theta}=\) odd number; the number of images \(=\frac{360^{\circ}}{\theta}-1\), if the object is placed on the angle bisector.
(3) If \(\frac{360^{\circ}}{\theta}=\) odd number; the number of images \(=\frac{360^{\circ}}{\theta}\), if the object is not placed on the angle bisector.
(4) If \(\frac{360^{\circ}}{\theta} \neq\) integer, then count the number of images as explained above.
Question 1. What should be the angle between two plane mirrors so that whatever the angle of incidence, the incident ray and the reflected ray from the two mirrors be parallel to each other
- 60°
- 90°
- 120°
- 175°
Answer: 1. 60°
Question 2. A light bulb is placed between two plane mirrors inclined at an angle of 60°. The number of images formed are
- 6
- 2
- 5
- 4
Answer: 2. 2
Question 3. A ray of light incidents on a plane mirror at an angle of 30°. The deviation produced in the ray is
- 30°
- 60°
- 90°
- 120°
Answer: 3. 90°
Question 4. A man runs towards the mirror at a speed of 15m/s. What is the speed of his image?
- 7.5 m/s
- 15 m/s
- 30 m/s
- 45 m/s
Answer: 4. 15 m/s
5. Spherical Mirrors
A Spherical Mirror is formed by polishing one surface of a part of a sphere. Depending upon which part is shining the spherical mirror is classified as (1) a Concave mirror, if the side towards the center of curvature is shining, and (2) a Convex mirror if the side away from the center of curvature is shining.
5.1 Important Terms Related With Spherical Mirrors:
A spherical shell with the center of curvature, pole aperture, and radius of curvature identified
Geometrical Optics NEET Physics Chapter 7 Key Concepts
- Center of Curvature (C): The center of the sphere from which the spherical mirror is formed is called the Center of curvature of the mirror. It is represented by C and is indicated in the figure.
- Pole (P): The center of the mirror is called the Pole. It is represented by the point P on the mirror APB in the figure.
- Principal Axis: The Principal Axis is a line that is perpendicular to the plane of the mirror and passes through the pole. The Principal Axis can also be defined as the line that joins the Pole to the Center of Curvature of the mirror.
- Aperture (A): The aperture is the segment or area of the mirror which is available for reflecting light. In figure. APB is the aperture of the mirror.
- Principle focus (F): It is the point of intersection of all the reflected rays for which the incident rays strike the mirror (with a small aperture) parallel to the principal axis. In the concave mirror it is real and in the convex mirror it is virtual. The distance from pole to focus is called focal length.
5.1 Important Terms Related With Spherical Mirrors Solved Examples
Example 1. Find the angle of incidence of the ray for which it passes through the pole, given that MI || CP.
Solution:
∠MIC = ∠CIP = θ
MI || CP ∠MIC = ∠ICP = θ
CI = CP
∠CIP = ∠CPI = θ
∴ In ΔCIP all angles are equal 3θ = 180°
⇒ θ = 60°
Example 2. Find the distance CQ if the incident light ray parallel to the principal axis is incident at an angle i. Also, find the distance CQ if i → 0.
Solution:
⇒ \(\cos \mathrm{i}=\frac{\mathrm{R}}{2 \mathrm{CQ}} \quad \Rightarrow \quad \mathrm{CQ}=\frac{\mathrm{R}}{2 \cos \mathrm{i}}\)
So, paraxial rays meet at a distance equal to R / 2 from the center of curvature, which is called the focus.
6. Ray tracing
6.1 Ray tracing: The following facts are useful in ray tracing.
1. If the incident ray is parallel to the principal axis, the reflected ray passes through the focus.
2. If the incident ray passes through the focus, then the reflected ray is parallel to the principal axis.
3. An incident ray passing through the center of curvature will be reflected back through the center of curvature (because it is a normal incident ray).
4. It is easy to make the ray tracing of a ray incident at the pole as shown below.
6.2 Sign Convention
We are using the coordinate sign convention.
Take origin at the pole (in case of the mirror) or at the optical center (in case of the lens)
Take the X axis along the principal axis, taking a positive direction along the incident light. u, v, R, and f indicate the x coordinate of the object, image, center of curvature, and focus respectively.
y-coordinate is taken positive above Principle Axis and negative below Principle Axis’ h1 and h2 denote the y coordinate of object and image respectively.
Note: This sign convention is used for reflection from mirror, and refraction through flat or curved surfaces or lenses.
6.3 Formulae For Reflection From Spherical Mirrors:
Mirror formula: \(\frac{1}{v}+\frac{1}{u}=\frac{2}{R}-\frac{1}{f}\)
The x-coordinate of the center of the Curvature and the focus of the Concave mirror are negative and those for the Convex mirror are positive. In the case of mirrors since light rays reflect back in the X-direction, therefore -ve sign of v indicates the real image and the +ve sign of v indicates the virtual image.
6.3 Formulae For Reflection From Spherical Mirrors Solved Examples
Example 1. The figure shows a spherical concave mirror with its pole at (0, 0) and principle axis along the x-axis. There is a point object at (–40 cm, 1cm), find the position of the image.
Solution:
According to sign convention, u = –40 cm
h1 = +1 cm
f = – 5 cm.
⇒ \(\frac{1}{v}+\frac{1}{u}=\frac{1}{f}\)
⇒ \(\frac{1}{v}+\frac{1}{-40}=\frac{1}{-5} ; v=\frac{-40}{7} \mathrm{~cm} . ; \frac{h_2}{h_1}=\frac{-v}{u}\)
⇒ \(h_2=-\frac{-v}{u} \times h_1=\frac{-\left(-\frac{40}{7}\right) \times 1}{-40}=-\frac{1}{7} \mathrm{~cm}\)
∴ The position of image is \(\left(\frac{-40}{7} \mathrm{~cm},-\frac{1}{7} \mathrm{~cm}\right)\)
Example 2. Converging rays are incident on a convex spherical mirror so that their extensions intersect 30 cm behind the mirror on the optical axis. The reflected rays form a diverging beam so that their extensions intersect the optical axis 1.2 m from the mirror. Determine the focal length of the mirror.
Solution:
In this case u = + 30
⇒ v = + 120
∴ \(\frac{1}{\mathrm{f}}=\frac{1}{v}+\frac{1}{u}=\frac{1}{120}+\frac{1}{30}\)
f = 24 cm
Example 3. Find the position of the final image after three successive reflections taking the first reflection on m1.
Solution:
I reflection: Focus of mirror = – 10 cm ⇒ u = – 15 cm
Applying mirror formula: \(\frac{1}{v}+\frac{1}{u}=\frac{1}{\mathrm{f}}\) v = -30 cm
For 2 Reflection On Plane Mirror: u = – 10 cm .-. v = 10 cm
For 3 Reflection On Curved Mirror Again: u = – 50 cm ; f = – 10 cm
Applying mirror formula: \(\frac{1}{v}+\frac{1}{u}=\frac{1}{f}\); v = -12.5 cm
Lateral Magnification (or transverse magnification) denoted by m is defined as m = and is related as \(\mathrm{m}=\frac{\mathrm{h}_2}{\mathrm{~h}_1}\).
From the definition of m positive sign of m indicates an erect image and a negative sign indicates an inverted image.
In the case of successive reflection from mirrors, the overall lateral magnification is given by m1 x m2 x m3 ……., where m1, m2, etc. are lateral magnifications produced by individual mirrors.
h1 and h2 denote the y coordinate of the object and image respectively.
Note: Using the following conclusions can be made.
we get m = \(\frac{f}{f-u}=\frac{f-v}{f}\)……(just a time saving formula)
Image Formed By The Concave Mirror
Focal Plane: A plane passing through focus and perpendicular to the principal axis is called the focal plane
Secondary Focus: Paraxial rays that are parallel to each other but not parallel to the principal axis will also meet at a single point in a focal plane after reflection from the spherical mirror (or refraction from the lens). That point is known as secondary focus.
NEET Physics Class 12 Chapter 7 Notes: Geometrical Optics
Example 1. An extended object is placed perpendicular to the principle axis of a concave mirror of a radius of curvature 20 cm at a distance of 15 cm from the pole. Find the lateral magnification produced.
Solution:
u = – 15 cm f = – 10 cm
Using \(\frac{1}{\mathrm{v}}+\frac{1}{\mathrm{u}}=\frac{1}{\mathrm{f}}\)we get, v = – 30 cm
∴ m = \(-\frac{v}{u}=-2 \text {. }\)
Aliter: \(\mathrm{m}=\frac{\mathrm{f}}{\mathrm{f}-\mathrm{u}}=\frac{-10}{-10-(-15)}=-2\)
Example 2. A person looks into a spherical mirror. The size of the image of his face is twice the actual size of his face. If the face is at a distance of 20 cm then find the nature of the radius of curvature of the mirror.
Solution:
The person will see his face only when the image is virtual. The virtual image of a real object is erect.
Hence m = 2
∴ \(\frac{-v}{u}=2 \quad \Rightarrow \quad v=40 \mathrm{~cm}\)
Applying \(\frac{1}{\mathrm{v}}+\frac{1}{\mathrm{u}}=\frac{1}{\mathrm{f}} ; \mathrm{f}=-40 \mathrm{~cm}\) or \(\mathrm{R}=80 \mathrm{~cm}\).
Aliter: \(m=\frac{f}{f-u} \quad \Rightarrow \quad 2=\frac{f}{f-(-20)}\)
f = \(-40 \mathrm{~cm}\) or \(R=80 \mathrm{~cm}\)
Example 3. An image of a candle on a screen is found to be double its size. When the candle is shifted by a distance of 5 cm then the image becomes triple its size. Find the nature and ROC of the mirror.
Solution:
Since the images formed on screen, it is real. Real object and real images imply concave mirrors.
Applying \(m=\frac{f}{f-u}\) or \(-2=\frac{f}{f-(u)}\)…….(1)
After shifting \(-3=\frac{f}{f-(u+5)}\)…..(2)
[Why u + 5 ?, why not u – 5: In a concave mirror are size of the real image will increase, only when the real object is brought closer to the mirror. In doing so, its x coordinate will increase]
From (1) and (2) we get, f = – 30 cm or R = 60 cm
Velocity Of Image
(1) Object Moving Perpendicular To Principal Axis: From the relation, we have \(\frac{\mathrm{h}_2}{\mathrm{~h}_1}=-\frac{\mathrm{v}}{\mathrm{u}} \text { or } \quad \mathrm{h}_2=-\frac{\mathrm{v}}{\mathrm{u}} \cdot \mathrm{h}_1\)
If a point object moves perpendicular to the principle axis, the x coordinate of both the object and the image becomes constant. On differentiating the above relation w.r.t. time , we get, \(\frac{\mathrm{dh}_2}{\mathrm{dt}}=-\frac{\mathrm{v}}{\mathrm{u}} \frac{\mathrm{dh}_1}{\mathrm{dt}}\)
Here, \(\frac{\mathrm{dh}_1}{\mathrm{dt}}\) denotes velocity of object perpendicular to the principle axis and \(\frac{\mathrm{dh}_2}{\mathrm{dt}}\) velocity of image perpendicular to the principle axis.
(2) Object Moving Along Principal Axis: On differentiating the mirror formula with respect to time we get \(\frac{d v}{d t}=-\frac{v^2}{u^2} \frac{d u}{d t}\) is the velocity of the image along Principle axis — and is the velocity of the object along principle axis.
A negative sign implies that the image, in the case of a mirror, always moves in the direction opposite to that of the object. This discussion is for velocity with respect to the mirror and along the x-axis.
(3) Object Moving At An Angle With The Principal Axis: Resolve the velocity of the object along and perpendicular to the principal axis find the velocities of the image in these directions separately and then find the resultant.
(4) Optical Power Of A Mirror (In Diopters) = \(\frac{1}{f}\)
f = focal length with a sign and in meters.
(5) If the object lying along the principle axis is not of very small size, the longitudinal magnification = \(\frac{v_2-v_1}{u_2-u_1}\) (it will always be inverted)
(6)If the size of the object is very small compared to its distance from Pole then
On differentiating the mirror formula we get \(\frac{d v}{d u}=-\frac{v^2}{u^2}\): Mathematically ‘du’ implies a small change in the position of the object and ‘dv’ implies corresponding small change in position of the image.
If a small object lies along the principal axis, du may indicate the size of the object and dv the size of its image along the principal axis (Note that the focus should not lie in between the initial and final points of the object).
In this case \(\frac{d u}{d v}\) is called longitudinal magnification. The negative sign indicates inversion of the image irrespective of the nature of the image and the nature of the mirror.
Example: A point object is placed 60 cm from the pole of a concave mirror of a focal length of 10 cm on the principle axis. Find
- The position of the image
- If the object is shifted 1 mm towards the mirror along the principle axis find the shift in the image. Explain the result.
Solution:
(1) u = – 60 cm; f = – 10cm
v = \(\frac{f u}{u-f}=\frac{-10(-60)}{-60-(-10)}=\frac{600}{-50}=-12 \mathrm{~cm}\).
(2) \(\frac{1}{v}+\frac{1}{u}=\frac{1}{\mathrm{f}}\)
Differentiating, we get \(d v=-\frac{v^2}{u^2} d u=-\left(\frac{-12}{-60}\right)^2[1 \mathrm{~mm}]=-\frac{1}{25} \mathrm{~mm}\)
[du = 1mm; sign of du is +ve because it is shifted in +ve direction defined by signconvention.]
-ve sign of dv indicates that the image will shift towards the negative direction.
The sign of v is negative. Which implies the image is formed on the negative side of the pole. (1) and (2) together imply that the image will shift away from the pole.
Note that differentials dv and du denote small changes only.
Newton’s Formula: XY = f²
X and Y are the distances (along the principal axis) of the object and image respectively from the principal focus. This formula can be used when the distances are mentioned or asked from the focus.
In the case of spherical mirrors if object distance (x) and image distance (y) are measured from focus instead of the pole, u = -(f + x) and v = – (f + y), by \(\frac{1}{v}+\frac{1}{u}=\frac{1}{f}\) we can write \(-\frac{1}{(f+y)}-\frac{1}{(f+x)}=-\frac{1}{f}\)
on solving xy = f²
This is Newton’s formula.
Question 1. A diminished virtual image can be formed only in
- Plane mirror
- A concave mirror
- A convex mirror
- Concave-parabolic mirror
Answer: 3. A convex mirror
Question 2. An object 5 cm tall is placed 1m from a concave spherical mirror which has a radius of curvature of 20 cm. The size of the image is
- 0.11 cm
- 0.50 cm
- 0.55 cm
- 0.60 cm
Answer: 3. 0.55 cm
Question 3. In a concave mirror experiment, an object is placed at a distance x1 from the focus and the image is formed at a distance x2 from the focus. The focal length of the mirror would be
- \(x_1 x_2\)
- \(\sqrt{x_1 x_2}\)
- \(\frac{x_1+x_2}{2}\)
- \(\sqrt{\frac{x_1}{x_2}}\)
Answer: 2. \(\sqrt{x_1 x_2}\)
Question 4. Given a point source of light, which of the following can produce a parallel beam of light
- Convex mirror
- Concave mirror
- Concave lens
- Two plane mirrors inclined at an angle of 90°
Answer: 2. Concave mirror
7. Refraction Of Light
When the light changes its medium some changes occur in its properties the phenomenon is known as refraction.
If the light is incident at an angle (0 < i < 90) then it deviates from its actual path.
It is due to changes in the speed of light as light passes from one medium to another medium.
If the light is incident normally then it goes to the second medium without bending, but still, it is called refraction.
The refractive index of a medium is defined as the factor by which the speed of light reduces as compared to the speed of light in a vacuum. μ= \(\frac{c}{v}\) = \(\frac{\text{speed of light in vacuum}}{\text{speed of light in medium}}\).
More (less) refractive index implies less (more) speed of light in that medium, which therefore is called a denser (rarer) medium.
7.1 Laws of Refraction
(1) The incident ray, the normal to any refracting surface at the point of incidence, and the refracted ray all lie in the same plane called the plane of incidence or plane of refraction.
(2) \(\frac{sin i}{sin r}\) = Constant for any pair of media and for light of a given wavelength. This is known as Snll’s.
Also, \(\frac{\text{Sin} i}{\text{Sin} r}=\frac{\mathrm{n}_2}{\mathrm{n}_1}=\frac{v_1}{v_2}=\frac{\lambda_1}{\lambda_2}\)
For applying in problems remember n1 sini = n2sinr
⇒ \(\frac{\mathrm{n}_2}{\mathrm{n}_1}={ }_1 \mathrm{n}_2\) = Refractive Index of the second medium with respect to the first medium.
C = speed of light in air (or vacuum) = 3 x 108 m/s.
Special Cases:
(1) Normal incidence: i = 0
From Snell’s law: r = 0
(2) When light moves from denser to rarer medium it bends away from normal.
(3) When light moves from rarer to denser medium it bends towards the normal.
Note:
(1) The higher the value of R.I., the denser (optically) is the medium.
(2) Frequency of light does not change during refraction.
(3) Refractive index of the medium relative to vacuum = \(\sqrt{\mu_r \epsilon_r}\) \(n_{\text {vaoum}}\)
= \(1 ; n_{\text {atr }}=51; n_{\text {mater }} \text { (average value }=4 / 3 ; n_{\text {glass }} \text { (average value) }=3 / 2\)
7.2 Deviation of a Ray Due to Refraction: Deviation (δ) of ray incident
at ∠i and refracted at ∠r is given by δ = |i− r|.
Geometrical Optics Notes for NEET Physics
Example 1. A light ray is incident on a glass sphere at an angle of incidence 60° as shown. Find the angles r, r’,e, and the total deviation after two refractions.
Solution:
Applying Snell’s law 1sin60° = sinr ⇒ r = 30°
From symmetry r’ = √3 r = 30°.
Again applying Snell’s law at the second surface 1sin e = sinr ⇒ e = 60°
Deviation at first surface = i – r = 60° – 30° = 30°
Deviation at second surface = e – r’ = 60° – 30° = 30°
Therefore total deviation = 60°.
Example 2. Find the angle θa made by the light ray when it gets refracted from water to air, as shown figure.
Solution:
Snell’s Law \(\mu_w \sin \theta_w=\mu_a \sin \theta_a ; \frac{4}{3} \times \frac{3}{5}=1 \sin \theta_a ; \sin \theta_a=\frac{4}{5} ; \theta_a=\sin ^{-1} \frac{4}{5}\)
Example 3. Find the speed of light in medium ‘a’ if the speed of light in medium ‘b’ is where c = speed of light in vacuum and light refracts from medium ‘a’ to medium ‘b’ making 45º and 60º respectively with the normal.
Solution:
Snell’s Law \(\mu_{\mathrm{a}} \sin \theta_{\mathrm{a}}=\mu_{\mathrm{b}} \sin \theta_{\mathrm{b}} \quad ; \quad \frac{\mathrm{c}}{\mathrm{v}_{\mathrm{a}}} \sin \theta_{\mathrm{a}}=\frac{\mathrm{c}}{\mathrm{v}_{\mathrm{b}}} \sin \theta_{\mathrm{b}}\)
∴ \(\frac{\mathrm{c}}{\mathrm{v}_{\mathrm{a}}} \sin 45^{\circ}=\frac{\mathrm{c}}{\mathrm{c} / 3} \sin 60^{\circ} . \quad ; \quad \mathrm{v}_{\mathrm{a}}=\frac{\sqrt{2} \mathrm{c}}{3 \sqrt{3}}\)
7.3 Principle of Reversibility of Light Rays
- A ray traveling along the path of the reflected ray is reflected along the path of the incident ray.
- A refracted ray reversed to travel back along its path will get refracted along the path of the incident ray. Thus the incident and refracted rays are mutually reversible.
8. Refraction Through A Parallel Slab
When light passes through a parallel slab, having the same medium on both sides, then
(1) The emergent ray is parallel to the incident ray.
Note: The emergent ray will not be parallel to the incident ray if the medium on both the sides of slab is different.
(2) Light is shifted laterally, given by (student should be able to derive it)
d = \(\frac{t \sin (i-r)}{\cos r}\); t = thickness of slab
Example 1. Find the lateral shift of the light ray while it passes through a parallel glass slab of thickness 10 cm placed in air. The angle of incidence in air is 60° and the angle of refraction in glass is 45°.
Solution:
d = \(\frac{\mathrm{t} \sin (\mathrm{i}-\mathrm{r})}{\cos \mathrm{r}}=\frac{10 \sin \left(60^{\circ}-45^{\circ}\right)}{\cos 45^{\circ}}\)
= \(\frac{10 \sin 15^{\circ}}{\cos 45^{\circ}}=10 \sqrt{2} \sin 15^{\circ}\).
8.1 Apparent Depth and Shift of Submerged Object
At near normal incidence (small angle of incidence i) apparent depth (d’) is given by:
d’ = \(\frac{d}{n_{\text {relative }}} \text { and } v^{\prime}=\frac{v}{n_{\text {reltive }}}\)
where \(n_{\text {rolethre }}=\frac{n_i \text { (R.I. of medium of incidence) }}{n_r \text { (R. I. of medium of refraction) }}\)
d = distance of the object from the interface = real depth
d’ = distance of the image from the interface = apparent depth
v = velocity of the object perpendicular to interface relative to the surface.
v´ = velocity of image perpendicular to interface relative to surface.
This formula can be easily derived using Snell’s law and applying the condition of nearly normal incidence…..
Apparent shift = \(\left(1-\frac{1}{\mathrm{n}_{\mathrm{rel}}}\right)\)
Example 1. An object lies 100 cm inside water. It is viewed from the air nearly normally. Find the apparent depth of the object.
Solution:
d’ = \(\frac{d}{n_{\text {relative }}}=\frac{100}{\frac{4 / 3}{1}}=75 \mathrm{~cm}\)
Example 2.
- Find the apparent height of the bird
- Find the apparent depth of fish
- At what distance will the bird appear to the fish?
- At what distance will the fish appear to the bird
- If the velocity of the bird is 12 cm/sec downward and the fish is 12 cm/sec in an upward direction, then find out their relative velocities with respect to each other.
Solution:
- \(d_B=\frac{36}{\frac{1}{\left(\frac{4}{3}\right)}}=\frac{36}{3 / 4}=48 \mathrm{~cm}\)
- \(d_F=\frac{36}{4 / 3}=27 \mathrm{~cm}\)
- For fish: \(\mathrm{d}_{\mathrm{B}}=36+48=84 \mathrm{~cm}\)
- For bird: \(\mathrm{d}_{\mathrm{F}}=27+36=63 \mathrm{~cm}\).
- Velocity of fish with respect to bird \(=12+\left(\frac{12}{4 / 3 / 1 / 1}\right)=21 \mathrm{~cm} / \mathrm{sec}\).
Velocity of bird with respect to fish \(=12+\left(\frac{12}{3 / 4 / 1 / 1}\right)=28 \mathrm{~cm} / \mathrm{sec}\).
Example 3. Find the distance of the final image formed by the mirror
Solution:
Shift = \(3\left(1-\frac{1}{3 / 2}\right)\)
For mirror object is at a distance \(=21-3\left(1-\frac{1}{3 / 2}\right)=20 \mathrm{~cm}\)
∴ The object is at the center of the curvature of the mirror. Hence the light rays will retrace and an image will form on the object itself.
8.2 Refraction through a composite slab (or refraction through a number of parallel media, as seen from a medium of R. I. n0)
Apparent depth (distance of final image from final surface) = \(\frac{t_1}{n_{1 \text { rel }}}+\frac{t_2}{n_{2 \text { rel }}}+\frac{t_3}{n_{3 \text { rel }}}+\ldots \ldots . .+\frac{t_n}{n_{n \text { rel }}}\)
Apparent shift = \(t_1\left[1-\frac{1}{n_{1 \mathrm{rcl}}}\right]+t_2\left[1-\frac{1}{n_{2 \text { rel }}}\right]+\ldots \ldots . .+\left[1-\frac{n}{n_{n \text { rel }}}\right] t_n\)
Where ‘t’ represents thickness and ‘n’ represents the R.I. of the respective media, relative to the medium of observer. (i.e. n1rel = n1/n0, n2rel = n2/n0 etc.)
Example: Find the apparent depth of the object seen below surface AB.
Solution:
∴ \(D_{\alpha e p}=\sum \frac{d}{\mu}=\frac{20}{\left(\frac{2}{1.8}\right)}+\frac{15}{\left(\frac{1.5}{1.8}\right)}=18+18=36 \mathrm{~cm}\)
Question 1. Monochromatic light is refracted from air into the glass of refractive index μ. The ratio of the wavelength of incident and refracted waves is
- 1: μ
- 1: μ²
- μ: 1
- 1: 1
Answer: 3. μ: 1
Question 2. The index of refraction of diamond is 2.0, the velocity of light in diamond in cm/second is approximately
- 6×1010
- 3.0×1010
- 2×1010
- 1.5×1010
Answer: 4. 1.5×1010
Question 3. A beam of light is converging towards a point I on a screen. A plane glass plate whose thickness in the direction of the beam = t, refractive index = μ, is introduced in the path of the beam. The convergence point is shifted by
- \(\mathrm{t}\left(1-\frac{1}{\mu}\right)\) away
- \(\mathrm{t}\left(1+\frac{1}{\mu}\right)\) away
- \(\mathrm{t}\left(1-\frac{1}{\mu}\right)\) nearer
- \(\mathrm{t}\left(1+\frac{1}{\mu}\right)\) nearer
Answer: 1. \(\mathrm{t}\left(1-\frac{1}{\mu}\right)\) away
Question 4. When light travels from air to water and from water to glass again from glass to CO2 gas and finally through air. The relation between their refractive indices will be given by
- \({ }_a n_a \times{ }_{\infty} n_{g 1} \times{ }_{g 1} n_{g a s} \times{ }_{g a s} n_a=1\)
- \({ }_a n_a \times{ }_{\infty} n_{g 1} \times{ }_{g a s} n_{g l} \times{ }_{g l} n_a=1\)
- \({ }_a n_0 \times{ }_{\infty} n_{g 1} \times{ }_{g 1} n_{\text {gess }}=1\)
- There is no such relation
Answer: 1. \({ }_a n_a \times{ }_{\infty} n_{g 1} \times{ }_{g 1} n_{g a s} \times{ }_{g a s} n_a=1\)
Question 5. When light enters from the air to water, then its
- Frequency increases and speed decreases
- Frequency is the same but the wavelength is smaller in water than in air
- Frequency is the same but the wavelength in water is greater than in air
- Frequency decreases and wavelength is smaller in water than in air
Answer: 2. Frequency is the same but the wavelength is smaller in water than in air
Question 6. A mark at the bottom of a liquid appears to rise by 0.1 m. The depth of the liquid is 1m. The refractive index of the liquid is
- 1.33
- 9/10
- 10/9
- 1.5
Answer: 3. 10/9
NEET Physics Class 12 Geometrical Optics Important Formulas
9 . Critical Angle And Total Internal Reflection ( T. I. R.)
The critical angle is the angle made in the denser medium for which the angle of refraction in the rarer medium is 90°.
When the angle in the denser medium is more than the critical angle the light ray reflects back in the denser medium following the laws of reflection and the interface behaves like a perfectly reflecting mirror.
In the figure
O = Object
NN’ = Normal to the interface
II’ = Interface
C = Critical angle;
AB = reflected ray due to T. I. R.
When i = C then r = 90°
∴ C = \(\sin ^{-1} \frac{n_r}{n_d}\)
9.1 Conditions of T. I. R.
- Light is incident on the interface from a denser medium.
- The angle of incidence should be greater than the critical angle (i > c).
The figure shows a luminous object placed in the denser medium at a distance of h from an interface separating two media of refractive indices μr and μa. Subscript r and d stand for rarer and denser medium respectively.
In the figure ray 1 strikes the surface at an angle less than critical angle C and gets refracted in rarer medium. Ray 2 strikes the surface at a critical angle and grazes the interface.
Ray 3 strikes the surface making an angle more than a critical angle and gets internally reflected. The locus of points where a ray strikes at a critical angle is a circle, called the circle of illuminance (C.O.I.).
All light rays striking inside the circle of illuminance get refracted in the rarer medium. If an observer is in a rarer medium, he/she will see light coming out only from within the circle of illuminance.
If a circular opaque plate covers the circle of illuminance, no light will get refracted in the rarer medium and then the object can not be seen from the rarer medium. Radius of C.O.I can be easily found.
Example 1. Find the max. angle that can be made in glass medium (μ = 1.5) if a light ray is refracted from glass to vacuum.
Solution:
1.5 sin C = 1 sin 90°, where C = critical angle.
sin C = 2/3
C = sin-1 2/3
Example 2. Find the angle of refraction in a medium (μ = 2) if the light is incident in a vacuum, making the angle equal to twice the critical angle.
Solution:
Since the incident light is in the rarer medium. Total Internal Reflection can not take place.
C = \(\sin ^{-1} \frac{1}{\mu}=30^{\circ}\)
∴ i = 2C = 60º
Applying Snell’s Law.
1 sin 60º = 2 sin r
⇒ \(\sin r=\frac{\sqrt{3}}{4} \quad \Rightarrow \quad r=\sin ^{-1}\left(\frac{\sqrt{3}}{4}\right)\)
Example 3. What should be the value of the angle δ so that light entering normally through the surface AC of a prism (n=3/2) does not cross the second refracting surface AB?
Solution:
Light rays will pass the surface AC without bending since it is incident normally. Suppose it strikes the surface AB at an angle of incidence i.
i = 90-θ
For the required conditions:
90° – θ > C
or sin (90° – θ) > sinC
Or, \(\cos \theta>\sin C=\frac{1}{3 / 2}=\frac{2}{3} \quad \text { or } \quad \theta<\cos ^{-1} \frac{2}{3} \text {. }\)
Example 4. What should be the value of refractive index n of a glass rod placed in air, so that the light entering through the flat surface of the rod does not cross the curved surface of the rod?
Solution:
It is required that all possible r’ should be more than a critical angle. This will be automatically fulfilled if minimum r’ is more than the critical angle ……….(A)
Angle r’ is minimum when r is maximum i.e. C( why ?). Therefore the minimum value of r’is 90-C.
From condition (A): 90° – C > C or C < 45°
sin C < \(\sin 45^{\circ} \quad ; \frac{1}{n} \quad<\frac{1}{\sqrt{2}} \text { or } n>\sqrt{2} \text {. }\)
Question 1. A cut diamond sparkles because of its
- Hardness
- High refractive index
- Emission of light by the diamond
- Absorption of light by the diamond
Answer: 2. Emission of light by the diamond
Question 2. The critical angle of light passing from glass to air is minimal for
- Red
- Green
- Yellow
- Violet
Answer: 4. Violet
Question 3. For total internal reflection to take place, the angle of incidence i and the refractive index μ of the medium must satisfy the inequality
- \(\frac{1}{\sin i}<\mu\)
- \(\frac{1}{\sin \mathrm{i}}>\mu\)
- \(\sin i <\mu\)
- \(\sin i>\mu\)
Answer: 1. \(\frac{1}{\sin i}<\mu\)
Question 4. The critical angle is the angle of incidence in the denser medium for which the angle of refraction in the rarer medium is
- 0°
- 57°
- 90°
- 180°
Answer: 3. 90°
10. Characteristics Of A Prism
(1) A homogeneous solid transparent and refracting medium bounded by two plane surfaces inclined at an angle is called a prism:
(2) PQ and PR are refracting surfaces.
(3) ∠QPR = A is called the refracting angle or the angle of the prism (also called the Apex angle).
(4) δ = angle of deviation
(5) For refraction of a monochromatic (single wavelength) ray of light through a prism; δ = (i + e) – (r1 + r2) and r1 + r2 = A
∴ δ = i + e – A.
(6) Variation of δ versus i (shown in diagram).
For one δ (except δ min) there are two values of angle of incidence.
If i and e are interchanged then we get the same value of 5 because of the reversibility principle of light
Note: (1) For the application of the above result medium on both sides of the prism must be the same.
(2) Based on the above graph we can also derive the following result, which says that i and e can be interchanged for a particular deviation in other words there are two angles of incidence for a given deviation (except minimum deviation).
(7)There is one and only one angle of incidence for which the angle of deviation is minimum.
(8)When δ = δmin, the angle of minimum deviation, then i = e and r1 = r2, the ray passes symmetrically w.r.t. the refracting surfaces. We can show by a simple calculation that δmin = 2imin – A
where imin = angle of incidence for minimum deviation, and r = A/2.
∴ \(\mathbf{n}_{\text {ret }}=\frac{\sin \left[\frac{A+\delta_m}{2}\right]}{\sin \left[\frac{A}{2}\right]}\), where \(\mathbf{n}_{\text {ret }}=\frac{n_{\text {prism }}}{n_{\text {surroundings }}}\)
Also \(\delta_{\min }=(n-1) \mathrm{A}\) (for small values of \(\angle \mathrm{A}\))
(9)For a thin prism \(\left(\mathrm{A} \leq 10^{\circ}\right)\) and for small value of i, all values of \(\delta=\left(n_{r e 1}-1\right) A \quad \text { where } n_{r e 1}=\frac{n_{\text {prism }}}{n_{\text {surrounding }}}\)
Example 1. The refracting angle of a prism A = 60° and its refractive index is, n = 3/2, what is the angle of incidence I to get minimum deviation? Also, find the minimum deviation. Assume the surrounding medium to be air (n = 1).
Solution:
For minimum deviation, r1 = r2 = = 30°.
Applying Snell’s law at I surface
1 × sin i = sin 30º \(\Rightarrow \quad \mathrm{i}=\sin ^{-1}\left(\frac{3}{4}\right) \Rightarrow \quad \delta_{\min }=2 \sin ^{-1}\left(\frac{3}{4}\right)-\frac{\pi}{3}\)
Example 2. Find the deviation caused by a prism having a refracting angle of 4º and a refractive
index of 3/2.
Solution: δ = (3/2 – 1) × 4° = 2°
NEET Physics Class 12 Geometrical Optics Important Formulas
11. Dispersion Of Light
The angular splitting of a ray of white light into a number of components and spreading in different directions is called Dispersion of Light. [It is for the whole Electro Magnetic Wave in totality]. This phenomenon is because waves of different wavelengths move with the same speed in a vacuum but with different speeds in a medium.
Therefore, the refractive index of a medium depends slightly on wavelength. This variation of refractive index with wavelength is given by Cauchy’s formula.
Cauchy’s formula n(λ) = \(\mathrm{n}(\lambda)=\mathrm{a}+\frac{\mathrm{b}}{\lambda^2}\) where a and b are positive constants of a medium.
Note: Such a phenomenon is not exhibited by sound waves.
The angle between the rays of the extreme colors in the refracted (dispersed) light is called the angle of dispersion. θ = δv – δr
For prism of small ‘A’ and with small ‘i’: θ = δv – δr = (nv – nr)A
Deviation of the beam (also called mean deviation) δ = δy = (ny – 1)A
nv, nr, and ny are R. I. of material for violet, red, and yellow colors respectively.
Example 1. The refractive indices of flint glass for red and violet light are 1.613 and 1.632 respectively. Find the angular dispersion produced by a thin prism of flint glass having a refracting angle of 5°.
Solution:
The deviation of the red light is δr = (μr – 1)A and the deviation of the violet light is δv = (μv – 1)A.
The dispersion = δv – δr = (K – K) A = (1.632 – 1.6 1 3) x 50 = 0.0950 .
Note: Numerical data reveals that if the average value of K is small K – K is also small and if the average value of K is large K – K is also large.
Thus, the larger the mean deviation, the larger will be the angular dispersion.
Dispersive power (ω) of the medium of the material of prism is given by: \(\boldsymbol{\omega}=\frac{\mathrm{n}_{\mathrm{v}}-\mathrm{n}_{\mathrm{r}}}{\mathrm{n}_{\mathrm{y}}-1}\)
ω is the property of a medium.
For small angled prism ( A ≤ 10° ) with a light incident at a small angle i: \(\frac{n_v-n_r}{n_y-1}=\frac{\delta_v-\delta_r}{\delta_y}=\frac{\theta}{\delta_y}=\frac{\text { angular dispersion }}{\text { deviation of mean ray (yellow) }}\)
∴ \(n_y=\frac{n_v+n_r}{2}\) if \(n_y\) is not given in the problem
n – 1 = refractivity of the medium for the corresponding color.
Example 2. The refractive index of glass for red and violet colors are 1.50 and 1.60 respectively. Find
- The ref. index for yellow color, approximately
- Dispersive power of the medium
Solution:
- \(\mu_r \simeq=\frac{\mu_v+\mu_{\mathrm{R}}}{2}=\frac{1.50+1.60}{2} 1.55\)
- \(\omega=\frac{\mu_v-\mu_R}{\mu_{\mathrm{r}}-1}=\frac{1.60-1.50}{1.55-1}=0.18\)
11. Direct Vision Combination
11.1 Dispersion without deviation (Direct Vision Combination)
The condition for direct vision combination is: \(\left[n_y-1\right] \quad A=\left[n_y^{\prime}-1\right] \quad A^{\prime} \Leftrightarrow\left[\frac{n_v+n_r}{2}-1\right] \quad A=\left[\frac{n_v^{\prime}+n_r^{\prime}}{2}-1\right] A^{\prime}\)
Two or more prisms can be combined in various ways to get different combinations of angular dispersion and deviation.
11.2 Deviation without dispersion (Achromatic Combination)
Condition for achromatic combination is (nv − nr) A = (n’v − nr) A’
Example 1. If two prisms are combined, as shown in the figure, find the total angular dispersion and angle of deviation suffered by a white ray of light incident on the combination.
Solution:
Both prisms will turn the light rays toward their bases and hence in the same direction. Therefore turnings caused by both prisms are additive.
Total angular dispersion
= \(\theta+\theta^{\prime}=\left(\mu_{\mathrm{V}}-\mu_{\mathrm{R}}\right) \mathrm{A}+\left(\mu_{\mathrm{V}}^{\prime}-\mu_{\mathrm{R}}^{\prime}\right) \mathrm{A}^{\prime}\)
= \((1.5-1.4) 4^{\circ}+(1.7-1.5) 2^{\circ}=0.8^{\circ}\)
Total deviation
= \(\delta+\delta^{\circ}\)
= \(\left(\frac{\mu_{\mathrm{V}}+\mu_{\mathrm{R}}}{2}-1\right) \mathrm{A}+\left(\frac{\mu_{\mathrm{V}}^{\prime}+\mu_{\mathrm{R}}^{\prime}}{2}-1\right) A^{\prime}=\left(\frac{1.5+1.4}{2}-1\right) 0.4^{\circ}+\left(\frac{1.7+1.5}{2}-1\right) 0.2^{\circ}\)
= \((1.45-1) 0.4^{\circ}+(1.6-1) 0.2^{\circ}\)
= \(0.45 \times 0.4^{\circ}+0.6 \times 0.2^{\circ}\)
= \(1.80+1.2=3.0^{\circ}\)
Example 2. Two thin prisms are combined to form an achromatic combination. For I prism A = 4, μR = 1.35, μr = 1.40, μv = 1.42. for II prism μ’R= 1.7, μ’r = 1.8 and μR = 1.9 find the prism angle of II prism and the net mean deviation.
Solution:
Condition for achromatic combination.
θ = 0’; \(\left(\mu_V-\mu_R\right) A=\left(\mu_V^{\prime}-\mu_R^{\prime}\right) A^{\prime}\)
∴ \(A^{\prime}=\frac{(1.42-1.35) 4^{\circ}}{1.9-1.7}=1.4^{\circ}\)
∴ \(\delta_{\text {Ne }}=\delta \sim \delta^{\prime}=\left(\mu_Y-1\right) A \sim\left(\mu_Y^{\prime}-1\right) A^{\prime}=(1.40-1) 4^{\circ} \sim(1.8-1) 1.4^{\circ}=0.48^{\circ}\).
Example 3. A crown glass prism of angle 50 is to be combined with a flint prism in such a way that the mean ray passes undeviated. Find (1) the angle of the flint glass prism needed and (2) the angular dispersion produced by the combination when white light goes through it. Refractive indices for red, yellow, and violet light are 1.5, 1.6, and 1.7 respectively for crown glass 1.8,2.0, and 2.2 for flint glass.
Solution:
The deviation produced by the crown prism is δ = (μ – 1)A
and by the flint prism is δ’ = (μ’ – 1)A’.
The prisms are placed with their angles inverted with respect to each other. The deviations are also in opposite directions.
Thus, the net deviation is: D = δ – δ’ = (μ – 1)A – (μ’ – 1)A’…..(1)
(1) If the net deviation for the mean ray is zero, (μ – 1)A = (μ’ – 1)A’.
or, \(A^{\prime}=\frac{(\mu-1)}{\left(\mu^{\prime}-1\right)} A=\frac{1.6-1}{2.0-1} \times 5^0=3^{\circ}\)
(2) The angular dispersion produced by the crown prism is: δv – δr = (μv – μr)A
and that by the flint prism is, δ’v – δ’r = (μ’v – μ’r)A
The net angular dispersion is, (μv – μr)A – (μ’v – μ’r)A
= (1.7 – 1.5) x 5° – (2.2 – 1.8) x 3° = – 0.2°
The angular dispersion has a magnitude of 0.2°
Question 1. The formula for dispersive power is (where symbols have their usual meanings) or If the refractive indices of crown glass for red, yellow, and violet colors are respectively and then the dispersive power of this glass would be
- \(\frac{\mu_u-\mu_y}{\mu_r-1}\)
- \(\frac{\mu_u-\mu_r}{\mu_y-1}\)
- \(\frac{\mu_v-\mu_y}{\mu_y-\mu_r}\)
- \(\frac{\mu_u-\mu_r}{\mu_y}-1\)
Answer: 2. \(\frac{\mu_u-\mu_r}{\mu_y-1}\)
Question 2. The angle of minimum deviation for a prism of refractive index 1.5 is equal to the angle of the prism. The angle of the prism is (cos 41° = 0.75)
- 62°
- 41°
- 82°
- 31°
Answer: 3. 82°
Question 3. In the formation of a primary rainbow, the sunlight rays emerge at minimum deviation from rain-drop after
- One internal reflection and one refraction
- One internal reflection and two refractions
- Two internal reflections and one refraction
- Two internal reflections and two refractions
Answer: 2. One internal reflection and two refractions
Question 4. Dispersive power depends upon
- The shape of a prism
- Material of prism
- Angle of prism
- Height of the prism
Answer: 2. Material of prism
12. Spectrum
The ordered pattern produced by a beam emerging from a prism after refraction is called a Spectrum.
12.1 Types Of Spectrum:
- Line spectrum: Due to source in atomic state.
- Band spectrum: Due to source in molecular state.
- Continuous spectrum: Due to white hot solid.
12.2 In Emission Spectrum: Bright colors or lines, emitted from the source are observed.
The spectrum emitted by a given source of light is called the emission spectrum. It is a wavelength-wise distribution of light emitted by the source.
The emission spectra are given by incandescent solids, liquids, and gases which are either burned directly as a flame (or a spark) or burnt under low pressure in a discharge tube.
12.3 In Absorption Spectrum: Dark lines indicate frequencies absorbed.
When a beam of light from a hot source is passed through a substance (at a lower temperature), a part of the light is transmitted but the rest of it is absorbed. With the help of a spectrometer, we can know the fraction of light absorbed corresponding to each wavelength.
The distribution of the wavelength absorption of light by a substance is called an absorption spectrum. Every substance has its own characteristic absorption spectrum.
12.4 Spectrometer: Consists of a collimator (to collimate light beam), prism, and telescope. It is used to observe the spectrum and also measure deviation.
13. Refraction At Spherical Surfaces
For paraxial rays incident on a spherical surface separating two media: \(\frac{\mathrm{n}_2}{\mathrm{v}}-\frac{\mathrm{n}_1}{\mathrm{u}}=\frac{\mathrm{n}_2-\mathrm{n}_1}{\mathrm{R}}\)….. (A) where light moves from the medium of refractive index n to the medium of refractive index n2.
Transverse magnification (m) (of dimension perpendicular to the principal axis) due to refraction at the spherical surface is given by \(\mathrm{m}=\frac{\mathrm{v}-\mathrm{R}}{\mathrm{u}-\mathrm{R}}=\left(\frac{\mathrm{v} / \mathrm{n}_2}{\mathrm{u} / \mathrm{n}_1}\right)\)
Example 1. Find the position, size, and nature of the image, for the situation shown in the figure. Draw a ray diagram.
Solution:
For refraction near point A, u = – 30 ; R = – 20; n1 = 2 ; n2 = 1
Applying refraction formula \(\frac{\mathrm{n}_2}{\mathrm{v}}-\frac{\mathrm{n}_1}{\mathrm{u}}=\frac{\mathrm{n}_2-\mathrm{n}_1}{\mathrm{R}} \Rightarrow \frac{1}{\mathrm{v}}-\frac{2}{-30}=\frac{1-2}{-20}\)
v = -60 cm
m = \(\frac{\mathrm{h}_2}{\mathrm{~h}_1}=\frac{\mathrm{n}_1 v}{\mathrm{n}_2 \mathrm{u}}=\frac{2(-60)}{1(-30)}=4\)
∴ \(\mathrm{~h}_2=4 \mathrm{~mm}\).
Special Case: Refraction at Plane Surfaces
Putting R = in the formula \(\frac{n_2}{v}-\frac{n_1}{u}=\frac{n_2-n_1}{R}\), we get;
v = \(\frac{\mathrm{n}_2 \mathrm{u}}{\mathrm{n}_1}\)
The same sign of v and u implies that the object and the image are always on the same side of the interface separating the two media. If we write the above formula as
v = \(\frac{u}{n_{\text {rel }}} \text {, }\).
it gives the relation between the apparent depth and real depth, as we have seen before.
Geometrical Optics Class 12 NEET Notes
Example 2. Using the formula of spherical surface or otherwise, find the apparent depth of an object placed 10 cm below the water surface if seen near normally from the air.
Solution:
Put R = in the formula of the Refraction at Spherical Surfaces we get, v = \(\frac{u_2}{n_1} \quad \Rightarrow \quad u=-10 \mathrm{~cm}\)
⇒ \(n_1=\frac{4}{3} \quad \Rightarrow \quad n_2=1 \quad \Rightarrow \quad v=-\frac{10 \times 1}{4 / 3}=-7.5 \mathrm{~cm}\)
negative sign implies that the image is formed in water.
Aliter: \(\mathrm{d}_{\text {esp }}=\frac{\mathrm{d}_{\text {real }}}{\mu_{\text {rel }}}=\frac{10}{4 / 3}=\frac{30}{4}=7.5 \mathrm{~cm}\).
14. Thin Lens
A thin lens is called convex if it is thicker at the middle and it is called concave if it is thicker at the ends. One surface of a convex lens is always convex. Depending on the other surface a convex lens is categorized as
- Biconvex or convexo-convex, if the other surface is also convex,
- Plano-convex if the other surface is plane and
- Concavo is convex if the other surface is concave.
Similarly, a concave lens is categorized as concavo-concave or biconcave, plano-concave, or convexo-concave.
For a spherical, thin lens having the same medium on both sides: \(\frac{1}{v}-\frac{1}{u}=\left(n_{n e 1}-1\right)\left(\frac{1}{R_1}-\frac{1}{R_2}\right) \cdots \ldots \ldots(1)\),
where \(\mathrm{n}_{\text {mel }}=\frac{\mathrm{n}_{\text {lens }}}{\mathrm{n}_{\text {medium }}}\) and \(\mathrm{R}_1\) and \(\mathrm{R}_2\) are \(\mathrm{x}\) coordinates of the center of curvature of the 1 st surface and \(2^{\text {nd }}\) surface respectively. \(\frac{1}{\mathrm{v}}-\frac{1}{\mathrm{u}}=\frac{1}{\mathrm{i}} \rightarrow\) Lens Maker’s Formula……..(2)
Lens has two Focii: If u = \(\infty\), then \(\frac{1}{v}-\frac{1}{\infty}=\frac{1}{f} \quad \Rightarrow \quad v=f\)
⇒ If incident rays are parallel to the principal axis then its refracted ray will cut the principal axis at ‘f’. It is called 2nd focus.
In the case of a converging lens, it is positive and in the case of a diverging lens, it is negative.
If v = ∞ that means \(\frac{1}{\infty}-\frac{1}{u}=\frac{1}{f}\) u = -f
If incident rays cut the principal axis at – f then its refracted ray will become parallel to the principal axis. It is called 1st focus. In the case of a converging lens, it is negative (f is positive) and in the case of a diverging lens it is positive (f is negative)
use of – f and + f in drawing the ray diagrams.
Notice that point B, its image B’, and the pole P of the lens are collinear. This is due to the parallel slab nature of the lens in the middle. This ray goes straight.
From the relation \(\frac{1}{\mathrm{f}}=\left(\mathrm{n}_{\mathrm{mel}}-1\right)\left(\frac{1}{\mathrm{R}_1}-\frac{1}{\mathrm{R}_2}\right)\) it can be seen that the second focal length depends on two factors.
(1) The factor \(\left(\frac{1}{R_1}-\frac{1}{R_2}\right)\) is
- Positive for all types of convex lenses and
- Negative for all types of concave lenses.
(2) The factor (nrel – 1) is
- Positive when the surrounding medium is rarer than the medium of the lens.
- Negative when the surrounding medium is denser than the medium of the lens.
So a lens is converging if f is positive which happens when both the factors (A) and (B) are of the same sign.
And a lens is diverging if f is negative which happens when the factors (A) and (B) are of opposite signs.
The focal length of the lens depends on the medium of the lens as well as its surroundings.
It also depends on the wavelength of incident light. The incapability of a lens to focus light rays of various wavelengths at a single point is known as chromatic aberration.
Example 1. Find the behavior of a concave lens placed in a rarer medium.
Solution:
Factor (A) is negative because the lens is concave.
Factor (B) is positive because the lens is placed in a rarer medium.
Therefore the focal length of the lens, which depends on the product of these factors, is negative and hence the lens will behave as a diverging lens.
Example 2. Show that the factor \(\left(\frac{1}{\mathrm{R}_1}-\frac{1}{\mathrm{R}_2}\right)\) (and therefore focal length) does not depend on which surface of the lens light strike first.
Solution:
Consider a convex lens of radii of curvature p and q as shown.
Case 1: Suppose light is incident from the left side and strikes the surface with a radius of curvature p, first.
Then \(R_1=+p ; R_2=-q\) and \(\left(\frac{1}{R_1}-\frac{1}{R_2}\right)=\left(\frac{1}{p}-\frac{1}{-q}\right)\)
Case 2: Suppose light is incident from the right side and strikes the surface with a radius of curvature q, first.
Then \(\mathrm{R}_1=+\mathrm{q} ; \mathrm{R}_2=-\mathrm{p}\) and \(\left(\frac{1}{\mathrm{R}_1}-\frac{1}{R_2}\right)=\left(\frac{1}{q}-\frac{1}{-p}\right)\)
Though we have shown the result for the biconvex lens, it is true for every lens.
Example 3. Find the focal length of the lens shown in the figure.
Solution:
⇒ \(\frac{1}{f}=\left(n_{r d}-1\right)\left(\frac{1}{R_1}-\frac{1}{R_2}\right)\)
⇒ \(\frac{1}{f}=(3 / 2-1)\left(\frac{1}{10}-\frac{1}{(-10)}\right) \quad \Rightarrow \quad \frac{1}{f}=\frac{1}{2} \times \frac{2}{10} \Rightarrow \quad f=+10 \mathrm{~cm}\)
Example 4. Find the focal length of the lens shown in the figure
Solution:
⇒ \(\frac{1}{f}=\left(n_{\text {rol }}-1\right)\left(\frac{1}{R_1}-\frac{1}{R_2}\right)=\left(\frac{3}{2}-1\right)\left(\frac{1}{-10}-\frac{1}{10}\right) f=-10 \mathrm{~cm}\)
Example 5. Find the focal length of the lens shown in the figure
- If the light is incident from the left side.
- If the light is incident from the right side.
Solution:
(1) \(\frac{1}{\mathrm{f}}=\left(\mathrm{n}_{\text {re1 }}-1\right)\left(\frac{1}{\mathrm{R}_1}-\frac{1}{\mathrm{R}_2}\right)=\left(\frac{3}{2}-1\right)\left(\frac{1}{-60}-\frac{1}{-20}\right) \quad \mathrm{f}=60 \mathrm{~cm}\)
(2) \(\frac{1}{\mathrm{f}}=\left(\mathrm{n}_{\text {me1 }}-1\right)\left(\frac{1}{\mathrm{R}_1}-\frac{1}{\mathrm{R}_2}\right)=\left(\frac{3}{2}-1\right)\left(\frac{1}{20}-\frac{1}{60}\right) \quad \mathrm{f}=60 \mathrm{~cm}\)
Example 6. The point object is placed on the principal axis of a thin lens with parallel curved boundaries i.e., having the same radii of curvature. Discuss the position of the image formed.
Solution:
⇒ \(\frac{1}{f}=\left(n_{r e d}-1\right)\left(\frac{1}{R_1}-\frac{1}{R_2}\right)=0\)
because \(R_1=R_2\)
∴ \(\frac{1}{v}-\frac{1}{u}=0\) or v = u i.e. rays pass without appreciable bending.
Example 7. The focal length of a thin lens in air is 10 cm. Now medium on one side of the lens is replaced by a medium of refractive index μ=2. The radius of curvature of the surface of the lens, in contact with the medium, is 20 cm. Find the new focal length
Solution:
Let the radius of the I surface be R1 and the refractive index of the lens be μ.
Let parallel rays be incident on the lens.
Applying refraction formula at first surface \(\frac{\mu}{V_1}-\frac{1}{\infty} \quad=\frac{\mu-1}{R_1}\)……(1)
At 2 surface \(\frac{2}{V}-\frac{\mu}{V_1}=\frac{2-\mu}{-20}\)……(2)
Adding (1) and (2) \(\frac{\mu}{V_1}-\frac{1}{\infty}+\frac{2}{V}-\frac{\mu}{V_1}=\frac{\mu-1}{R_1}+\frac{2-\mu}{-20}\)
= \((\mu-1)\left(\frac{1}{R_1}-\frac{1}{-20}\right)-\frac{\mu-1}{20}-\frac{2-\mu}{20}=\frac{1}{\mathrm{f}} \text { (in air) }+\frac{1}{20}-\frac{2}{20}\)
v = 40cm
f = 40 cm
Example 8. Shown a point object and a converging lens. Find the final image formed.
Solution:
⇒ \(\frac{1}{v}-\frac{1}{u}=\frac{1}{\mathrm{f}} \Rightarrow \frac{1}{v}-\frac{1}{-15}=\frac{1}{10} \quad \Rightarrow \quad \frac{1}{v}=\frac{1}{10}-\frac{1}{15}=\frac{1}{30}\)
v = +30 cm
Example 9. Find the position of the final image formed.
Solution:
For converging lens
u = –15 cm, f = 10 cm v = \(\frac{f u}{f+u}=30 \mathrm{~cm}\)
For diverging lens u = 5 cm
f = –10 cm; v = \(\frac{f u}{f+u}=10 \mathrm{~cm}\)
Example 10. Shows two converging lenses. Incident rays are parallel to the principal axis. What should be the value of d so that the final rays are also parallel?
Solution:
The final rays should be parallel. For this, the 2 focus of L1 must coincide with the I focus of L2
d = 10 + 20 = 30 cm
14.2 Transverse Magnification (m)
Transverse magnification (m) of (of dimension perpendicular to the principal axis) is given by m = \(\frac{v}{u}\)
If the lens is thick or/and the medium on both sides is different, then we have to apply the formula given for refraction at spherical surfaces step by step.
Example 1. An extended real object of size 2 cm is placed perpendicular to the principal axis of a converging lens of focal length 20 cm. The distance between the object and the lens is 30 cm.
- Find the lateral magnification produced by the lens.
- Find the height of the image.
- Find the change in lateral magnification, if the object is brought closer to the lens by 1 mm along the principal axis.
Solution:
Using \(\frac{1}{\mathrm{v}}-\frac{1}{\mathrm{u}}=\frac{1}{\mathrm{f}}\) and \(\mathrm{m}=\frac{\mathrm{v}}{\mathrm{u}}\)
we get \(m=\frac{f}{f+u} \pm\)……(A)
m = \(\frac{+20}{+20+(-30)}=\frac{+20}{-10}=-2\)
-ve sign implies that the image is inverted.
(2) \(\frac{\mathrm{h}_2}{\mathrm{~h}_1}=\mathrm{m}\)
∴ \(\mathrm{h}_2=\frac{-\mathrm{f}}{(\mathrm{f}+\mathrm{u})^2} \mathrm{mh}_1=(-2)(2)=-4 \mathrm{~cm}\)
(3) Differentiating (A) we get \(\mathrm{dm}=\frac{-\mathrm{f}}{(\mathrm{f}+\mathrm{u})^2} \mathrm{du}=\frac{-(20)}{(-10)^2}(0.1)=\frac{-2}{100}=-.02\)
Note that the method of differential is valid only when changes are small.
Alternate method: u(after displacing the object) = -(30 + 0.1) = – 29.9 cm
Applying the formula \(\mathrm{m}=\frac{\mathrm{f}}{\mathrm{f}+\mathrm{u}} ; \quad \mathrm{m}=\frac{20}{20+(-29.9)}=-2.02\)
∴ change in \(m \text { ‘ }=-0.02 \text {. }\)
Since in this method differential is not used, this method can be used for changes, small or large.
14.3 Displacement Method To Find Focal Length Of Converging Lens
Fix an object of small height H and a screen at a distance of D from an object (as shown in the figure).
Move a converging lens from the object towards the screen. Let a sharp image form on the screen when the distance between the object and the lens is ‘a. From the lens formula we have
⇒ \(\frac{1}{D-a}-\frac{1}{-a}=\frac{1}{f} \quad \text { or } \quad a^2-D a+f D=0 \ldots(A)\)
This is a quadratic equation and hence two values of ‘a’ are possible. Call them a1 and a2.
Thus a1 and a2 are the roots of the equation. From the properties of roots of a quadratic equation,
∴ a1 + a2 = D ⇒ a1a2 = f D
Also \(\left(a_1-a_2\right)=\sqrt{\left(a_1+a_2\right)^2-4 a_1 a_2}=\sqrt{D^2-4 f D}=d \text { (suppose). }\)
‘d’ physically means the separation between the two positions of the lens.
The focal length of the lens in terms of D and d.
so, \(a_1-a_2=\sqrt{\left(a_1+a_2\right)^2-4 a_1 a_2}\)
= \(\sqrt{\mathrm{D}^2-4 \mathrm{fD}}=\mathrm{d} \quad \Rightarrow \quad f=\frac{\mathrm{D}^2-\mathrm{d}^2}{4 \mathrm{D}}\)
condition, d = 0, i.e. the two position coincide f = \(\frac{\mathrm{D}^2}{4 \mathrm{D}}\)
∴ D = 4f
Roots of the equation a² – Da + f D = 0, become imaginary if b² – 4ac < 0. = D² -4fD < 0 = D (D- 4f) < 0 = D – 4f = 0
for real value of a in equation a² – Da + f D = 0
b² — 4ac ≥ 0. = D² — 4f D ≥ 0.
so D ≥ 4f ⇒ Dmin = 4f
Lateral Magnification In Displacement Method: if m1 and m2 be two magnifications in two positions (In the displacement method)
⇒ \(m_1=\frac{v_1}{u_1}=\frac{\left(D-a_1\right)}{-a_1} \quad m_2=\frac{v_2}{u_2}=\frac{D-a_2}{-a_2}=\frac{a_1}{-\left(D-a_1\right)}\)
So \(m_1 m_2=\frac{\left(D-a_1\right)}{-a_1} \times \frac{a_1}{-\left(D-a_1\right)}=1\).
If image length is h1 and h2 in the two cases, then \(m_1=-\frac{h_1}{H}; m_2=-\frac{h_2}{H}; m_1 m_2=1\)
∴ \(\frac{h_1 h_2}{H^2}=1 ; h_1 h_2=H^2 ; H=\sqrt{h_1 h_2}\)
Question 1. The radius of curvature for a convex lens is 40 cm, for each surface. Its refractive index is 1.5. The focal length will be
- 40 cm
- 20 cm
- 80 cm
- 30 cm
Answer: 1. 40 cm
Question 2. A thin lens focal length of f1 and an aperture has a diameter of d. It forms an image of intensity I. Now the central part of the aperture up to diameter d/2 is blocked by an opaque paper. The focal length and image intensity will change to
- \(\frac{\mathrm{f}}{2}\) and \(\frac{\mathrm{I}}{2}\)
- \(\mathrm{f}\) and \(\frac{\mathrm{I}}{4}\)
- \(\frac{3 \mathrm{f}}{4}\) and \(\frac{\mathrm{I}}{2}\)
- \(\mathrm{f}\) and \(\frac{3 \mathrm{I}}{4}\)
Answer: 4. \(\mathrm{f}\) and \(\frac{3 \mathrm{I}}{4}\)
Question 3. A lens of power +2 diopters is placed in contact with a lens of power –1 diopter. The combination will behave like
- A convergent lens of focal length 50 cm
- A divergent lens of focal length 100 cm
- A convergent lens of focal length 100 cm
- A convergent lens of focal length 200 cm
Answer: 3. A convergent lens of focal length 100 cm
Geometrical Optics Class 12 NEET Notes
Question 4. If in a plano-convex lens, the radius of curvature of the convex surface is 10 cm and the focal length of the lens is 30 cm, then the refractive index of the material of the lens will be
- 1.5
- 1 .66
- 1.33
- 3
Answer: 3. 1.33
Question 5. The silt of a collimator is illuminated by a source as shown in the adjoining figures. The distance between the silt S and the collimating lens L is equal to the focal length of the lens. The correct direction of the emergent beam will be as shown in the figure
- 1
- 3
- 2
- None of the figures
Answer: 3. 2
Question 6. A convex lens makes a real image 4 cm long on a screen. When the lens is shifted to a new position without disturbing the object, we again get a real image on the screen which is 16 cm tall. The length of the object must be
- 1/4 cm
- 8 cm
- 12 cm
- 20 cm
Answer: 2. 8 cm
Question 7. A thin convex lens of refractive index 1.5 has a focal length of 15 cm in air. When the lens is placed in a liquid of refractive index 4/3, its focal length will be
- 15 cm
- 10 cm
- 30 cm
- 60 cm
Answer: 4. 60 cm
Image Formation By A Lens
15. Combination Of Lenses
The equivalent focal length of thin lenses in contact is given by \(\frac{1}{F}=\frac{1}{f_1}+\frac{1}{f_2}+\frac{1}{f_3} \ldots\)
where f1, f2, f3 are the focal lengths of individual lenses.
If two converging lenses are separated by a distance d and the incident light rays are parallel to the common principal axis , then the combination behaves like a single lens of focal length given by the relation \(\frac{1}{F}=\frac{1}{f_1}+\frac{1}{f_2}-\frac{d}{f_1 f_2}\) and the position of equivalent lens is \(\frac{-\mathrm{d} \quad \mathrm{F}}{\mathrm{f}_1}\) with respect to 2nd lens
Example 1. Find the lateral magnification produced by the combination of lenses shown in the figure.
Solution:
⇒ \(\frac{1}{\mathrm{f}}=\frac{1}{\mathrm{f}_1}+\frac{1}{\mathrm{f}_2}=\frac{1}{10}-\frac{1}{20}=\frac{1}{20} \quad \Rightarrow \quad \mathrm{f}=+20\)
∴ \(\frac{1}{v}-\frac{1}{-10}=\frac{1}{20} \quad \Rightarrow \quad \frac{1}{v}=\frac{1}{20}-\frac{1}{10}=\frac{-1}{20}=-20 \mathrm{~cm}\)
∴ m = 2
Example 2. Find the focal length of the equivalent system.
Solution:
⇒ \(\frac{1}{f_1}=\left(\frac{3}{2}-1\right)\left(\frac{1}{10}+\frac{1}{10}\right)=\frac{1}{2} \times \frac{2}{10}=\frac{1}{10}\)
⇒ \(\frac{1}{f_2}=\left(\frac{6}{5}-1\right)\left(\frac{-1}{10}-\frac{1}{20}\right)=\frac{1}{5} \times\left(\frac{-30}{10 \times 20}\right) \quad=\frac{-3}{100}\)
⇒ \(\frac{1}{f_3}=\left(\frac{8}{5}-1\right)\left(\frac{1}{20}+\frac{1}{20}\right)=\frac{3}{50}\)
⇒ \(\frac{1}{\mathrm{f}}=\frac{1}{\mathrm{f}_1}+\frac{1}{\mathrm{f}_2}+\frac{1}{\mathrm{f}_3}=\frac{1}{10}+\frac{-3}{100}+\frac{3}{50} \mathrm{f}=\frac{100}{13}\)
Question 1. Two similar plano-convex lenses are combined together in three different ways as shown in the adjoining figure. The ratio of the focal lengths in three cases will be
- 2 : 2: 1
- 1: 1: 1
- 1 : 2: 2
- 2: 1: 1
Answer: 2. 1: 1: 1
Question 2. A concave and convex lens have the same focal length of 20 cm and are put into contact to form a lens combination. The combination is used to view an object of 5 cm length kept at 20 cm from the lens combination. As compared to the object, the image will be
- Magnified and inverted
- Reduced and erect
- Of the same size as the object and erect
- Of the same size as the object but inverted
Answer: 3. Of the same size as the object and erect
16. Combination Of Lens And Mirror
The combination of lens and mirror behaves like a mirror of focal length ‘f given by \(\frac{1}{\mathrm{f}}=\frac{1}{\mathrm{~F}_{\mathrm{m}}}-\frac{2}{\mathrm{~F}_{\ell}}\)
If lenses are more than one, ‘f’ is given by \(\frac{1}{\mathrm{f}}=\frac{1}{\mathrm{~F}_{\mathrm{m}}}-2\left(\sum \frac{1}{\mathrm{f}_{\ell}}\right)\)
For the following figure ‘f’ is given by \(\frac{1}{\mathrm{f}}=\frac{1}{\mathrm{~F}_{\mathrm{m}}}-2\left(\frac{1}{\mathrm{f}_1}+\frac{1}{\mathrm{f}_2}\right)\)
Example: Find the position of the final image formed. (The gap shown in the figure is of negligible width)
Solution:
⇒ \(\frac{1}{f_{e q}}=\frac{1}{10}-\frac{2}{10}=\frac{-1}{10} \Rightarrow f_{e q}=-10 \mathrm{~cm}\)
⇒ \(\frac{1}{v}+\frac{1}{-20}=\frac{1}{-10} \quad \Rightarrow v=-20 \mathrm{~cm}\)
Hence image will be formed on the object itself
Some Interesting Facts About Light:
(1)The Sun Rises Before It Actually Rises And Sets After It Actually Sets: The atmosphere is less dense as its height increases, and it is also known that the index of refraction decreases with a decrease in density. So, there is a decrease in the index of refraction with height. Due to this, the light rays bend as they move in the earth’s atmosphere
(2)The Sun Is Oval Shaped At The Time Of Its Rise And Set: The rays diverging from the lower edge of the sun have to cover a greater thickness of air than the rays from the upper edge. Hence the former are refracted more than the latter, and so the vertical diameter of the sun appears to be a little shorter than the horizontal diameter which remains unchanged.
(3)The Stars Twinkle But Not The Planets: The refractive index of the atmosphere fluctuates by a small amount due to various reasons. This causes a slight variation in the bending of light due to which the apparent position of the star also changes, producing the effect of twinkling.
(4)Glass Is Transparent, But Its Powder Is White: When powdered, light is reflected from the surface of innumerable small pieces of glass and so the powder appears white. Glass transmits most of the incident light and reflects very little hence it appears transparent.
(5)Greased Or Oiled Paper Is Transparent, But Paper Is White: The rough surface of paper diffusely reflects incident light so it appears white. When oiled or greased, very little reflection takes place and most of the light is allowed to pass hence it appears transparent.
(6)An Extended Water Tank Appears Shallow At The Far End:
(7)A Test Tube Or A Smoked Ball Immersed In Water Appears Silvery White When Viewed From The Top: This is due to Total internal reflection
(8)Ships Hang Inverted In The Air In Cold Countries And Trees Hang Inverted Underground In Deserts:
17. Structure Of Eye
Light enters the eye through a curved front surface, the corner. It passes through the pupil which is the central hole in the iris. The size of the pupil can change under the control of muscles.
The light is further focussed by the eye lens on the retina. The retina is a film of nerve fibers covering the curved back surface of the eye. The retina contains rods and cones that sense light intensity and color, respectively, and transmit electrical signals via the optic nerve to the brain which finally processes this information.
The shape (curvature) and therefore the focal length of the lens can be modified somewhat by the ciliary muscles. For example, when the muscle is released, the focal length is about 2.5 cm and (for a normal eye) objects at infinity are in sharp focus on the retinas.
When the object is brought closer to the eye, in order to maintain the same image-lens distance (2.5 cm), the focal length of the eye-lens becomes shorter by the action of the ciliary muscles. This property of the eye is called accommodation. If the object is too close to the eye, the lens cannot curve enough to focus the image onto the retina, and the image is blurred.
The closest distance for which the lens can focus light on the retina is called the least distance of distinct vision, or the near point. The standard value (for normal vision) taken here is 25 cm. (Often the near point is given the symbol D.)
Geometrical Optics NEET Physics Chapter 7 Study Material
18. Defects Of Vision
Regarding the eye, it is nothing that:
(1)In eye convex, the eye lens forms real inverted and diminished images at the retina by changing its convexity (the distance between the eye lens and the retina is fixed)
(2)The human eye is most sensitive to yellow-green light having wavelength 5550 A° and least to violet (4000 A°) and red (7000 A°)
(3)The size of an object as perceived by the eye depends on its visual angle when the object is distant its visual angle and hence image L, at the retina is small (it will appear small), and as it is brought near to the eye its visual angle and hence the size of the image I2 will increase.
(4)The far and near points for the normal eye are usually taken to be infinity and 25 cm respectively ie., the normal eye can see very distant objects clearly but near objects only if they are at a distance greater than 25 cm from the eye. The ability of the eye to see objects from an infinite distance to 25 cm from it is called the Power of accommodation.
(5)If the object is at infinity i.e., a parallel beam of light enters the eye is least strained and said to be relaxed or unstrained. However, if the object is at least a distance of distinct vision (L.D.D.V] i.e., D (=25 cm) eye is under maximum strain, and the visual angle is maximum.
The limit of resolution of the eye is one minute ie., two objects will not be visible distinctly to the eye if the angle subtended by them on the eye is less than one minute.
The persistance of vision is (1/10) sec i.e., If the time interval between two consecutive light pulses is lesser than 0.1 sec eye cannot distinguish them separately. This fact is taken into account in motion pictures.
In the case of the eye following are the common defects of vision.
18.1 Myopia
[or short-sightedness or near – near-sightendness]
In it, distant objects are not clearly visible. i.e. Far Point is at a distance lesser than Infinity and hence the image of a distant object is formed before the retina.
This defect is (i.e., negative focal length or power) which forms the image of a distant object at the far point of the patient – eye [which is lesser than ] so that in this case from lens formula we have \(\frac{1}{-F. P}-\frac{1}{-(\text { distance of object })}=\frac{1}{f}=P\) And if the object is at \(\quad P=\frac{1}{f}=\frac{1}{-F . P .} \quad \ldots\)
where F.P is the farther point of the eye
18.2 Hypermetropia
[Or Long-sightedness or far-sightendness]
In it near objects are not clearly visible i.e., Near Point is at a distance greater than 25 cm and hence the image of the near object is formed behind the retina.
This defect is remedied by using spectacles having a convergent lens (i.e., positive focal length of power) which is the image of near objects at the Near Point of the Patient-eye (which is more than 25 cm).
So in this case from the lens formula we have \(\frac{1}{- \text { N.P. }} \frac{1}{- \text { F.P. }}-\frac{1}{-(\text { distance of object })}=\frac{1}{f}=P\)
If object is placed at D = 25 cm = 0.25 cm
P = \(\frac{1}{f}=\left[\frac{1}{0.25}-\frac{1}{N . P .}\right]\)…..(2)
where N.P is near the point of the eye
18.3 Presbyopia
In this both near and far objects are not clearly visible i.e., the far point is lesser than infinity and the near point is greater than 25 cm. It is an old age disease as in old age ciliary muscles lose their elasticity and so can not change the focal length of the eye-lens effectively hence eye loses its power of accommodation.
18.4 Astigmatism
Due to the imperfect spherical nature of the eye-lens, the focal length of the eye lens in two orthogonal directions becomes different so the eye cannot see objects in two orthogonal directions clearly simultaneously.
This defect is directional and is remedied by using the cylindrical lens in a particular direction. If in the spectacle of a person suffering from astigmatism, the lens is slightly rotated the arrangement will get spoiled.
Question 1. For a normal eye, the least distance of distinct vision is
- 0.25 m
- 0.50 m
- 25 m
- Infinite
Answer: 1. 0.25 m
Question 2. For the myopic eye, the defect is cured by
- Convex lens
- Concave lens
- Cylindrical lens
- Toric lens
Answer: 2. Concave lens
Question 3. Lens used to remove long-sightedness (hypermetropia ) is or A person suffering from hypermetropia requires which type of spectacle lenses
- Concave lens
- Plano- concave lens
- Convexo- concave lens
- Convex lens
Answer: 4. Convex lens
Question 4. The image formed on the retina is
- Real and inverted
- Virtual and erect
- Real and erect
- Virtual and inverted
Answer: 1. Real and inverted
Chromatic Aberration
The image of a white object in white light formed by a lens is usually colored and blurred. This defect of the image is called chromatic aberration and arises due to the fact that the focal length of a lens is different for different colors. As the R.I. μ of the lens is maximum for violet while minimum for red, violet is focused nearest to the lens while red is farthest from it as shown in the figure.
As a result of this, in the case of the convergent lens if a screen is placed at Fv center of the image will be violet and focused while the sides are red and blurred. While at FR, the reverse is the case, i.e., the center will be red and focused while the sides violet and blurred. The difference between fv and fR is a measure of the longitudinal chromatic aberration (L.C.A), i.e., L.C.A = \(f_R-f_V=-d f \text { with } d f=f_v-f_R\)……(1)
However, as for a single lens, \(\frac{1}{\mathrm{f}}=(\mu-1)\left[\frac{1}{\mathrm{R}_1}-\frac{1}{\mathrm{R}_2}\right]\)……(1)
– \(\frac{\mathrm{df}}{\mathrm{f}^2}=\mathrm{d} \mu\left[\frac{1}{\mathrm{R}_1}-\frac{1}{\mathrm{R}_2}\right]\)…….(2)
Dividing Equation (3) by (2);
– \(\frac{\mathrm{df}}{\mathrm{f}}=\frac{\mathrm{d} \mu}{(\mu-1)}=\omega \quad\left[\omega=\frac{\mathrm{d} \mu}{(\mu-1)}\right]\) = dispersive power…….(4)
And hence, from Eqns. (1) and (4),
L.C.A. = –df = ωf
Now, as for a single lens neither f nor can be zero, we cannot have a single lens free from chromatic aberration.
Condition Of Achromatism: In case of two thin lenses in contact \(\frac{1}{\mathrm{~F}}=\frac{1}{\mathrm{f}_1}+\frac{1}{\mathrm{f}_2}\)
i.e., \(-\frac{d F}{F^2}=-\frac{d f_1}{f_1^2}-\frac{d f_2}{f_2^2}\)
The combination will be free from chromatic aberration if dF = 0
i.e., \(\frac{\mathrm{df}_1}{\mathrm{f}_1^2}+\frac{\mathrm{df}_2}{\mathrm{f}_2^2}=0\)
which with the help of Eqn. (4) reduces to \(\frac{\omega_1 f_1}{f_1^2}+\frac{\omega_2 f_2}{f_2^2}=0 \quad \text { i.e., } \frac{\omega_1}{f_1}+\frac{\omega_2}{f_2}=0\)……(5)
This condition is called the condition of achromatism (for two thin lenses in contact) and the lens combination which satisfies this condition is called achromatic lens, from this condition, i.e., from Eqn.
(5) it is clear that in the case of achromatic doublet:
(1) The two lenses must be of different materials.
Since, if \(\omega_1=\omega_2, \quad \frac{1}{f_1}+\frac{1}{f_2}=0 \quad \text { i.e., } \quad \frac{1}{F}=0\) or \(F=\infty\)
i.e., combination will not behave as a lens, but as a plane glass plate.
(2) As ω1 and ω2 are positive quantities, for equation (5) to hold, f1 and f2 must be of opposite nature, i.e. if one of the lenses is converging the other must be diverging.
(3) If the achromatic combination is convergent, \(\mathrm{f}_{\mathrm{c}}<\mathrm{f}_{\mathrm{D}} \quad \text { and as } \quad-\frac{\mathrm{f}_{\mathrm{C}}}{\mathrm{f}_{\mathrm{D}}}=\frac{\omega_{\mathrm{C}}}{\omega_{\mathrm{D}}}, \quad \omega_{\mathrm{C}}<\omega_{\mathrm{D}}\)
i.e., in a convergent achromatic doublet, a convex lens has lesser focal length and dispersive power than the divergent one.
Question 1. Chromatic aberration in the formation of images by a lens arises because:
- Of non-paraxial rays.
- The radii of curvature of the two sides are not the same.
- Of the defect in grinding.
- The focal length varies with wavelength.
Answer: 4. The focal length varies with wavelength.
Question 2. Chromatic aberration of a lens can be corrected by:
- Providing different suitable curvatures of its two surfaces.
- Proper polishing of its two surfaces.
- Suitably combining it with another lens.
- Reducing its aperture.
Answer: 3. Suitably combining it with another lens.
Question 3. A combination is made of two lenses of focal lengths f and f’ in contact; the dispersive powers of the materials of the lenses are 0 and 0′. The combination is achromatic when:
- \(\omega=\omega_0, \omega^{\prime}=2 \omega_0, f^{\prime}=2 f\)
- \(\omega=\omega_0, \omega^{\prime}=2 \omega_0, f^{\prime}=f / 2\)
- \(\omega=\omega_0, \omega^{\prime}=2 \omega_0, f^{\prime}=-f / 2\)
- \(\omega=\omega_0, \omega^{\prime}=2 \omega_0, f^{\prime}=-2 f\)
Answer: 4. \(\omega=\omega_0, \omega^{\prime}=2 \omega_0, f^{\prime}=-2 f\)
From points (2) and (3) of passage: f and f’ must be of opposite sign.
Also ωc < ωo and fc < fD which is satisfied only by (4).
Question 4. The dispersive power of crown and flint glasses are 0.02 and 0.04 respectively. An achromatic converging lens of focal length 40 cm is made by keeping two lenses, one of crown glass and the other of flint glass, in contact with each other. The focal lengths of the two lenses are:
- 20 cm and 40 cm
- 20 cm and –40 cm
- –20cm and 40 cm
- 10 cm and –20cm
Answer: 2. 20 cm and –40 cm
⇒ \(\frac{\omega_1}{f_1}+\frac{\omega_2}{f_2}=0 \Rightarrow \frac{\omega_1}{\omega_2}=-\frac{f_1}{f_2}=\frac{1}{2}\)…..(1)
⇒ \(\frac{1}{F}=\frac{1}{f_1}+\frac{1}{f_2}=\frac{1}{40}\)…..(2)
After solving (1)and (2) \(f_1=20 \mathrm{~cm} \quad f_2=-40 \mathrm{~cm} \text {. }\)
Question 5. Chromatic aberration in a spherical concave mirror is proportional to :
- f
- f²
- 1/f
- None of these
Answer: 4. None of these
Chromatic aberration doesn’t occur in the case of spherical mirrors
Chapter 7 Geometrical Optics Optical Instruments
Definition: Optical instruments are used primarily to assist the eye in viewing an object.
Types of Instruments: Depending upon the use, optical instruments can be categorized in the following way
1. Microscope
It is an optical instrument used to increase the visual angle of neat objects that are too small to be seen by the naked eye.
1.1 Simple Microscope: It is also known as a magnifying glass or simply magnifier and consists of a convergent lens with the object between its focus and optical center and the eye close to it. The image formed by it is erect, virtually enlarged, and on the same side of the lens between the object and infinity.
The magnifying power (MP) or angular magnification of a simple microscope (or an optical instrument) is defined as the ratio of the visual angle with an instrument to the maximum visual angle for clear vision when the eye is unadded (i.e. when the object is at least distance of distinct vision)
i.e., magnifying power is M.P
MP = \(\frac{\text { Visual angle with instrument }}{\text { Max.visual angle for unadded eye }}=\frac{\theta}{\theta_0}\)
If an object of size h is placed at a distance u (< D) from the lens and its image size h’ is formed at a distance V (D) from the eye \(\theta=\frac{\mathrm{h}^{\prime}}{\mathrm{v}}=\frac{\mathrm{h}}{\mathrm{u}} \quad \text { with } \theta_0=\frac{\mathrm{h}}{\mathrm{D}}\)
So magnifying power MP = \(\frac{\theta}{\theta_0}=\frac{h}{u} \times \frac{D}{h}=\frac{D}{u}\)…..(1)
Now there are two possibilities
(a1) If their image is at infinity [Far point] In this situation from lens formula \(\frac{1}{v}-\frac{1}{u}=\frac{1}{f}\) we have \(\frac{1}{\infty}-\frac{1}{-u}=\frac{1}{f}\) i.e., u=f
So \(M P=\frac{D}{u}=\frac{D}{f}\)…..(2)
As here u is maximum [as the object is to be within focus], MP is minimum and as in this situation parallel beam of light enters the eye, the eye is least strained and is said to be normal, relaxed, or unstrained.
(a2) If the image is at D [Near point] In this situation as v=D, from lens formula \(\frac{1}{v}-\frac{1}{u}=\frac{1}{f}\) we have \(\frac{1}{-D}-\frac{1}{-u}=\frac{1}{f}\)
i.e., \(\quad \frac{D}{u}=1+\frac{D}{f} \quad So \quad M P=\frac{D}{u}=\left[1+\frac{D}{f}\right]\)…..(3)
As the minimum value of v for clear vision is D, in this situation u is the minimum and hence this is the maximum possible MP of a simple microscope in this situation final image is closest to the eye, eye is under maximum strain.
Special Points
- A simple magnifier is an essential part of most optical instruments (such as a microscope or telescope) in the form of an eyepiece or an ocular.
- The magnifying power (MP) has no unit. It is different from the power of a lens which is expressed in diopter (D) and is equal to the reciprocal of focal length in metre.
- With the increase in the wavelength of light used, the focal length of the magnifier will increase and hence its MP will decrease.
Example: A man with a normal near point (25 cm) reads a book with small print using a magnifying a thin convex lens of focal length 5 cm. (1) What is the closest farthest distance at which he can read the book when viewing through the magnifying glass? (2) What is the maximum and minimum MP possible using the above simple microscope?
Solution:
(1)As for the normal eye far and near points are 25 cm respectively, so for the magnifier and
However, for a lens as \(\frac{1}{v}-\frac{1}{u}=\frac{1}{f} \quad \text { i.e., } \quad u=\frac{f}{\left(\frac{f}{v}\right)-1}\)
So u will be minimum when
i.e., \((\mathrm{u})_{\min }=\frac{5}{\left(\frac{-5}{25}\right)-1}=-\frac{25}{6}=-4.17 \mathrm{~cm}\)
And u will be maximum when
So, the closest and furthest distance of the book from the magnifier (or eye) for clear viewing is 4.17 cm and 5 cm respectively.
(2)An in case of simple magnifier MP = (\(\frac{D}{u}\)). So MP will be minimum when u = max = 5 cm
i.e., \((M P)_{\min }=\frac{-25}{-5}=5 \quad\left[=\frac{D}{f}\right]\)
And MP will be maximum when \(\mathrm{u}=\min =\left(\frac{25}{6}\right) \mathrm{cm}\)
i.e., \((M P)_{\max }=\frac{-25}{-\left(\frac{25}{6}\right)}=6\left[=1+\frac{D}{f}\right]\)
1.2 Compound-Microscope
Construction: It consists of two convergent lenses of short focal lengths and apertures arranged co-axially lens (of focal length f0) facing the object is called an objective or field lens while the lens (of focal length f0) facing the eye, eye-piece or ocular. The objective has a smaller aperture and smaller focal length than the eyepiece. The separation between objective and eye-piece can be varied.
Image Formation: The object is placed between F and 2F of the objective so the image IM formed by the objective (called intermediate image) is inverted, real enlarged, and at a distance greater than f0 on the other side of the lens. This image IM acts as an object for an eye-piece and is within its focus. So eye-piece forms the final image I which is erect, virtual, and enlarged with respect to the intermediate image IM.
So the final image I with respect to the object is inverted, virtual, enlarged, and at a distance D from the eye on the same side of the eye-piece as IM. This all is shown in the figure.
Magnifying power (MP)
The magnifying power of an optical instrument is defined as \(\mathrm{MP}=\frac{\text { Visual angle with instrument }}{\text { Max. Visual angle for unadded eye }}=\frac{\theta}{\theta_0}\)
If the size of the object is h and the least distance of distinct vision is D.
⇒ \(\theta_0=\left[\frac{\mathrm{h}}{\mathrm{u}_{\mathrm{e}}}\right] \times\left[\frac{\mathrm{D}}{\mathrm{h}}\right]=\left[\frac{\mathrm{h}^{\mathrm{\prime}}}{\mathrm{h}}\right]\left[\frac{\mathrm{D}}{\mathrm{u}_{\mathrm{e}}}\right]\)
But for objective \(\mathrm{m}=\frac{\mathrm{l}}{\mathrm{O}}=\frac{\mathrm{v}}{\mathrm{u}} \text { i.e., } \frac{\mathrm{h}^{\prime}}{\mathrm{h}}=-\frac{v}{u} \text { [as } \mathrm{u} \text { is positive] }\)
So, MP = \(-\frac{v}{u}\left[\frac{D}{u_c}\right]\) with length of tube \(\mathrm{L}=\mathrm{v}+\mathrm{u}_{\mathrm{e}^{\prime}}\)……(1) now there are two possibilities
(b1) If the final image is at infinity (far point):
This situation is called normal adjustment as in this situation eye is least strained or relaxed. In this situation as for eye-piece V = ∞
⇒ \(\frac{1}{-\infty}-\frac{1}{-u_e}=\frac{1}{f_e} \text { i.e., } \quad u_e=f_e=\text { maximum }\)
Substitution this value of in equation (1), we have \(M P=-\frac{v}{u}\left[\frac{D}{f_e}\right] \text { with } L=v+f_e\)
MP = \(-\frac{v}{u}\left[\frac{D}{f_e}\right] \text { with } L=v+f_e\)…….(2)
A microscope is usually considered to operate in this mode unless stated otherwise. In this mode, as ue is maximum MP is minimum for a given microscope.
(b2) If The Final Image Is At D (Near Point): In this situation as for eye-piece v = D
⇒ \(\frac{1}{-D}-\frac{1}{-u_e}=\frac{1}{f_e} \quad \text { i.e., } \quad \frac{1}{u_e}=\frac{1}{D}\left[1+\frac{D}{f_e}\right]\)
Substituting this value of \(u_d\) in equation (1), we have
MP = \(-\frac{v}{u}\left[1+\frac{D}{f_e}\right] \quad \text { with } \quad L=v+\frac{f_e D}{f_e+D}\)…..(3)
In this situation as ue is minimum MP is maximum and the eye is most strained.
Example: The focal length of the objective and eyepiece of a microscope are 2 cm and 5 cm respectively and the distance between them is 20 cm. Find the distance of the object from the objective, when the final image seen by the eye is 25 cm from the eyepiece. Also, find the magnifying power.
Solution:
Given \(\mathrm{f}_0=2 \mathrm{~cm}, \mathrm{f}_{\mathrm{e}}=5 \mathrm{~cm}\)
⇒ \(\left|v_0\right|+\left|u_e\right|=20 \mathrm{~cm}\)
∴ \(\mathrm{v}_{\mathrm{e}}=-25 \mathrm{~cm}\)
From lens formula \(\frac{1}{\mathrm{f}_{\mathrm{c}}}=\frac{1}{\mathrm{v}_{\mathrm{o}}}-\frac{1}{\mathrm{u}_{\mathrm{c}}} \quad \Rightarrow \quad \frac{1}{\mathrm{u}}=\frac{1}{\mathrm{v}_{\mathrm{e}}}-\frac{1}{\mathrm{f}_{\mathrm{e}}}=-\frac{1}{25}-\frac{1}{5}\)
⇒ \(u_0=-\frac{25}{6} \mathrm{~cm}\)
Distance of real image from objective \(v_0=20-\frac{25}{6}\left|u_0\right|=20-=\frac{120-25}{6}=\frac{95}{6} \mathrm{~cm}\)
Now \(\frac{1}{f_0}=\frac{1}{v_0}-\frac{1}{u_0}\) given \(\frac{1}{u_0}=\frac{1}{v_0}-\frac{1}{f_0}=\frac{1}{(95 / 6)}-\frac{1}{2}\)
i.e., \(\frac{1}{u_0}=\frac{6}{95}-\frac{1}{2}=\frac{12-95}{190}=-\frac{83}{190}\)
∴ \(u_o=-\frac{190}{83}=-2.3 \mathrm{~cm}\)
Magnifying power M = \(-\frac{\mathrm{v}_0}{\mathrm{u}_{\mathrm{o}}}\left(1+\frac{\mathrm{D}}{\mathrm{f}_e}\right)=-\frac{95 / 6}{(190 / 83)}\left(1+\frac{25}{3}\right)=-41.5\)
Question 1. The focal lengths of the objective and eye lens of a microscope are 1 cm and 5cm respectively If the magnifying power for the relaxed eye is 45, then the length of the tube is
- 30 cm
- 25 cm
- 15cm
- 12 cm
Answer: 3. 15cm
Question 2. In a compound microscope magnification will be large, if the focal length of the eyepiece is
- Large
- Smaller
- Equal to that of objective
- Less than that of objective
Answer: 2. Smaller
Question 3. The magnifying power of a simple microscope is (when the final image is formed at D= 25 cm from the eye)
- \(\frac{D}{f}\)
- 1 + \(\frac{D}{f}\)
- 1 + \(\frac{f}{D}\)
- 1- \(\frac{D}{f}\)
Answer: 2. 1 + \(\frac{D}{f}\)
Geometrical Optics NEET Physics Chapter 7 Study Material
Question 4. If in a compound microscope m1 and m2 be the linear magnification of the objective lens and eye lens respectively then the magnifying power of the compound microscope will be
- \(m_1-m_2\)
- \(\sqrt{m_1+m_2}\)
- \(\left(m_1+m_2\right) / 2\)
- \(m_1 \times m_2\)
Answer: 4. \(m_1 \times m_2\)
2. Telescope
2. 1 Astronomical Telescope It is an optical instrument used to increase the visual angle of distant large objects such as a star a planet a cliff etc. An astronomical telescope consists of two converging lenses.
The one facing the object is called objective or field-lens and has large focal length and aperture. The distance between the two lenses is adjustable.
As a telescope is used to see distant objects, it object is between and 2F of the objective, and hence image formed by the objective is real, inverted, and diminished and is between F and 2F on the other side of it.
This image (called intermediate image) acts as an object for the eye-piece and shifting the position of the eyepiece is brought within its focus. So final image I, with respect to the intermediate image is erect, virtual, enlarged, and at a distance D from the eye.
This in turn implies that the final image with respect to the object is inverted, enlarged, and at a distance D from the eye.
Magnifying Power (MP)
Magnifying Power of a telescope is defined as
MP = \(\frac{\text { Visual angle with instrument }}{\text { Visual angle for unadded eye }}=\frac{\theta}{\theta_0}\)
But from the figure. \(\theta_0=\left(\frac{y}{f_0}\right) \text { and } \theta=\left(\frac{y}{-u_0}\right)\)
So MP \(=\frac{\theta}{\theta_0}=-\left[\frac{f_0}{u_e}\right]\) with length of tube
L = \(\left(\mathrm{f}_0+\mathrm{U}_e\right)\)……..(1)
(d1) If the final image is at infinity (far point) This situation is called normal adjustment as in this situation eye is least strained or relaxed. In this situation as for eye-piece v = ∞
⇒ \(\frac{1}{-\infty}-\frac{1}{u_e}=\frac{1}{f_e} \quad \text { i.e., } \quad u_e=f_e\)
So, substituting this value of u e in equation (1) we have \(M P=-\left(\frac{f_0}{f_e}\right) \text { and } L=\left(f_0+f_e\right)\)
Usually, a telescope operates in this mode unless stated otherwise. In this mode, as ue is the maximum for a given telescope MP is the minimum while the length of the tube maximum.
(d2) If the final image is at D (near point)
In this situation as for eye-piece v = D
⇒ \(\frac{1}{-D}-\frac{1}{-u_e}=\frac{1}{f_e} \quad \text { i.e., } \quad \frac{1}{-u_e}=\frac{1}{f_e}\left[1+\frac{f_e}{D}\right]\)
So substituting this value of u e in Equation (1), we have \(M P=\frac{f_0}{f_e}\left[1+\frac{f_e}{D}\right] \text { with } L=f_0+\frac{f_e D}{f_e+D}\)…..(3)
In this situation, ue is minimum so for a given telescope MP is maximum while the length of the tube is minimum and the eye is most strained. In the case of a telescope if the object and final image are at infinity and total light entering the telescope leaves it parallel to its axis as shown in the figure.
⇒ \(\frac{f_0}{f_e}=\frac{\text { Aperture of object }}{\text { Aperture of eyepiece }} \quad \text { i.e., } \quad M P=\frac{f_0}{f_e}=\frac{D}{d}\)……(4)
2.2 Terrestrial Telescope
Uses a third lens in between the objective and eyepieces so as to form the final image. This lens simply inverts the image formed by the objective without affecting the magnification.
Length of tube L = f0 + fe + 4f
2.3 Galileo’s Telescope
Convex lens as objective.
The concave lens is as eyepiece.
The field of view is much smaller
∴ eyepiece lens in concave.
(1) \(M=\frac{f_0}{f_e}\left[1-\frac{f_c}{v_e}\right]\)
(2) \(\mathrm{M}=\frac{\mathrm{f}_0}{\mathrm{f}_e}\)
Final image is at \(\alpha \quad L=f_0-f_0\)
(3) \(M=\frac{f_0}{f_e}\left[1-\frac{f_e}{D}\right]\)
The final image is at D. \(L=f_0-u_e\)
2.4 Binocular
In this telescope, as the intermediate image is outside the tube, the telescope cannot be used for making measurements. If two telescopes are mounted parallel to each other so that an object can be seen by both eyes simultaneously, the arrangement is called ‘binocular’
In a binocular, the length of each tube is reduced by using a set of totally reflecting prisms which provide an intense, erect image free from lateral inversion. Through a binocular, we get two images of the same object from different angles at the same time.
Their superposition gives the perception of depth also along with length and breadth, i.e., a binocular vision given a proper three-dimensional (3-D) image.
Example 1. A telescope consists of two convex lenses of focal length 16 cm and 2 cm. What is the angular magnification of a telescope for a released eye? What is the separation between the lenses? If an object subtends an angle of 0.5º on the eye, what will be the angle subtended by its image?
Solution:
Angular magnification M = \(\frac{\alpha}{\beta}=\frac{F}{f}=\frac{16}{2}=8 \mathrm{~cm}\)
Separation between lenses = F + f = 16 + 2 = 18 cm
Here α = 0.5°
∴ Angular subtended by image β= M α = 8 x 0.5° = 4°
Example 2. The magnifying power of the telescope is found to be 9 and the separation between the lenses is 20 cm for related eye. What are the focal lengths of component lenses?
Solution:
Magnification M = \(\frac{F}{f}\)
Separation between lenses d = F + f
Given \(\frac{F}{f}\) = 9 i.e., F = 9f …..(1)
and F + f = 20 …..(2)
Putting the value of F from (1) in (2), we get 9f + f = 20 ⇒ 10 f = 20
⇒ \(\frac{20}{10}\) = 2cm
∴ F = 9f = 9 x 2 = 18 cm .
∴ F = 18 cm, f = 2 cm
Question 1. The magnifying power of a telescope can be increased by
- Increasing the focal length of the system
- Fitting eyepiece of high-power
- Fitting eyepiece of low-power
- Increasing the distance between objects
Answer: 2. Fitting eyepiece of high-power
Question 2. A simple telescope, consisting of an objective of focal length 60 cm and a single eye lens of focal length 5 cm is focussed on a distant object in such a way that parallel rays come out from the eye lens. If the object subtends an angle of 2° at the objective, the angular width of the image
- 10°
- 24°
- 50°
- 1/6°
Answer: 2. 24°
Question 3. If the telescope is reversed i.e., seen from the objective side
- The object will appear very small
- The object will appear very large
- There will be no effect on the image formed by the telescope
- The image will be slightly greater than the earlier one
Answer: 1. The object will appear very small
Question 4. The aperture of a telescope is made large, because
- To increase the intensity of the image
- To decrease the intensity of the image
- To have greater magnification
- To have a lesser resolution
Answer: 1. To increase the intensity of the image
Question 5. The magnifying power of a telescope is 9. When it is adjusted for parallel rays, the distance between the objective and the eye-piece is found to be 20 cm. The focal lengths of the two lenses are
- 18 cm, 2 cm
- 11 cm, 9 cm
- 10 cm, 10 cm
- 15 cm, 5 cm
Answer: 1. 18 cm, 2 cm
Question 6. A reflecting telescope utilizes
- A concave mirror
- A convex mirror
- A prism
- A plano-convex lens
Answer: 1. A concave mirror
Geometrical Optics NEET Physics Chapter 7 Study Material
3. Comparison Between Compound – Microscope And Astronomical-Telescope
Scattering Of Light
When light from some source (ga, sun, stars) enters the earth’s atmosphere then it gets reflected in various directions by the particles of dust, smoke, and gas molecules. The phenomenon of this diffuse reflection is known as scattering. This was initially suggested by Tindal
According to Rayleigh, the intensity (I) of scattered light is inversely proportional to the fourth power of wavelength of light. i.e. λ
i.e. \(I \propto \frac{1}{\lambda^4}\)
That is the reason why red light (λ more) gets scattered minimum and violet light (λ less) gets scattered maximum (IR = 16IV)
(3) Consequences of scattering of light
- The appearance of the blue color of the sky.
- The danger signals are made red.
- The appearance of the black color of the sky in the absence of atmosphere.
- The appearance of the red color of the sun at sunrise and sunset.
Luminous bodies — The bodies which emit light themselves are known as luminous bodies.
Rainbow
- The seven-coloured curved strip formed as a result of dispersion of light through water droplets which keep suspended in the atmosphere after rains, is known as a rainbow.
- The rainbow is of two types –
- Primary rainbow
- Secondary rainbow
Difference between primary and secondary rainbow
Limit Of Resolution And Resolving Power Of Optical Instruments
- Resolving power
- The ability of an optical instrument to produce separate diffraction patterns of two nearby objects is known as resolving power.
- The ability of an optical instrument to show two closely lying objects or spectral lines as separate is known as its resolving power.
- Limit of resolution – The reciprocal of resolving power is defined as the limit of resolution.
- Rayleigh’s limit of resolution – The distance between two object points, when the central maximum of the diffraction pattern of one coincides with the first minimum of the diffraction pattern of another, is defined as Rayleigh’s limit of resolution.
Resolving Power (R.P.) Of Telescope
R . P = \(\frac{a}{1.22 \lambda}=\frac{1}{\Delta \theta}=\frac{D}{d}\)
a = diameter of the aperture of the objective
λ = wavelength of light used
Δθ = Limit of resolution
d = distance of two objects
D = distance of objects from the objective lens
The resolving power of an electron microscope is 4 x 103 times that of an ordinary microscope.
Resolving power (R.P.) of microscope
Rp = \(\frac{2 \mu \sin \theta}{1.22 \lambda}\)
μ = refractive index of the medium.
Solved Miscellaneous Problems
Problem 1. See the following figure. Which of the object(s) shown in the figure will not form its image in the mirror?
Solution:
No ray from O3 is incident on the reflecting surface of the mirror, so its image is not formed.
Problem 2. The figure shows an object AB and a plane mirror MN placed parallel to the object. Indicate the mirror length required to see the image of the object if the observer’s eye is at E.
Solution:
Required length of Mirror = MN
∴ \(\triangle M N E \& \triangle A^{\prime} B^{\prime} E\) are similar
∴ \(\frac{M N}{O E}=\frac{A^{\prime} B^{\prime}}{\mathrm{C}^{\prime} \mathrm{E}} \Rightarrow \quad M N=\frac{A^{\prime} B^{\prime}}{2}=\frac{A B}{2}\)
Problem 3. An object is kept fixed in front of a plane mirror which is moved by 10 m/s away from the object, to find the velocity of the image.
Solution:
⇒ \(\overrightarrow{\mathrm{V}}_{\mathrm{IM}}=-\overrightarrow{\mathrm{V}}_{\mathrm{OM}}\)
⇒ \(\overrightarrow{\mathrm{V}}_{\mathrm{I}, \mathrm{G}}-\overrightarrow{\mathrm{V}}_{\mathrm{M}, \mathrm{G}}=-\overrightarrow{\mathrm{V}}_{\mathrm{O}, \mathrm{G}}+\overrightarrow{\mathrm{V}}_{\mathrm{M}, \mathrm{G}}\)
⇒ \(\overrightarrow{\mathrm{V}}_{\mathrm{M}, \mathrm{G}}=\frac{\overrightarrow{\mathrm{V}}_{\mathrm{L}, \mathrm{G}}+\overrightarrow{\mathrm{V}}_{\mathrm{O}, \mathrm{G}}}{2}\)
(because \(\overrightarrow{\mathrm{V}}_{\mathrm{O}, \mathrm{G}}=0 \quad \Rightarrow \quad \frac{\overrightarrow{\mathrm{V}}_{\mathrm{I}, \mathrm{G}}}{2}=10 \hat{\mathrm{i}} \mathrm{m}\)
⇒ \(\overrightarrow{\mathrm{V}}_{\mathrm{I}, \mathrm{G}}=20 \hat{\mathrm{i}} \mathrm{m} / \mathrm{s}\)
Problem 4. Find the position of the final image after three successive reflections taking the first reflection on m1.
Solution:
1st reflection:
u = – 15cm
f = -10 cm
⇒ \(\frac{1}{v}+\frac{1}{u}=\frac{1}{f}=\frac{-3+2}{30}=-\frac{1}{30}\)
v = – 30 cm
2nd reflection at the plane mirror:
u = 5 cm
v = – 5 cm
for 3rd reflection on curved mirror again: u = – 20 cm
v = \(\frac{u f}{u-f}=\frac{(-20) \times(-10)}{-20+10}=\frac{200}{-10}=-20 \mathrm{~cm}\)
NEET Physics Class 12 Chapter 7 Geometrical Optics Notes
Problem 5. A coin is placed 10 cm in front of a concave mirror. The mirror produces a real image that has a diameter of 4 times that of the coin. What is the image distance?
Solution:
m = \(\frac{d_2}{d_1}=-\frac{v}{u}\)
We have, u = 10 cm (virtual object) as the real image is formed
v = – mu = – 4 ×10 cm = – 40 cm
Problem 6. A small statue has a height of 1 cm and is placed in front of a spherical mirror. The image of the statue is inverted and is 0.5cm tall and located 10 cm in front of the mirror. Find the focal length and nature of the mirror.
Solution:
We have \(\mathrm{m}=\frac{\mathrm{h}_2}{\mathrm{~h}_1}=-\frac{0.5}{1}=-0.5\)
v = -10 cm (real image)
But m = \(\frac{f-v}{f}-0.5=\frac{f+10}{f} \quad \Rightarrow \quad f=\frac{-20}{3} \mathrm{~cm}\)
So, a concave mirror
Problem 8. A light ray deviates by 300 (which is one-third of the angle of incidence) when it gets refracted from a vacuum to a medium. Find the refractive index of the medium.
Solution:
δ = i – r \(\Rightarrow \quad \frac{\mathrm{i}}{3}=\mathrm{i}-\mathrm{r}=30^{\circ} . \quad \Rightarrow \quad \mathrm{i}=90^{\circ} \quad \Rightarrow \quad 2 \mathrm{i}=3 \mathrm{r}\)
∴ r = \(\frac{2 \mathrm{i}}{3}=60^{\circ}\)
So, \(\mu=\frac{\sin 90^{\circ}}{\sin 60^{\circ}}=\frac{1}{\sqrt{3 / 2}}=\frac{2}{\sqrt{3}}\)
Problem 8. A coin lies on the bottom of a lake 2m deep at a horizontal distance x from the spotlight (a source of a thin parallel beam of light) situated 1 m above the surface of a liquid of refractive index μ= √2 and height 2m. Find x.
Solution:
⇒ \(\sqrt{2}=\frac{\sin 45^{\circ}}{\sin r} \quad \Rightarrow \sin r=\frac{1}{2} \quad \Rightarrow \quad r=30^{\circ}\)
x = \(R Q+Q P=1 m+2 \tan 30^{\circ} \mathrm{m}=\left(1+\frac{2}{\sqrt{3}}\right) \mathrm{m} \text { Ans. }\)
Problem 9. A ray of light falls at an angle of 30º onto a plane-parallel glass plate and leaves it parallel to the initial ray. The refractive index of the glass is 1.5. What is the thickness d of the plate if the distance between the rays is 3.82 cm?
(Given: \(\sin ^{-1}\left(\frac{1}{3}\right)=19.5^{\circ} ; \cos 19.5^{\circ}=0.94 ; \sin 10.5^{\circ}=0.18\))
Solution:
Using \(s=\frac{d \sin (i-r)}{\cos r} \quad \Rightarrow \quad d=\frac{3.82 \times \cos r}{\sin \left(30^{\circ}-r\right)}\)……(1)
Also, \(\quad 1.5=\frac{\sin 30^{\circ}}{\sin \mathrm{r}} \quad \Rightarrow \quad \sin \mathrm{r}=\frac{1}{3}\)
so, \(r=19.5^{\circ}\)
So, \(d=\frac{3.82 \times \cos 19.5^{\circ}}{\sin \left(30^{\circ}-19.5^{\circ}\right)}=\frac{3.82 \times 0.94}{\sin 10.5^{\circ}}=\frac{3.82 \times 0.94}{0.18}=19.948 \mathrm{~cm} \approx 0.2 \mathrm{~cm}\)
Problem 10. Light passes through many parallel slabs one by one as shown in the figure.
Prove that n1sini1, = n2sini2 = n3sini3 = n4sini4 = ……[Remember this]. Also, prove that if n1 = n4 then light rays in medium n1 and in medium n4 are parallel.
Solution:
We have, \(\frac{\sin i_1}{\sin i_2}=\frac{n_2}{n_1}\)
⇒ \(n_1 \sin i_1=n_2 \sin i_2\)…..(1)
Similarly \(n_2 \sin i_2=n_3 \sin i_3\) so on
so, \(\quad n_1 \sin i_1=n_2 \sin i_2=n_3 \sin i_3=\)
⇒ \(n_1 \sin i_1=n_4 \sin i_4 \Rightarrow \quad \sin i_1=\sin i_4\)
(\(n_1=n_2\))
so, \(i_1=i_4\)
Hence, light rays in medium n1 and in medium n4 are parallel.
Problem 11. An object lies 90 cm in the air above the water’s surface. It is viewed from the water nearly normally. Find the apparent height of the object
Solution:
d’ = \(\frac{d}{n_{r e 1}}=\frac{d}{n_i / n_r}=\frac{90 \mathrm{~cm}}{\frac{1}{4 / 3}}=\frac{90 \times 4}{3} \mathrm{~cm}=120 \mathrm{~cm} \text { Ans. }\)
Problem 12. Prove that the shift in position of object due to parallel slab is given by shift = \(\mathrm{d}\left(1-\frac{1}{\mathrm{n}_{\mathrm{rcl}}}\right)\) where \(\mathrm{n}_{\mathrm{rat}}=\frac{\mathrm{n}}{\mathrm{n}^{\prime}}\)
Solution:
Because of the ray refraction at the first surface, the image of O is formed at I1. For this refraction, the real depth is AO = x and the apparent depth is AI1
Thus: \(\mathrm{AI}_1=\frac{A O}{n_1 / n_r}=\frac{A O}{n^{\prime} / n}=\frac{n(A O)}{n^{\prime}}\)
The point \(\mathrm{I}_1\) acts as the object for the refraction of the second surface.
Due to this refraction, the image of \(\mathrm{I}_1\) is formed at \(\mathrm{I}_2\).
Thus, \(\mathrm{BI}_2=\frac{\left(B I_1\right)}{\left(n / n^{\prime}\right)}=\frac{n^{\prime}}{n}\left(B I_1\right)=n^{\prime} / n\left(A B+\mathrm{AI}_1\right)=\frac{n^{\prime}}{n}\left[d+\frac{n}{n^{\prime}}(A O)\right]=\frac{n^{\prime}}{n} d+A O\).
Net shift \(=\mathrm{OI}_2=\mathrm{BO}-\mathrm{BI}_2\)
= \(d+(A O)-\frac{n^{\prime}}{n} d-A O=d\left(1-\frac{n^0}{n}\right)=d \quad\left(1-\frac{1}{n_{\text {rel }}}\right)\) where \(n_{\text {rel }}=n_{\text {rel }}=\frac{n}{n^{\prime}}\).
Problem 13. Find the apparent depth of object O below surface AB, seen by an observer in the medium of refractive index μ2
Solution:
∴ \(\mathrm{d}_{\text {esp. }}=\frac{\mathrm{t}_1}{\mu_1 / \mu_2}\)
Problem 14. In the above question, what is the depth of the object corresponding to incident rays striking on surface CD in medium μ2
Solution:
Depth of the object corresponding to incident ray striking on the surface CD in medium µ2 = t2
Problem 15. In the above question if the observer is in medium μ3, what is the apparent depth of the object seen below the surface CD?
Solution:
If the observer is in medium µ3. apparent depth below surface CD = QI2.
= \(\sum \frac{t_i}{\left(n_{\text {rel }}\right)_{\mathrm{i}}}=\frac{\mathrm{t}_2}{\mu_2 / \mu_3}+\frac{t_1}{\mu_1 / \mu_3}\)
Problem 16. Find the radius of the circle of illuminance, If a luminous object is placed at a distance of h from the interface in the denser medium.
Solution:
tan c = \(\frac{r}{h}\)
∴ r = h tan c
circle of illuminance
But \(C=\sin ^{-1} \frac{1}{\left(\mu_d / \mu_r\right)} \quad \text { so, } \quad r=h \tan \left[\sin ^{-1} \frac{1}{\left(\mu_d / \mu_r\right)}\right]=h \cdot \frac{\mu_r}{\sqrt{\mu_d^2-\mu_r^2}}\)
Problem 17. A ship is sailing in a river. An observer is situated at a depth of h in water (μw). If x >> h, find the angle made from the vertical, of the line of sight of the ship.
Solution:
C = \(\sin ^{-1}\left(\frac{\mu_a}{\mu_w}\right)\)
Problem 18. Find r, r´, e, δ for the case shown in the figure.
Solution:
Here \(\theta=180^{\circ}-75^{\circ}=105^{\circ}\)
sin\(45^{\circ}=\sqrt{2} \sin r\)
∴ r = \(\sin ^{-1} \frac{1}{2}=30^{\circ}\)
r’ \(=180^{\circ}-(r+\theta)\)
= \(180^{\circ}-30^{\circ}-105^{\circ}=45^{\circ}\)
sin \(e=\sqrt{2} \sin r^{\prime}\)
∴ sin \(e=\sqrt{2} \times \sin 45^{\circ}=1\)
∴ e = \(90^{\circ}\)
So, \(\delta=\mathrm{i}+\mathrm{e}-\mathrm{A}=45^{\circ}+90^{\circ}-75^{\circ}=60^{\circ}\)
NEET Physics Class 12 Chapter 7 Geometrical Optics Notes
Problem 19. From the graph of angle of deviation δ versus angle of incidence i, find the prism angle
Solution:
From the graph
δ = i + e – A.
30° = 30° + 60° – A
A = 60°
use the result If i and e are interchanged then we get the same value of δ
Problem 20. Find the focal length of a plano-convex lens with R1 = 15 cm and R2 = ∞. The refractive index of the lens material n = 1.5.
Solution:
∴ \(\frac{1}{\mathrm{f}}=(\mathrm{n}-1)\left(\frac{1}{\mathrm{R}_1}-\frac{1}{\mathrm{R}_2}\right)=(1.5-1)\left(\frac{1}{15}-\frac{1}{\infty}\right)=0.5 \times \frac{1}{15}\)
∴ f = 30 cm
Problem 21. Find the focal length of a concavo-convex lens (positive meniscus) with R1 = 15 cm and R2 = 25 cm. The refractive index of the lens material n = 1.5.
Solution:
⇒ \(\frac{1}{\mathrm{f}}=(1.5-1)\left(\frac{1}{15}-\frac{1}{25}\right)=0.5\left(\frac{10-6}{150}\right)\)
∴ f = \(\frac{300}{4}=75 \mathrm{~cm}\)
Problem 22. The figure shows a point object and a diverging lens. Find the final image formed.
Solution:
∴ \(\frac{1}{v}-\frac{1}{u}=\frac{1}{f} \Rightarrow \quad \frac{1}{v}=\frac{1}{-10}+\frac{1}{(-10)} \quad=-\frac{2}{10}=v=-5 \mathrm{~cm}\)
Problem 23. Find the equivalent focal length of the system for paraxial rays parallel to the axis.
Solution:
∴ \(\frac{1}{f_{e q}}=\frac{1}{f_1}+\frac{1}{f_2}-\frac{d}{f_1 f_2}=\frac{1}{10}+\frac{1}{-10}-\frac{20}{10(-10)}=\frac{1}{5}\)
⇒ \(f_{o q}=5 \mathrm{~cm}\)
Problem 24. See the figure Find the equivalent focal length of the combination shown in the figure and the position of the image.
Solution:
For the women lens \(\frac{1}{f}=(n-1)\left(\frac{1}{R_1}-\frac{1}{R_2}\right)=\left(\frac{3}{2}-1\right)\left(\frac{1}{-10}-\frac{1}{10}\right)=-\frac{1}{2} \times \frac{2}{10}=\frac{1}{10}\)
Ago, Fm \(=\mathrm{R} / 2=\frac{10}{2}=5 \mathrm{~cm}\)
∴ \(\frac{1}{f_{e q}}=\frac{1}{F_m}-2 \frac{1}{\mathrm{f}}=\frac{1}{5}+2 \times \frac{1}{10}=\frac{2}{5}\)
∴ \(f_{\mathrm{eq}}=2.5 \mathrm{~cm}\)
Chapter 7 Geometrical Optics Questions and Answers
Question 1. The sky appears to be red in color at the time the sun sat. The reason is
- The blue color of sun rays is scattered away by the atmosphere
- As the sun emits only red color in the morning
- White light is made to appear red by the Atmosphere
- None of the above
Answer: 1. The Blue color of sun rays is scattered away by the atmosphere
Question 2. A light ray traveling in a glass medium is incident on the glass-air interface at an angle of incidence θ. The reflected (R ) and transmitted (T) intensities, both as a function of θ, are plotted. The correct sketch is
Answer: 3
Initially, most of the parts will be transmitted.
When θ > ic, all the light rays will be total internal reflected.
So transmitted intensity = 0
So the correct answer is (3)
Question 3. The image of an object, formed by a plano-convex lens at a distance of 8 m behind the lens, is real and is one-third the size of the object. The wavelength of light inside the lens is 2/3 times the wavelength in free space. The radius of the curved surface of the lens is :
- 1 m
- 2 m
- 3 m
- 6 m
Answer: 3. 3 m
v = \(8 \mathrm{~m} \quad \text { (magnification }=-\frac{1}{3}=\frac{v}{u} \text { ) }\)
u = -24 m
⇒ \(\frac{1}{f}=\left(\frac{3}{2}-1\right)\left(\frac{1}{\infty}+\frac{1}{R}\right)\)
R = 3m
Question 4. I is the image of a point object O formed by a spherical mirror, then which of the following statements is incorrect:
- If O and I are on the same side of the principal axis, then they have to be on opposite sides of the mirror.
- If O and I are on opposite sides of the principal axis, then they have to be on the same side of the mirror.
- If O and I are on opposite sides of the principal axis, then they can be on opposite sides of the mirror as well.
- If O is on the principal axis then I has to lie on the principal axis only.
Answer: 3. If O and I are on opposite sides of the principal axis, then they can be on opposite sides of the mirror as well.
⇒ \(\frac{\mathrm{I}}{\mathrm{O}}=-\frac{v}{u}\)
If O and I are on the same side of PA. \(\frac{\mathrm{I}}{\mathrm{O}}\) will be positive which implies v and u will be of opposite signs.
Similarly, if O and I are on opp. sides, \(\frac{\mathrm{I}}{\mathrm{O}}\) will be -ve which implies v and u will have the same sign.
If O is on PA, \(\mathrm{I}=\left(-\frac{v}{u}\right)(\mathrm{O})=0 \Rightarrow \mathrm{I}\) will also be on. P.A.
Question 5. The diamond shines because-
- It absorbs maximum light from the sun
- Of the nature of diamond
- Of total internal reflection
- Of refraction
Answer: 3. Of total internal reflection
NEET Physics Class 12 Chapter 7 Geometrical Optics Notes
Question 6. A transparent cube contains a small air bubble. Its apparent distance is 2 cm when seen through one face and 5 cm when seen through another face. If the refractive index of the material of the cube is 1.5, the real length of the edge of the cube must be :
- 7 cm
- 7.5 cm
- 10.5 cm
- 3.5 cm
Answer: 3. 10.5 cm
Refractive index \((\mu)=\frac{\text { Real depth }}{\text { Apparent depth }}\)
Refractive index \((\mu)=1.5\)
Apparent depth \(=2+5=7 \mathrm{~cm}\)
So, 1.5 = \(\frac{\text { Real depth }}{7}\)
∴ Real depth \(=1.5 \times 7=10.5 \mathrm{~cm}\)
Question 7. The maximum refractive index of a material, of a prism of apex angle 90°, for which light may be transmitted is:
- √3
- 1.5
- √2
- None of these
Answer: 3. √2
For transmission \(r_2 \leq \sin ^{-1}(1 / \mu)\) and \(r_1 \leq \sin ^{-1}(1 / \mu)\)
⇒ \(r_1+r_2 \quad \leq \quad 2 \sin ^{-1}(1 / \mu) \quad \mathrm{A} \leq 2 \sin ^{-1}(1 / \mu)\)
⇒ \(\sin ^{-1}(1 / \mu) \geq 45^{\circ} \Rightarrow \frac{1}{\mu} \geq \frac{1}{\sqrt{2}} \Rightarrow \mu \leq \sqrt{2}\)
Question 8. The refractive index of the material of a prism is and its refracting angle is 30°. One of the refraction surfaces of the prism is made a mirror inwards. A beam of monochromatic light entering the prism from the other face will retrace its path after reflection from the mirrored surface if its angle of incidence on the prism is :
- 45°
- 60°
- 0°
- 30°
Answer: 1. 45°
According to the given condition, the beam of light will retrace its path after reflection from BC.
So \(\angle \mathrm{CPQ}=90^{\circ}\)
Thus, the angle of refraction at surface AC \(\angle \mathrm{PQN}=\angle \mathrm{r}=90^{\circ}-60^{\circ}=30^{\circ}\)
By Snell’s law \(\mu=\frac{\sin \mathrm{i}}{\sin \mathrm{r}} \Rightarrow \sqrt{2}=\frac{\sin \mathrm{i}}{\sin 30^{\circ}}\)
x \( \sin 30^{\circ}=\sin \mathrm{i} \Rightarrow =\sin \mathrm{i}\)
sin \(\mathrm{i}==\sin 45^{\circ}\)
∴ \(\mathrm{i}=45^{\circ}\)
Question 9. Prism of which material is used for the study of the infrared spectrum :
- Rock salt
- Flint glass
- Crown glass
- Quartz
Answer: 1. Rock salt
Rock salt prism is used for studying of infrared spectrum.
Question 10. The ray diagram could be correct
- If n1 = n2 = ng
- If n1 = n2 and n1 < ng
- If n1 = n2 and n1 > ng
- Under no circumstances
Answer: 3. If a, = n2 and a, > n
Question 11. A biconvex lens of focal length 15 cm is in front of a plane mirror. The distance between the lens and the mirror is 10 cm. A small object is kept at a distance of 30 cm from the lens. The final image is
- Virtual and at a distance of 16 cm from the mirror
- Real and at a distance of 16 cm from the mirror
- Virtual and at a distance of 20 cm from the mirror
- Real and at a distance of 20 cm from the mirror
Answer: 2. Real and at a distance of 16 cm from the mirror
First image, \(\frac{1}{v}-\frac{1}{u}=\frac{1}{f}\)
⇒ \(\frac{1}{v}-\frac{1}{-30}=\frac{1}{15}\)
v = 30, the image in formed 20 cm behind the mirror.
The second image, by a plane mirror, will be 20 cm in front of the plane mirror.
For third image \(\frac{1}{v}-\frac{1}{10}=\frac{1}{15}\)
⇒ \(\frac{1}{v}=\frac{1}{10}+\frac{1}{15}=\frac{3+2}{30}=\frac{5}{30}\)
v = 6 cm
Question 12. A far-sighted man who has lost his spectacles reads a book by looking through a small hole (3-4 mm) in a sheet of paper, The reason will be
- Because the hole produces an image of the letters at a longer distance
- Because in doing so, the focal length of the eye lens is effectively increased
- Because in doing so, the focal length of the eye lens is effectively decreased
- None of these
Answer: 3. Because in doing so, the focal length of the eye lens is effectively decreased
Question 13. If in a compound microscope, m1 and m2 is the linear magnification of the objective lens and eye lens respectively, then the magnifying power of the compound microscope will be
- \(m_1-m_2\)
- \(\sqrt{m_1+m_2}\)
- \(\left(m_1+m_2\right) / 2\)
- \(m_1 \times m_2\)
Answer: 4. \(m_1 \times m_2\)
Magnification of a compound microscope is given by \(\mathrm{m}=\frac{v_0}{u_0} \times \frac{D}{u_c} \Rightarrow|\mathrm{m}|=\mathrm{m}_0 \times \mathrm{m}_{\mathrm{e}}\)
Question 14. A telescope has an objective lens of 10 cm diameter and is situated at a distance of one kilometer from two objects. The minimum distance between these two objects, which can be resolved by the telescope, when the mean wavelength of light is 5000 Å, of the order of :
- 0.5 m
- 5 m
- 5 mm
- 5 cm
Answer: 3. 5 mm
Resolving limit of telescope is
θ \( \propto \frac{x}{D}=\frac{\lambda}{d}\)
x = \(\frac{\lambda D}{d}\)
Given, \(\lambda=5000\) Å = \(5000 \times 10^{-10} \mathrm{~m}\),
D = \(1 \mathrm{~km}=1000 \mathrm{~m}\)
d = \(10 \mathrm{~cm}=0.1 \mathrm{~m}\)
Hence, \(x=\frac{5000 \times 10^{-10} \times 1000}{0.1}\)
= \(5 \times 10^{-3} \mathrm{~m}\)
= \(5 \mathrm{~mm}\)
NEET Physics Class 12 Chapter 7 Geometrical Optics Notes
Question 15. The angular resolution of a 10 cm diameter telescope at a wavelength of 5000 A is of the order of :
- 108 rad
- 10-2 rad
- 10-4 rad
- 10-6 rad
Answer: 4. 10-6 rad
Angular resolution = \(\frac{1.22 \lambda}{\mathrm{d}}\)
= \(\frac{1.22 \times 5000 \times 10^{-10}}{10 \times 10^{-2}}\)
= \(6.1 \times 10^{-6}\)
\(10^{-6} \mathrm{rad}\)
Question 16. A ray of light traveling in air is incident on a medium of refractive index μ. If the angle of refraction is twice the incident angle, the incident angle is
- \(\sin ^{-1}\left(\frac{1}{\mu}\right)\)
- \(\sin ^{-1}\left(\frac{1}{2 \mu}\right)\)
- \(\cos ^{-1}\left(\frac{1}{\mu}\right)\)
- \(\cos ^{-1}\left(\frac{1}{2 \mu}\right)\)
Answer: 4. \(\cos ^{-1}\left(\frac{1}{2 \mu}\right)\)
Using Snell’s law,
sini = μ sin 2i
sini = 2 μ sini cosi
i = \(\cos ^{-1}\left(\frac{1}{2 \mu}\right)\)
However, the given data are not appropriate since when light ray goes from rarer to denser medium then light moves towards the normal.
Question 17. A point source S is placed at the bottom of a transparent block of height 10 mm and a refractive index of 2.72. It is immersed in a lower refractive index liquid as shown in the figure. It is found that the light emerging from the block to the liquid forms a circular bright spot of diameter 11.54 mm on the top of the block. The refractive index of the liquid is
- 1.21
- 1.30
- 1.36
- 1.42
Answer: 3. 1.36
Sin \(\mathrm{i}_{\mathrm{c}}=\frac{r}{\sqrt{r^2+h^2}} \Rightarrow \frac{n_f}{n_B}=\frac{r}{\sqrt{r^2+h^2}}\)
= \(\frac{r}{\sqrt{r^2+h^2}} \times 2.72=\frac{5.77}{11.54} \times 2.72=1.36\)
Question 18. An experiment is performed to find the refractive index of glass using a traveling microscope. In this experiment distances are measured by –
- A standard laboratory scale
- A meter scale provided on the microscope
- A screw gauge provided on the microscope
- A vernier scale provided on the microscope
Answer: 4. A vernier scale provided on the microscope
A traveling microscope moves horizontally on a main scale provided with a vernier scale, provided with the microscope
by the electrical equivalence of a loop.
= 10, so 5 pictures indicate 5 x 10 people].
