NCERT Solutions For Class 10 Science Chapter 10 Light-Reflection And Refraction

NCERT Solutions For Class 10 Science Chapter 10 Light-Reflection And Refraction

Question 1. Which lens can be used as a magnifying glass?

For which position  of an object does a convex lens form:

1. A virtual and erect image?
2. A real and inverted image of the same size as that of an object? Dlabeled ray diagrams to show the formation of the required image in each of the above two cases.

Answer:

A convex lens can be used as a magnifying glass.

1. When an object is between optical center C and focus F₁.

2. Wan a hen object is at 2F₁.

Question 2.

1. Wrthe the ite relation between u, v, and f for lens and for mirrors, where u, v, f are object distance, image dist, ance, and focal length respectively.
2. The magnification produced by a plane mirror is m = + 4. Write the information about the image given by this statement.
3. Draw a ray diagram for the following and show the formation of the images in case of
   concave mirror when the object is placed in:
   1. Between the pole and focus point the center of curvature

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Answer:

1. Mirror formula \(\frac{1}{f}=\frac{1}{v}+\frac{1}{u}\)
   1. Lens formula \(\frac{1}{f}=\frac{1}{v}-\frac{1}{u}\)
2. It means the height of the image is 4 times the height of the object of the image is virtual.
3. (1) Between the pole and focus point

2. The center of curvature

Question 3.

1. Under what condition, does a concave produce a virtual and magnified image? Dralabeledlled ray diagram to show the formation of image in the above case. Also, state the position of the object to provide a magnified and real image.
2. A ray of light moving along one principal axis is falling on a concave mirror. Draw the path the of reflected ray. Also, state the values of the angle of incidence and reflection in this case.

Answer:

1. The hen object lies between the focus and the pole of a mirror or with the focus of a concave mirror case.

The image formed will be real and magnified when the hen object is placed between C and F.

2. The diagram is as shown:

3. i = 0°, r = 0°

Question 4. A 2.0 cm tall object is placed perpendicular to the principal axis of a convex lens of focal length 10 cm. The distance of the object from the lens is 15 cm. Find the proposition and the size of the image forms.
Answer:

Given O = 2.0 cm, f= + 10 cm, u = – 15 cm, v = ?, I = ?

Using the lens formula, \(\frac{1}{v}-\frac{1}{u}=\frac{1}{f}\), we have

⇒ \(\frac{1}{v}=\frac{1}{f}+\frac{1}{u}=\frac{1}{10}+\frac{1}{-15}=\frac{3-2}{30}=\frac{1}{30} \quad \text { or } \quad v=30 \mathrm{~cm}\)

The image is formed on the other side of the lens at 30 cm from it.

⇒ m = \(=\frac{\mathrm{I}}{\mathrm{O}}=\frac{v}{u} \quad \text { or } \quad \mathrm{I}=\frac{v}{u} \times \mathrm{O}=\frac{30}{-15} \times 20=-4 \mathrm{~cm}\)

The negative sign shows that the image is real and inverted.

Question 5.

1. A thin converging lens forms a:
   1. Real magnified image.
   2. Virtual magnified image of an object placed in front of it. Write the positions of the objects in each case.
2. Dlabeledlled ray diagrams to show the image formation in each case.
3. How will the following be affected on cutting this lens the halves along the principal axis:
   1. Focal length intensity of the image formed by the half lens.

Answer:

1. (1) Between F and 2F of a lens.

(2) Within the focal length.

2.

3. No effect on focal length, but intensity decreases.

Question 6. An object 3.0 cm high is placed perpendicular to the principal axis of a concave lens of focal length 7.5 cm. The image is formed at a distance of 5.0 cm from the lens. CalculThe distance

1. ance at which the object is placed, and
2. Size and nature of image formed.

Answer:

Given O = 3.0 cm, f = −7.5 cm, v = 5.0 cm, u = ?, I = ?

Using lens formula, \(\frac{1}{v}-\frac{1}{u}=\frac{1}{f}\) we have

⇒ \(\frac{1}{u}=\frac{1}{v}-\frac{1}{f}=\frac{1}{-5}-\frac{1}{-7.5}=-\frac{1}{15} \quad \text { or } \quad u=-15 \mathrm{~cm}\)

The object is placed 15 cm from the concave lens.

Also, m = \(m=\frac{\mathrm{I}}{\mathrm{O}}=\frac{v}{u}\)

⇒ \(\mathrm{I}=\frac{v}{u} \times \mathrm{O}=\frac{-5.0}{-15.0}\) x 3.0

= 1 cm.

The image is virtual and erect and has a size of 1 cm.

Question 7.

1. A concave mirror produces a three-times enlarged image of an object at 10 cm in front of it. Calculate the focal length of the mirror.
2. Show the formation of the image with the help of a ray diagram when the object is placed 6 cm away from the pole of a convex mirror.

Answer:

1. Given m=-3 (Real image), u = — 10 cm, f=?

Now,

m =\(m=\frac{-v}{u}\)

Therefore, \(-3=\frac{-v}{u} \quad \text { or } \quad v=3 u=-30 \mathrm{~cm}\)

Using the mirror formula, we have

⇒ \(\frac{1}{f}=\frac{1}{-30}+\frac{1}{-10}=\frac{1}{7.5}\)

Therefore, f= -7.5 cm

2. The ray diagram is as shown:

Question 8.

1. State laws of refraction.
2. A ray of light enters from medium A into a slab made up of a transparent substance B (as shown in the figure). Refractive indices of medium A and B are 2.42 and 1.65 respectively. Complete the path the of ray of light till it emerges out of the slab.

Answer:

1. The two laws are:
   1. The incident ray, the refracted, ray, and the normal to the interface of two transparent media at the point of incidence, all lie in the same plane.
   2. The ratio of the sine of the angle of incidence to the there other t angle of refraction is constant, for the light of a color and the given pair of media. This law is also known as Snell’s law of refraction.
2. The path of rays is as shown the in figure:

Question 9.

1. Write one use of a concave mirror as well a as convex mirror.
2. Draw ray diagrams for the following cases when a ray of light:
   1. Passing through the center of curvature of a concave mirror is incident on it. Parallel thee the principal axis is incident an on a convex minor. an an 
   2. Is an incident at the pole of a convex minor.
   3. Passing through the ugh focus of a concave mirror incident on it.

Answer:

1. Concave mirrors are commonly used in torches, search, gets, and vehicle and es, headlights to a get powerful parallel beam of light. Convex mirrors are commonly used rear-view view mirrors in vehicles.
2. The ray diagrams are shown below:

Question 10.

1. Draw a ray diagram in each of the following cases to show the position and nature of the image formed with the hen object is placed:
   1. Between the focus centimeter of curvature of a concave mirror.
   2. Between the een focus the pole of a concave mirror.
   3. BeBetweenhe een center curvature and infinity for a concave mirror.
2. Give mathematical formulae for determining magnification produced by a spherical mirror. What does m = 1 signify? Identify the mirror that can produce it.

Answer:

1. The ray diagrams are as shown:

2. Magnification (m) = height of image, height of object= \(\frac{v}{u}\) m = -1 signifies:

1. size of image = size of objeThe negative sign indicates that the image is real and inverted. A concave mirror ran an or when that is the the the the at center of curvature.

Question 11.

1. The principal focus of a concave mirror
2. Why do we prefer a convex mirror as a rearview mirror in vehicles?
3. We wish to obtain an erect image of an object using a convex lens. Draw a ray diagram to show the image formation in this case.

Answer:

1. Principal focSeveralr of r parallel to the principal axis, after reflection from a concave mirror, meets at a point on the principal axis which is called the principal focus.
2. Conmirrorsrror always gives an erect though diminished image. They have a wider field of view as they are curved outwards and hence the driver can view a much larger area.
3. The diagram is as shown:

Question 12.

1. Describe an activity to find the approximate value of the focal length of a concave mirror.
2. What happens to the size of the image of an object when it is moved gradually away from a convex minor?
3. In an experiment to study refraction through a glass slab, it is observed that a ray of light undergoing refraction emerges parallel to the direction the of incident ray. Why does it happen so? Explain with the help of a diagram.

Answer:

1. Activity: Hold a concave mirror in your hand and direct it towards the screen. Direct the light reflected by the mirror onto a sheet of paper held close to the mirror.

Move the sheet back and forth slowly until a bright, sharp spot of light is seen on the paper. The distance of this image from the position of the mirror gives the approximate value of the focal length of the mirror.

2. The size of the image becomes smaller.

3. The extent of bending of the ray of light at the opposite parallel faces PQ (air-glass interface) and SR (glass air interface) of the rectangular glass slab is equal and opposite.

Question 13. Draw ray diagrams showing the image formation by a concave mirror when an object is placed

1. Betwthe een pole the and focus of the mirror
2. Between the focus centimeter of curvature of the mirror the 
3. At the centre of curvature of the mirror
4. A little better center of curvature of the mirror
5. At infinity

Answer:

The enlarged, vir, virtual, and erect image forms behind the mirror when the object is placed between the pole the and focus of the mirror.

The enlarged, real, and inverted image forms by the center of curvature when the object is placed between the eye focus and the centimeter of curvature of the mirror.

The real and inverted images equal to the size the the of object form the the center of curvature when the object is placed at centimeters of curvature of the mirror.

The diminished, real, and inverted image forms the center of curvature and focuses when the object is placed a little bit on the center of curvature of the mirror.

The real, I’ve rated, and highly reduced image forms at focus F when the object is placed at infinity.

Question 14. Draw ray diagrams showing the image formation by a convex lens when an object is placed

1. Betwthe een optical center and focus of the lens
2. Between focus and twice the focal length of the lens
3. At twice the focal length of the lens
4. At infinity
5. At the focus of the lens

Answer:

The enlarged, virtual, and erect image forms beyond 2F₁ on the same side of the object when the object is placed between the optical center and focus F₁ of the lens.

The enlarged, real, and inverted image forms beyond focus 2F₂ on the other side of the object, when the object is placed between focus F₁ and twice the focal length of the lens.

The real and inverted images equal to the size of the object format focus 2F₂ on the other side of the object when the object is placed at twice the focal length of the lens.

The real, rated, and highly reduced image forms at focus F₂ on the other side of the object when the object is placed at infinity.

The real, rated, and highly magnified image forms at infinity on the other side of the object when the object is placed at the focus of the lens.

Question 15. Write laws of refraction. Explain the same with the help of a ray diagram, when a ray of light passes through a rectangular glass slab.
Answer:

The following are the laws of refraction of light:

1. The incident ray, the refracted, ray, and the normal to the interface of two transparent media at the point of incidence, all lie in the same plane.
2. The ratio of the NE of the angle of incidence to the sine of the angle of refraction is constant, for the light of a colorlour for the given pair of media. This law is also known as Snell’s law of refraction the constant is called the refractive index. The ray diagram. is as shown.

In a rectangular glass slab, the emergent rays are parallel to the incident ray because the extent of bending of the ray of light at the opposite face of the rectangular glass slab is equal and opposite so that the emergent ray is parallel to the incident ray.

Question 16. Draw ray diagrams showing the image formation by a concave lens when an object is placed

1. Between focus and twice the focal length of the lens
2. Beyond twice the focal length of the lens.

Answer:

The image formed is virtual, erect, diminished in, tween optical, and focus F when the object is placed between focus and twice the focal length of the lens.

The image formed is virtual, erect, diminished n,    and between the een optical cent center focus F when the object is placed beyond twice the focal length of the lens.

Question 17. Draw ray diagrams showing the image formation by a convex mirror when an object is placed

1. At infinity a 
2. At a finite distance from the mirror

Answer:

The virtual, erect, and highly diminished image of the object forms at focus F behind the mirror when the object is placed at infinity.

The virtual, erect, and diminished image forms between focus F and pole P behind the mirror when the object is placed a at finite distance from the mirror.

Question 18. The image of a candle flame formed by a lens is obtained on a screen placed on the other side of the lens. If the image is three times the size of the flame and the distance between the lens and the image is 80 cm, at what distance should the candle be placed from the lens? What is the nature of the image at a distance of 80 cm from the lens?
Answer:

The image is real as only the real image can be taken on the screen

Here, image distance, v = 80 cm

Magnification, m = −3

Object distance, u =?

∵ Magnification, m = \(=\frac{v}{u}\)

⇒ -3 \(=\frac{80}{u} \Rightarrow u=-\frac{80}{3} \mathrm{~cm}\)

Nature of image: Real, inverted, magnified, formed beyond 2F

Using the lens formula, we have

⇒ \(\frac{1}{v}-\frac{1}{u}=\frac{1}{f}\)

⇒ \(\frac{1}{f}=\frac{1}{80}-\frac{3}{-80}=\frac{4}{80}=\frac{1}{20}\)

f = 20 cm

Positive focal length denotes that the hat lens is convex.

Question The sizeSizethe the of image of an object by a mirror having a focal length of 20 cm is observed to be reduced to 1/ofofd to its size. At what distance, the object has been placed from the mirror? What is the nature of the image and the mirror?
Answer:

For concave mirror:

Focal length,

f = -20 cm

Magnification, \(m=-\frac{1}{3}\)

∵ \(m=-\frac{v}{u}\)

⇒ \(m=-\frac{1}{3}=-\frac{v}{u}\)

⇒ \(v=\frac{u}{3}\)

Use the ing mirror formula,

⇒ \(\frac{1}{v}+\frac{1}{u}=\frac{1}{f}\)

⇒ \(\frac{1}{f}=\frac{1}{u}+\frac{3}{u}=\frac{4}{u}\)

u = 4f = 4(-20)

= – 80 cm

The object should be placed at a distance of 80 cm from the concave mirror.

For convex mirror:

Focal length, f = + 20 cm

Magnification, \(m=-\frac{v}{u}\)

⇒ \(m=+\frac{1}{3}=-\frac{v}{u}\)

⇒ \(v=-\frac{u}{3}\)

Mirror formula:

⇒ \(\frac{1}{v}+\frac{1}{u}=\frac{1}{f}\)

⇒ \(\frac{1}{f}=-\frac{3}{u}+\frac{1}{u}=-\frac{2}{u}\)

u = − 2f = − 2(20)

= – 40 cm

The object should be placed at a distance of a nice 40 cm from the convex mirror to face an orm virtual, erect, and diminished image.

Question 20. Define the ine power of a lens. What is its unit? One student uses a lens with an h focal length of 50 cm and another of -50 cm. What is the nature of the lens and its power used by each of them?
Answer:

Power of lens: It is defined as the ability of a lens to bend the rays of light. It is given by the reciprocal of focal length. Its unit is dioptre.

If focal length, ƒ = 50 cm, then

⇒ \(\mathrm{P}=\frac{100}{f}=\frac{100}{50}\) = 2 D, lens is convex.the

If the focal length, f= -50 cm, then

⇒ \(\mathrm{P}=\frac{100}{f}=\frac{100}{-50}\) = -2 D, lens is concave.

Question 21. A student focussed the image of a candle flame on a white screen using a convex lens. He noted down the position of the candle, the lens as under:

Position of candle = 12.0 cm

Position of convex lens = 50.0 cm

Position of the screen = 88.0 cm

1. What is the focal length of the convex lens?
2. Where will the image be formed, if he shifts the candle towards the lens at a position of 31.0 cm?
3. What will be the nature of the image formed if he further shifts the candle towards the lens?
4. Draw a ray diagram to show the formation of the image in the case as said above.

Answer:

Object distance, u = Position of the convex lens – Position of the candle = 50 cm – 12 cm = 38 cm

By sign convention, v = +38 cm

1. Using lens formula, \(\frac{1}{v}-\frac{1}{u}=\frac{1}{f}\)

⇒ \(\frac{1}{f}=\frac{1}{38}-\frac{1}{-38}=\frac{2}{38}=\frac{1}{19}\)

The focal length of the convex lens is 19 cm.

2. After shifting the candle towards the lens at a position of 31 cm, then

Object distance, u = Position of convex lens – position of candle = 50 cm -31 cm = 19 cm

By sign convention, u = -19 cm

So, the candle lies at the focus of the lens, hence its image forms at infinity.

3. When the student further shifts the candle towards the lens, the lens foan enlarges, a viral and erect image of the candle.

4. The ray diagram showing the formation of the

Question 22. The angle between an incident ray and the mirror is shown below. The total angle turned by the ray of light is 70°. What is the value of θ?

Answer:

The ray diagram is redrawn as shown below. Since the angle of incidence reflection, therefore, θ = 90° – 35° = 55°.

Question 23. What are two types of reflection?
Answer:

1. Regular
2. Irregular

Question 24. The characteristics of the image formed by a plane mirror.
AnswThe image

1. image is virtual.
2. The size is the same as that of the object.
3. It is formed at the same distance.

Question 25. Give uses of concave mirrors.
Answer:

1. Used as reflectors in car headlights, search, rights, etc.
2. Used as a shaving mirror.
3. Used the solar cooker to focus the sunlight on one point.

Question 26. the  In case of a convex mirror, if the object is moved away from the surface of the mirror, how does the position and size of the image change?
Answer:

As the object is moved away from a convex mirror, the distance of the virtual image, formed behind it, from the mirror increases (between pole and focus) i.e., the image shifts from the pole towards the focus, and the size the of image gradually decreases. When the object is at infinity (very far), the image is at its focus.

Question 27. An object is brought towards a concave mirror. How does the position and size of the image change?
Answer:

When the object is at infinity from the concave mirror, the image is at the focus and it is a diminished and inverted image.

• As the object is brought towards the mirror the image shifts away from the mirror and its size increases.
• When the object is the centimeter of curvature of the mirror, the image is also at cent centimeter curvature and it is of size equal to the size the of object.
• By further bringing the object toward the mirror, the image gets magnified and it moves away from centimeters of curvature.
• When the object is at the focus of the mirror, the image is at infinity. If the object is further moved towards the mirror, the image now becomes erect and magnified and it is formed behind the mirror.

Question 28. Which is a better reflector: A plane mirror right-angled prism?
Answer: A right-angled prism is a better reflector because light incident on its hypotenuse is reflected without absorption of light. On the other hand, in the case of a plane mirror, there is a possibility of partial absorption of light.

Question 29. A convex mirror used for rear-view on an automobile has a radius of curvature of 3.00 m. If a bus is located 5.00 m from this mirror, find the position, nature, and size of the image.
Answer:

Radius of curvature, R = +3.00 cm, (R is +ve for a convex mirror)

Object distance, u = -5.00 m

Image distance, v =?

Height of the image, h′ =?

Focal length, \(f=\frac{\mathrm{R}}{2}=+\frac{3.00 \mathrm{~m}}{2}=+1.50 \mathrm{~m}\)

As \(\frac{1}{v}+\frac{1}{u}=\frac{1}{f}\)

∴ \(\frac{1}{v}=\frac{1}{f}-\frac{1}{u}=+\frac{1}{1.50}-\frac{1}{(-5.00)}=\frac{1}{1.50}+\frac{1}{5.00}=\frac{5.00+1.50}{7.50}\)

⇒ \(v=\frac{+7.50}{6.50}\) = +1.15 m

The image is 1.15 m at the back of the mirror.

Magnification,

m = \(\frac{h^{\prime}}{h}=-\frac{v}{u}=-\frac{1.15 \mathrm{~m}}{-5.00 \mathrm{~m}}\) = +0.23

The image is virtual, erect, and smaller in size by a factor of 0.23.

Question 30. An object, 4.0 cm in size, is placed at 25.0 cm in front of a concave mirror of a focal length of 15.0 cm. At what distance from the mirror should a screen be placed to obtain a sharp image? Find the nature and the size of the image.
Answer:

Object size, h = +4.0 cm

Object distance, u = -25.0 cm

Focal length, f= -15.0 cm

Image distance, v =?

Image size, h’ =?

As \(\frac{1}{u}+\frac{1}{v}=\frac{1}{f}\)

∴ \(\frac{1}{v}=\frac{1}{f}-\frac{1}{u}=\frac{1}{-15}-\frac{1}{-25}=-\frac{1}{15}+\frac{1}{25}\)

or  \(\frac{1}{v}=\frac{-5+3}{75}=\frac{-2}{75}\)

or V =-37.5 cm

The screen should be placed 37.5 cm from the mirror on the object side of the mirror to obtain a sharp image.

Magnification, \(m=\frac{h^{\prime}}{h}=-\frac{v}{u}\)

Image size, h’= \(-\frac{v h}{u}\)

⇒ \(\frac{(-37.5 \mathrm{~cm})(+4.0 \mathrm{~cm})}{(-25 \mathrm{~cm})}\)

= -6.0 cm

The image is real, inverted, and enlarged in size.

Question 31. A concave lens has a focal length of 15 cm. At what distance should the object from the lens be placed so that it forms an image 10 cm from the lens? Also, find the magnification produced by the lens.
Answer:

A concave lens always forms a virtual, erect image on the same side of the object.

Image distance, v = -10 cm

Focal length, f = −15 cm

Object distance u =?

If is -ve for a concave lens

By lens formula, \(\frac{1}{v}-\frac{1}{u}=\frac{1}{f}\)

∴ \(\frac{1}{u}=\frac{1}{v}-\frac{1}{f}=\frac{1}{-10}-\frac{1}{-15}=\frac{-3+2}{30}=-\frac{1}{30}\)

or u = -30 cm

Thus, the object should be placed 30 cm from the lens on the left side.

Magnification, \(m=\frac{v}{u}=\frac{-10}{-30}=+\frac{1}{3}\)= +0.33

The positive sign shows that the image is erect and virtual. The image is reduced to one-third in size than the object.

Question 32. A 2.0 cm tall object is placed perpendicular to the principal axis of a convex lens of focal length 10 cm. The distance of the object from the lens is 15 cm. Find the nature, position, and size of the image. Also find its magnification.
Answer:

Height of the object, h = +2.0 cm

Focal length, f= +10 cm

Object distance, u = -15 cm

If is +ve for a convex lens]

Image distance, v =?

Height of the image, h’=?

By lens formula; \(\frac{1}{v}-\frac{1}{u}=\frac{1}{f}\)

∴ \(\frac{1}{v}=\frac{1}{u}+\frac{1}{f}\)

or \(\frac{1}{v}=\frac{1}{(-15)}+\frac{1}{10}\)

⇒ \(\frac{1}{v}=-\frac{1}{15}+\frac{1}{10}\)

⇒ \(\frac{1}{v}=\frac{-2+3}{30}=\frac{1}{30}\)

or v = +30 cm

The positive sign of u shows that the image is formed at a distance of 30 cm on the other side of the optical center. The image is real and inverted.

Magnification, \(m=\frac{h^{\prime}}{h}=\frac{v}{u}\)

∴ \(m=\frac{v}{u}=\frac{+30 \mathrm{~cm}}{-15 \mathrm{~cm}}=-2\)

Also image size,

h’= \(h^{\prime}=\frac{v h}{u}\)

⇒ \(=\frac{(+30)(+2.0)}{-15}\)

= -4.0 cm

The negative sign of m shows that the image is inverted and real.

It is formed below the principal axis. Thus, a real, inverted image, 4 cm tall, is formed at a distance of 30 cm on the other side of the lens. The image is two times enlarged.