NEET Physics Class 12 Chapter 7 Wave Optics Notes

Wave Optics

1. Wavefronts

Consider a wave spreading out on the surface after a stone is thrown in. Every point on the surface oscillates. At any time, a photograph of the surface would show circular rings on which the disturbance is maximum.

• All points on such a circle oscillate in phase because they are at the same distance from the source. Such a locus of points that oscillate in phase is an example of a wavefront.
• A wavefront is defined as a surface of constant phase. The speed with which the wavefront moves outwards from the source is called the phase speed.
• The energy of the wave travels in a direction perpendicular to the wavefront.
• shows light waves from a point source forming a spherical wavefront in three-dimensional space.
• The energy travels outwards along straight lines emerging from the source. i.e. radii of the spherical wavefront. These lines are the rays.
• Notice that when we measure the spacing between a pair of wavefronts along any ray, the result is a constant. This example illustrates two important general principles that we will use later:
• Rays are perpendicular to wavefronts.

The time taken by light to travel from one wavefront to another is the same along any ray.

• If we look at a small portion of a spherical wave, far away from the source, then the wavefronts are like parallel planes. The rays are parallel lines perpendicular to the wavefronts.
• This is called a plane wave and is also sketched in  A linear source such as a slit Illuminated by another source behind it will give rise to cylindrical wavefronts.
• Again, at a larger distance from the source, these wavefronts may be regarded as planar.

Figure: Wavefronts and the corresponding rays in. two cases: (1) diverging spherical wave. (2) plane wave. The figure on the left shows a wave (e.g…….. e.g.ht) in three dimensions. The figure on the right shows a wave in two dimensions (a water surface).

2. Principle Of Superposition :

When two or more waves simultaneously pass through a point, the disturbance of the point is given by the sum of the disturbances each wave would produce in the absence of the other wave(s). In the case of a wave on a string, a disturbance means displacement, in the case of a sound wave it means pressure change, in the case of E.M.W. it is an electric field or magnetic field. The superposition of two lights traveling in almost the same direction results in modification in the distribution of intensity of light in the region of superposition. This phenomenon is called interference.

2.1 Superposition Of Two Sinusoidal Waves :

Consider the superposition of two sinusoidal waves (having the same frequency), at a particular point.

Let, x₁(t) = a₁sin ωt

and, x₂(t) = a₂sin (ωt + φ)

represent the displacement produced by each of the disturbances. Here we are assuming the displacements to be in the same direction. Now according to the superposition principle, the resultant displacement will be given by,

x(t) = x₁(t) + x₂(t) = a₁sin ωt + a₂sin (ωt + φ) = A sin (ωt + φ0)

where A2 = a₁₁ + a₂₂ + 2a₁. a₂cos φ ……. (1)

and tan φ0 \(=\frac{a_2 \sin \phi}{a_1+a_2 \cos \phi}\)……. (2)

Solved Examples

Example 1. If i1= 3sin ωt and i2= 4 cos ωt, find i3.

Solution:

From Kirchhoff’s current law,

i₃= i₁+ i₂

= 3 sin ωt + 4 sin (ωt + \(\frac{\pi}{2}\)) = 5 sin (ωt + tan–1\(\left(\frac{4}{3}\right)\))

Example 2. S₁ and S₂ are two sources of light that produce individual disturbance at point P given by E₁= 3sin ωt, E₂= 4 cos ωt. Assuming \(\overrightarrow{\mathrm{E}}_1 \& \overrightarrow{\mathrm{E}}_2\)to be along the same line, find the result of their superposition.
Solution :

3. Superposition Of Progressive Waves; Path Difference :

Let and S₂ be two sources producing progressive waves (disturbance traveling in space given by and y₂)
At point P,

y₁= a₁sin (ωt – kx₁+ θ₁)

y₂= a₂sin (ωt – kx₂+ θ₂)

y = y₁+y₂= A sin(ωt + Δφ)

Here, the phase difference,

Δφ = (ωt – kx₁+ θ₁) – (ωt – kx₂+ θ₂)

= k(x₂– x₁) + (θ₁– θ₂) = kΔp + Δθ

Here Δp = Δx is the path difference

Clearly, phase dPhasPhaseto path difference = k (path difference)

where k = \(\frac{2 \pi}{\lambda}\)

⇒ Δφ = kΔp =2π/λΔx ….. (1)

For Constructive Interference :

Δφ = 2nπ, n = 0, 1, 2 ……..

or, Δx = nλ

Amax = A₁+ A₂

Intensity, \(\sqrt{I_{\max }}=\sqrt{I_1}+\sqrt{I_2} \quad \Rightarrow \quad I_{\max }=\left(\sqrt{I_1}+\sqrt{I_2}\right)^2\)… (2)

For Destructive interference :

Δφ = (2n + 1)π, n = 0, 1, 2 …….

or, Δx = (2n + 1)λ/2

Amin = |A₁– A₂|

Intensity, \(\sqrt{I_{\min }}=\sqrt{I_1}-\sqrt{I_2} \quad \Rightarrow \quad I_{\max }=\left(\sqrt{I_1}-\sqrt{I_2}\right)^2\)… (3)

Solved Examples

Example 3. Light from two sources, each of the same frequency and traveling in the same direction, but with intensity in the ratio 4: 1 interfered and the ratio maximum he to minimum intensity.
Solution :

⇒ \(\frac{I_{\max }}{I_{\min }}=\left(\frac{\sqrt{I_1}+\sqrt{I_2}}{\sqrt{I_1}-\sqrt{I_2}}\right)^2=\left(\frac{\sqrt{\frac{I_1}{I_2}}+1}{\sqrt{\frac{I_1}{I_2}}-1}\right)^2=\left(\frac{2+1}{2-1}\right)^2=9: 1\)

4. Coherence :

Two sources that vibrate with a fixed phase difference between them are said to be coherent. The phase differences between light come from sucfromurces do not depend on time.

In a conventional light source, however, light comes from a large number of individual atoms, each atom emitting a pulse lasting for about 1 ns. Even if atoms were emitting under similar conditions, waves from different atoms would differ in their initial phases.

Consequently, the only light coming fr, om two such sources have a fixed phase relationship for about 11n hence interference pattern will keep changing every billionth of a second. The eye can notice intensity changes that last at least to third of a second.

Hence we will observe uniform intensity on the screen which is the sum of the two individual intensities. Such sources are said to be incoherent. Light beams coming light two such independent sources do not have any fixed phase relationship and they do not produce any stationary interference pattern. For such sources, the result at any point is given by

Ι = Ι₁+ Ι₂…… (1)

5. Young’S Young’sSlit Experiment (Y.D.S.E.)

In 1802 Thomas Young devised a method to produce a stationary interference pattern. This was based upon the division of a site the single wavefront into two; these two wavefronts acted as if they emanated from two sources having a fixed phase relationship. Hence when they were allowed to interfere, stationary intera a ference pattern was observed.

Figure: YounArrangementment to produce stationary interference pattern by division of work-wavefront front S₁ and S₂

5.1 Analysis Of Interference Pattern

We ensured the above arrangement that the light wave passing through S₁ is in a phase that passes through S₂. However, ever the wreachingachi, ng P from S₂ may not be in phase with the wave reaching P from S₁ because the latter must travel a longer path to reach P than the former. We have already discussed phase-phasehase differences due to differences. If the path difference is equal to zero or is an integral multiple of wavelengths, the arriving waves are exactly in phase and undergo constructive interference.

If the path difference is an odd multiple of half a wavelength, the arriving waves are out of phase and undergo fully destructive interference. Thus, it is the path difference Δx, which determines the intensity at a point P.

Path difference Δp = S₁P – S₂P = \(\sqrt{\left(y+\frac{d}{2}\right)^2+D^2}-\sqrt{\left(y-\frac{d}{2}\right)^2+D^2}\) …(1)

Approximation I :

For D >> d, we can approximate rays \(\vec{r}_1 \text { and } \vec{r}_2\) as being approximately parallel, at angle θthe principle axis principle axis.

Now, S₁P – S₂P = S₁A = = S₁S₂sin θ

⇒ path difference = d sin θ …(2)

Approximation II :

Further if θ is small, i.e.y << D, sin θ = tan θ = \(\frac{y}{D}\)

and hence, path difference = \(\frac{dy}{D}\)…(3)

For maxima (constructive interference),

Δp = \(\frac{\text { d.y }}{D}=n \lambda\)

⇒ y = \(\frac{\mathrm{n} \lambda \mathrm{D}}{\mathrm{d}}\)n = 0, ± 1, ± 2, ± 3 …..(4) d

Here n = 0 corresponds to the central maxima

n = ±1 correspond to the 1st maxima

n = ±2 correspond to the 2nd maxima and so on.

For minima (destructive interference).

⇒ \(\Delta \mathrm{p}= \pm \frac{\lambda}{2}, \pm \frac{3 \lambda}{2} \pm \frac{5 \lambda}{2}\)

⇒ \(\Delta p= \begin{cases}(2 n-1) \frac{\lambda}{2} & n=1,2,3 \ldots \ldots \ldots \ldots \\ (2 n+1) \frac{\lambda}{2} & n=-1,-2,-3 \ldots \ldots \ldots\end{cases}\)

⇒ \(y= \begin{cases}(2 n-1) \frac{\lambda D}{2 d} & n=1,2,3 \ldots \ldots \ldots \ldots \\ (2 n+1) \frac{\lambda D}{2 d} & n=-1,-2,-3 \ldots \ldots\end{cases}\)

Here n = ± 1 corresponds  to the first minima,

n the = ± 2 corresponds to second minima and so on.

5.2 Fringe Width :

It is the distance between two maxima of successive order on one side of the central maxima. This is also equal to distabetween the two successive minima.

fringe width β = \(\frac{\lambda D}{d}\) … (1)

Notice that it is directly proportional to wavelength and inversely proportional to the distance between the two slits.

5.3 Intensity :

Suppose the electric field components of the light waves arriving at point P (in the Figure) from the two slits S₁ and S₂ vary with time as

E₁= E₀sin ωt

and E₂= E₀sin (ωt + φ)

Here φ = kΔx = \(\frac{2 \pi}{\lambda}\)Δx

and we have assumed that intensity of the two slits S₁ and S₂ are the same (say Ι0); hence waves have the same amplitude E0.

then the resultant electric field at point P is given by,

E = E₁+ E₂= E₀sin ωt + E₀sin (ωt + φ) = E₀´ sin (ωt +φ´)

where E₀´2 = E₀2 + E₀2 + 2E₀. E₀cos φ = 4 E₀ 2cos2 φ/2

Hence the resultant intensity at point P,

Ι = 4Ι₀cos2 \(\frac{\phi}{2}\)…….(2) 2

Ιmax = 4Ι₀ when \(\frac{\phi}{2}\) = nπ , n = 0, ±1, ±2,……., 2

Ιmin = 0 when \(\frac{\phi}{2}\) = \(\left(n-\frac{1}{2}\right) \pi\) = 0,±1,±2 ………. 2

Here φ = kΔx = \(\frac{2 \pi}{\lambda} \Delta x\)

If D >> d, φ = \(\frac{2 \pi}{\lambda}\)d sin θ

If D >> d & y << D, φ = \(\phi=\frac{2 \pi}{\lambda} d \frac{y}{D}\)

However if the two slits were of different intensities Ι₁and Ι₂,

say E₁= E₀1, sin ωt

and E₂= E₀2, sin (ωt + φ)

then resultant field at point P,

E = E₁+ E₂= E₀sin (ωt + φ)

where E₀2 = E₀12 + E₀2 + 2E₀. E₀₂ cos φ = 4E₀2cos₂φ/2

Hence resultant intensity at point P,

Ι = Ι₁+ Ι₂+ 2\(\sqrt{I_1 I_2}\)cos φ ………… (3)

Solved Examples

Example 4. In a YDSE, D = 1m, d = 1mm and λ = 1/2 mm

1. Find the distance between the first and central maxima on the screen.
2. Find the no of maxima and minima obtained on the screen.

Solution :

D >> d

Hence ΔP = d sin θ

⇒ \(\frac{\mathrm{d}}{\lambda}\)= 2,

clearly, n << \(\frac{\mathrm{d}}{\lambda}\)= 2 is not possible for any value of n.

Hence Δp = \(\frac{\mathrm{dy}}{D}\) cannot be used for Ist maxima,

Δp = d sin θ = λ

⇒ sin θ = \(\frac{\lambda}{d}=\frac{1}{2}\) θ = 30º

Hence, y = D tan θ = \(\frac{1}{\sqrt{3}}\)

Maximum path difference ΔPmax = d = 1 mm

⇒ Highest order maxima  \(n_{\max }=\left[\frac{d}{\lambda}\right]=2 \text { and highest order minima } n_{\min }=\left[\frac{d}{\lambda}+\frac{1}{2}\right]=2\)

Total no. of maxima = 2n_max + 1* = 5 *(central maxima).

Total no. of minima = 2n_min = 4

Example 5. Monochromatic light of wavelength 5000 Aº is used in Y.D.S.E., with slit-width, d = 1mm, distal between the  screen and slits, D = 1m.the the  If intensity these  two slis are, Ι₁= 4Ι₀, Ι₂sin0, find

1. Fringe width β
2. Distance of 5th minima from the central maxima on the screen
3. Intensity at y = \(\frac{1}{3} \mathrm{~mm}\)
4. Distance of the 1000th maxima
5. Distance of the 5000th maxima λ= 10

Solution :

β = \(\frac{\lambda D}{d}=\frac{5000 \times 10^{-10} \times 1}{1 \times 10^{-3}}=0.5 \mathrm{~mm}\)

y = (2n – 1) \(\frac{\lambda D}{2 d}\) , n = 5 ⇒ y = 2.25 mm

At y = 1/3 mm, y << D

Hence Δp = \({d. y}{D}\)Δφ = \(\frac{2 \pi}{\lambda} \Delta \mathrm{p}=2 \pi \frac{\mathrm{dy}}{\lambda \mathrm{D}}=\frac{4 \pi}{3}\) Now resultant intensity

Ι = \(\mathrm{I}_1+\mathrm{I}_2+2 \sqrt{\mathrm{I}_1 \mathrm{I}_2} \cos \Delta \phi=4 \mathrm{I}_0+\mathrm{I}_0+2 \sqrt{4 \mathrm{I}_0^2} \cos \Delta \phi=5 \mathrm{I}_0+4 \mathrm{I}_0 \cos \frac{4 \pi}{3}=3 \mathrm{I}_0\)

⇒ \(\frac{d}{\lambda}=\frac{10^{-3}}{0.5 \times 10^{-6}}=2000\)

n = 1000 is not << 2000

Hence now Δp = d sin θ must be used Hence, d sin θ = nλ = 1000 λ

⇒ sin θ = 1000 \(\frac{\lambda}{d}=\frac{1}{2}\) ⇒ θ = 30º

y = D tan θ =\(\frac{1}{\sqrt{3}}\)Meter

Highest order max max max = \(\left[\frac{d}{\lambda}\right]\) = 2000

Hence, n = 5000 is not possible.

5.4 Shape Of Interference Fringes In Ydse :

We discuss the shape of fringes when two pinholes are used instead of the two slits in YDSE.

Fringes are the locus of a point that moves in such a way that its path differs from the two slits and remains constant.

S₂P – S₁P = Δ = constant ….(1)

If Δ = ± \(\frac{\lambda}{2}\), the fringe represents 1st minima. 2

If Δ = ± \(\frac{3\lambda}{2}\) it represents 2nd minima

If Δ = 0 it represents central maxima, Y

If Δ = ± λ, it represents 1st ma,  time, etc.

Equation (1) represents a hyperbola with its two foci S₁ and S₂

The interference pattern that we get on screen is the section of a hyperboloid of ra evolution when we revolve the hyperbola about the axis S₁S₂.

1. If the screen is ⊥er to the X axis, i.e. in the YZ plane, as is generally the case, fringes are hyperbolic with a straight central section.
2.  If the screen is in the XY plane, again fringes are hyperbolic.
3. The If screen is tooth the  Y axis (along S₁S₂), ie in the XZ plane, fringes are concentric circles with λ. center on the axis S₁S₂; the central fringe is bright if S₁S₂= nλ and dark if S₁S₂= (2n – 1)\(\frac{\lambda}{2}\)

5.5 Ydse With White Light :

The central maxima will be white because all wavelengths will constructively interfere interferometer slightly below (or above) the position of central maxima fringes will colored red. or exacoForred is a point on the screen such that

S₂P – S₁P = \(\frac{\lambda_{\text {violet }}}{2}=190 \mathrm{~nm} \text {, }\)

completely destructive interference will occur for violet light. Hence we will have a line devoid of violet color that will color reddish. And if
λ= 350 nm,

S₂P–S₁P = \(\frac{\lambda_{\text {red }}}{2}=350 \mathrm{~nm} \text {, }\)

completely destructive interference for red light results and the line at this position will be violet. The colored fringes are colored at points far away from the central white fringe; for these points, there are so many wavelengths that interfere constructively, that we obtain a uniform white illumination. for example if

S₂P – S₁ P = 3000 nm,

then constructive interference will occur for wavelengths λ = \(\frac{3000}{n}\) nnm. In the visible region this wavelength, h are 750 nm (red), 600 nm (yellow), and 500 nm (greenish-yellow) greenish-yellowSuchsuch a lSuchpea,r white to the unaided eye.

Thus with white light w, we get a white central fringe at the point of zero path difference, followed by a few colored fringes on its sides, the color soon fading off to a uniform white.

In the usual interference pattern with a monochromatic source, a large number of identical interference fringes are obtained and it is usually not possible to determine the position of central maxima. Interference with white light is used to determine the position of central maxima in such cases.

Solved Examples

Example 6. A beam of light consisting of wavelengths 6000Å and 4500Å is used in a YDSE with D = 1m and d = 1 mm. Find the least distance from the central maxima, where bright fringes due to the two wavelengths coincide.
Solution:

β₁= \(\frac{\lambda_1 D}{d}=\frac{6000 \times 10^{-10} \times 1}{10^{-3}}=0.6 \mathrm{~mm}\)

β₂= \(\frac{\lambda_2 \mathrm{D}}{\mathrm{d}}=0.45 \mathrm{~mm}\)

Let n1th maxima of λ₁and n2th maxima of λ₂coincide at a position y then, y = n₁P₁= n₂P₂= LCM of β₁and β₂

⇒ y = LCM of 0.6 cm and 0.45 mm ⇒ y = 1.8 mm Ans. At this point, 3rd maxima for 6000 Å and 4th maxima for 4500 Å coincide

Example 7. White light is used in a YDSE with D = 1m and d = 0.9 mm. Light reaching the screen at position y = 1 mm is passed through a prism and its spectrum is obtained. Find the missing lines in the visible region of this spectrum.
Solution :

Δp = \(\frac{\mathrm{yd}}{\mathrm{D}}\) = 9 × 10^-4 × 1 × 10^-3 m = 900 nm

for minima Δp = (2n – 1)λ/2

⇒ λ = \(\frac{2 \Delta P}{(2 n-1)}=\frac{1800}{(2 n-1)}=\frac{1800}{1}, \frac{1800}{3}, \frac{1800}{5}, \frac{1800}{7}\) ……..

of these 600 nm and 360 nm lie in the visible range. Hence these will be missing lines in the visible spectrum.

6. Geometrical Path & Optical Path :

Actual distance The acThe actuated by light in a medium is called geometrical path (Δx). Consider a light wave given by the equation

E = E₀sin (ωt – kx + φ)

If the light travels by Δx, its phase changes by kΔx = \(\frac{\omega}{v}\) Δx, where ω, the frequency of light does not depend on the medium, but v, the speed of light depends on the medium as v = \(\frac{\mathrm{c}}{\mu} \text {. }\)

Consequently, change in phase

Δφ = kΔx = \(\frac{\omega}{\mathrm{C}}\)(μΔx) c

A distance Δx in a medium of refractive index μ suffers the same phase change as when it travels a distance μΔx in a vacuum. i.e. an ath length of Δx in the medium of refracted the tive index μ is equivalent to a path length of μΔx in vacuum.

The quanta a tity μΔx is called the optical path length of light, Δxopt. In terms of InInl path length, the phase difference would be given by,

Δφ = \(\frac{\omega}{c} \Delta x_{\text {opt }}=\frac{2 \pi}{\lambda_0} \Delta x_{\mathrm{opt}}\)Δxopt …. (1) 0

where λ0= wavelength of light in vacuum. However in terms of the geometrical path length Δx,

Δφ =\( \frac{\omega}{c}(\mu \Delta x)=\frac{2 \pi}{\lambda} \Delta x \)…..(2)

where λ = wavelength of light in the medium (λ = \(\)

6.1 Displacement Of Fringe :

On the production of a glass slab in the path of the light coming out of the slits–

On the introduction of the thin glass-glass slackness t and refractive index μ, the optical path of the ray S₁P increases by t(μ – 1).

Now the path difference between waves coming from S1afromfrom2at any point P is

Δp = S₂P – (S₁P + t (μ – 1))

= (S₂P –S₁P) – t(μ – 1)

⇒ Δp = d sin θ – t (μ – 1) if d << D

and Δp =\(\frac{y d}{D}\)– t(μ – 1) If y << D as well.

for central bright the the rangp = 0 ⇒\(\frac{y d}{D}\)= t(μ – 1).

⇒ y = oo’ = (μ – 1)t \(\frac{\mathrm{D}}{\mathrm{d}}=(\mu-1) \mathrm{t} \cdot \frac{\beta}{\lambda}\)

The whole fringe pattern gets shifted by the same distance

Δ = (μ – 1).\(\frac{D{d}=(\mu-1) t \cdot \frac{\beta}{\lambda}\)

Notice that this shift is in the direction of the slit before which the glass slab is placed. If the glass slab is placed before the upper slit, the fringe pattern gets shifted upwards and if the glass slab is placed before the lower slit the fringe pattern gets shifted downwards.

Example 8. In a YDSE with d = 1mm and D = 1m, slabs of (t = 1μm, μ = 3) and (t = 0.5 μm, μ = 2) are introduced in front of upper and lower slit respectively. Find the shift in the fringe pattern.
Solution :

OpThe optical path for light coming from upper slit S₁ is

S₁P + 1μm (2 – 1) = S₂P + 0.5 μm

Similarly o,  the optical path for light coming from S₂ is

S₂P + 0.5 μm (2 – 1) = S₂P + 0.5 μm

Path difference : Δp = (S₂P + 0.5 μm) – (S₁P + 2μm) = (S₂P – S₁P) – 1.5 μm.

= \(\frac{y d}{D}-1.5 \mu \mathrm{m}\)

for central bright fringe Δp = 0 1.5 m

⇒ y = \(\frac{1.5 \mu \mathrm{m}}{1 \mathrm{~mm}}\) × 1m = 1.5 mm. 1mm

The whole pattern is shifted by 1.5 mm upwards. Ans.

7. YDSE With Oblique Incidence :

In YDSE, the ray is incident on the slit at an inclination of θ0to the axis of symmetry of the experimental set-up

for points above the central point on the screen, (say for P₁)

Δp =d sinθ₀+ (S₂ P₁– S₁P₁)

⇒ Δp = d sinθ₀+ dsinθ1(If d << D)

and for points below O on the screen, (say for P₂)

Δp = |(dsin θ₀+ S2P₂) – S₁P₂|

= |d sin θ0– (S₁P₂– S₂P₂)|

⇒ Δp = |d sin θ0– d sinθ₂| (if d << D)

We obtain central maxima at a point where Δp = 0.

(d sinθ₀– d sinθ₂) = 0

or θ₂= θ₀.

This corresponds to the point O’ in the diagram.

Hence we have finally for path difference.

⇒ \(\Delta p=\left\{\begin{array}{l}
d\left(\sin \theta_0+\sin \theta\right)-\text { for points above } O \\
d\left(\sin \theta_0-\sin \theta\right)-\text { for points between O\&O’ } \\
d\left(\sin \theta-\sin \theta_0\right)-\text { for points below } O^{\prime}
\end{array}\right.\)… (1)

Solved Examples

Example 9. In YDSE with D = 1m, d = 1mm, light of wavelength 500 nm is incident at an angle of 0.57º w.r.t. the axis of symmetry of the experimental setup. If the center is set up Symmetry of the screen is O as shown.

1. Find the position of the central maxima
2. The intensity at point O in terms of intensity of central maxima Ι0.
3. Number of maxima lying between O and the central maxima.

Solution :

θ = θ0= 0.57º

⇒ y = –D tan θ ~ _ – Dθ = – 1 meter ×\(\left(\frac{0.57}{57} \mathrm{rad}\right)\)

⇒ y = – 1cm.

For point 0, θ = 0

Hence, Δp = d sin θ₀; dθ₀= 1 mm × (10–2rad)

= 10,000 nm = 20 × (500 nm)

⇒ Δp = 20 λ

Hence point O corresponds to the 20th maxima

⇒ in the tensity at 0 = Ι₀

19 maxima lie between central maxima and 0, excluding maxima at O and central maxima.

8. Thin-Film Interference :

In YDSE we obtained two coherent sources from a  sources incoherent source by division of wave-front. Here we do the same by division of Amplitude (into reflected and refracted waves).

When a plane wave (parallel rays) is incident normally on a thin film of uniform thickness d then waves reflected from the upper surface interfere with waves reflected from the lower surface.

The wave reflected from the lower surface travels an extra optical path of 2μd, where μ is the refractive index of the film.
Further, if the film is placed in air the wave reflected from the upper surface (from a denser medium) suffers a sudden phase change of π, while the wave reflected from the lower surface (from a rarer medium) suffers no such phase change.

Consequently condition for constructive and destructive interference in the reflected light is given by,

2μd = nλ for destructive interference

and 2μd = (n + 1/2)λ for constructive interference ….(1)

where n = 0, 1, 2 …………..

and λ = wavelength in free space.

Interference will also occur in the transmitted light and here condition of constructive and destructive interference will be the reverse of (9.1)

⇒ \(2 \mu d= \begin{cases}(n \lambda) \text {for constructive interference } \\ \left(n+\frac{1}{2} \lambda\right) \text { for destructive interference }\end{cases}\) ….(1)

This can easily be explained by energy conservation (when intensity is maximum in reflected light it has to be minimum in transmitted light) However the amplitude of the directly transmitted wave and the wave transmitted after one reflection differ substantially and hence the fringe contrast in transmitted light is poor. It is for this reason that thin film interference is generally viewed only in the reflected light.

In deriving equation (1) we assumed that the medium surrounding the thin film on both sides is rarer compared to the medium of thin film.

If the medium on both sides is denser, then there is no sudden phase change in the wave reflected from the upper surface, but there is a sudden phase change of π in waves reflected from the lower surface. The conditions for constructive and destructive interference in reflected light would still be given by equation (1).

However, if the medium on one side of the film is denser and that on the other side is rarer, then either there is no sudden phase in any reflection, or there is a sudden phase change of π in both reflections from an upper and lower surface. Now the condition for constructive and destructive interference in the reflected light would be given by equation 2 and not equation (1).

Solved Examples

Example 10. White light, with a uniform intensity across the visible wavelength range 430–690 nm, is perpendicularly incident on a water film, with an index of refraction μ = 1.33 and thickness d = 320 nm, that is suspended in the air. At what wavelength λ is the light reflected by the film brightest to an observer?
Solution :

This situation is like that of Figure, for which the equation gives the interference maxima. Solving for λ and inserting the given data, we obtain

λ = \(\frac{2 \mu \mathrm{d}}{\mathrm{m}+1 / 2}=\frac{(2)(1.33)(320 \mathrm{~nm})}{m+1 / 2}=\frac{851 \mathrm{~nm}}{m+1 / 2}\)

for m = 0, this gives us λ = 1700 nm, which is in the infrared region. For m = 1, we find l = 567 nm, which is yellow-green light, near the middle of the visible spectrum. For m = 2, λ = 340 nm, which is in the ultraviolet region. So the wavelength at which the light seen by the observer is brightest is

λ = 567 nm.

Example 11. A glass lens is coated on one side with a thin film of magnesium fluoride (MgF₂) to reduce reflection from the lens surface (figure). The index of refraction of MgF₂ is 1.38; that of the glass is 1.50. What is the least coating thickness that eliminates (via interference) the reflections at the middle of the visible spectrum (λ = 550 nm)? Assume the light is approximately perpendicular to the lens surface.
Solution :

The situation here differs from the figure in that n₃> n₂> n₁. The reflection at point a still introduces a phase difference of π but now the reflection at point b also does the same ( see figure 9.2). Unwanted reflections from glass can be, suppressed (at a chosen wavelength) by coating the glass with a thin transparent film of magnesium fluoride of a properly chosen thickness which introduces a phase change of half a wavelength. For this, the path length difference of 2L within the film must be equal to an odd number of half wavelengths:

2L = (m + 1/2)λn₂,

or, with λn₂= λ/n₂,

2n₂L = (m + 1/2)λ.

We want the least thickness for the coating, that is, the smallest L. Thus we choose m = 0, the smallest value of m. Solving for L and inserting the given data, we obtain

L = \(\frac{\lambda}{4 n_2}=\frac{550 \mathrm{~nm}}{(4)(1.38)}=96.6 \mathrm{~nm}\)

9. Fresnel’s Birpism Experiment

It is an optical device to obtain two coherent sources by refraction of lights.

The angle of biprism is 179º and the refracting angle is α = 1/2º. (3) Distance between source and screen D = a + b.

Distance between two coherent source = d = 2a (μ –1)α

Where a = distance between source and Biprism

b = distance between screen and Biprism

μ = refractive index of the material of the prism

⇒ \(\lambda=\frac{d \beta}{D}=\frac{2 a(\mu-1) \alpha \beta}{(a+b)}=\frac{\sqrt{d_1 d_2} \cdot \beta}{(a+b)}\)

Note- α is in radian α = α ×\(\frac{3.14}{180}\) Suppose the refracting angle and refractive index are not known then d can be calculated by a convex lens.

One convex lens whose focal length (f) and 4f < D.

A first convex lens is kept near biprism and d₁ is calculated then it is kept near the eyepiece and d2is cald₂ is d₂ listed. d =\(\sqrt{\mathrm{d}_1 \mathrm{~d}_2}\)

Application :

With the help of this experiment the wavelength of monochromatic light, the thickness of thin films, and their refractive index and distance between apparent coherent sources can be determined.

When Fresnel’s arrangement is immersed in water

Effect on d dwater < dair. Thus when the Fress biprism experiment is immersed in water, then the separation between the two virtual sources decreases but in Young’s double slit experiment, it doe not change.

In Young’s double slit experiment, β decreases, and in Fresnel’ss biprism experiment β increases.

Solved Examples

Example 12 In Fresnel’s biprism experiment the width of 10 fringes is 2cm which is a formalised distance of two 2 meters from the slit. If the wavelength of light is 5100 Å then the distance between two coherent sources.
Solution:

⇒ \(d=\frac{D \lambda}{\beta}\)……..(1)

According to question λ = 5100 × 10–10 m

β =\(\frac{2}{10} \times 10^{-2} \mathrm{~m}\) ……..(2)

D = 2m

d = ?

From eqs. (1) and (2) \(\mathrm{d}=\frac{2 \times 51 \times 10^{-8}}{2 \times 10^{-3}}\) = 5.1 × 10^-4 m

10. Diffraction

10.1. Meaning Of Diffraction

It is the spreading of waves around the corners of an obstacle, of the order of wavelength.

10.2. Definition Of Diffraction

The phenomenon of bending of light waves around the sharp edges of opaque obstacles or aperture and their encroachment in the geometrical shadow of obstacle or aperture is defined as the diffraction of light.

10.3. Necessary Conditions Of Diffraction Of Waves

The size of the obstacle (a) must be of the order of the wavelength of the waves (λ).

⇒ \(\frac{\mathrm{a}}{\lambda} \approx 1\)

Note: The greater the wavelength wave higher the degree of diffraction. This is the reason that the diffraction of sound and radio waves is easily observed but for the diffraction of light, additional arrangements are made.

λsound > λlight

The wavelength of sound is nearly equal to the size of the obstacle. If the size of the obstacle is a and the wavelength of light is λ then,

S.No. a V/S λ Diffraction

[1] a << λ Not possible

[2] a >> λ Not possible

[3] a ~ λ Possible

10.4. Interpretation Of Diffraction

As a result of diffraction, maxima, and minima of light intensities are found which are unequal intensities. Diffraction is the result of the processing of waves from an infinite number of coherent sources on the same wavefront after the wavefront has been distorted by the obstacle.

10.5. Example Of Diffraction

When an intense source of light is viewed as viewed tia open, colors are colored in the light.

The sound produced in one room can be heard in the nearby room.

The appearance of a shining circle around the section of the sun just before sunrise colored red spectrum is observed if a light source at a far distance once seen through a thin cloth.

10.6. Two Type Of Diffraction

Fresnel Diffraction: Fresnel diffraction involves non-plane (spherical) wavefronts so that the sources and the point p where the diffraction effect is to be observed) are to be at a finite distance from the diffracting obstacle.

FraunhofFraunhofertion: Fraunhofer diffraction deals with wavefronts that are plane on arrival and an effective viewing distance of infinity.Itf follows that fraunhofFraunhofertion is an important special case ofFresnell diffraction. In youngs double slit experiment, we assume the screen to be relatively distant, that we have fraunhofFraunhoferons.

Fresnel Diffraction: According to the toFreprinciplencipal to determine the intensity of light at any point, a wavefront can be divided into a number several, which are knoFresnel’ssnel’s half period point on the wavefront is a source of secondary wavelets, so that the wave from two consecutive zones reach the point of observation in opposite phase corresponding to a path difference of λ/2.

10.7. Difference Between Interference and Diffraction Of Light

10.8. Fraunhofer Diffraction For Single Slit

In this diffraction pattern, the central max area is bright on the side of it, and maxim minima occur symmetrically

For Diffraction Maxima :

a sin θ = (2n + 1) λ/2

For Diffraction Minima :

a sin θ = nλ

The maxima or minima are observed due to the superposition of waves emerging from infinite secondary sources between the A and B points of the slit.

Fringe width :

The distance between two secondary minima formed on two sides of central the maximum is known as the width of central the maximum

W= \(\frac{2 \mathrm{f} \lambda}{\mathrm{a}}\)

f = focal distance of convenience

a = width of slit

Angular width = Wθ \(\frac{2 \mathrm{f} \lambda}{\mathrm{a}}\)

10.9. Resolving Power (R.P.)

A large number of images are formed as a consequence of light diffraction from a source. If two sources are separated such that their central maxima do not overlap, their images can be distinguished and are said to be resolved R.P. of an optical instrument is its ability to distinguish two neighbouring points.

Linear R.P. = d/λD Angular R.P. = d/λ

D = Observed distance

d = Distance between two points

Telescope :

Limit of resolution = θ = sin–1 \(\frac{1.22 \lambda}{\mathrm{a}}\); For small angles θ = \(\)

Resolving power = \(\frac{1}{\text { limit of resolution }}\)

Microscope :

Limit of resolution (the smallest distance between two object) = x_min = \(x_{\min }=\frac{1.22 \lambda}{2 \mu \sin \theta}\)

Prism : R.P. = t (dµ/dλ) = λ/dλ.

Diffraction Grating :

R.P. = λ/dλ = N × n (N is the total number of lines and n is the order of spectrum)

Eye: The limit of resolution of the human eye is 1′ of arc (One minute of arc)

10.10. Difference Between Fresnel and Fraunhofer Diffraction

10.11. Comparative Study Of Diffraction Of Light and Sound

Sound travels in the form of waves, that’s why it is also diffracted. Generally, diffraction of sound waves is easily observed rather than light because the wavelength of sound waves is the order of obstacle, but the wavelength of light is very small in comparison to the obstacle.

Ordinary audible sound has a wavelength of the order of 1m and the size of ordinary obstacles has the same order. That is why diffraction is easily observed.

Ordinary light has a wavelength of 10–7 m and ordinary obstaobstaclesa have a greater size in comparison to their length which is why a diffraction pattern is not observed.

Generally,tion in ultrasonic waves is not observed because their wavelength has an order of 1 cm.

10.12 Rectlinear Motion Of Light

The rectilinear motion of light can be explained by the diffraction of light. the

If the size of an obstacle is the order of the wavelength of light, then diffraction of light takes place and its rectilinear motion of light is not possible. the

If the size of the obstacle is a much greater wavelength length of light, then the rectilinear motion of light is observed.

11. Polarisation Of Light

Light propagates as transverse EM waves. The magnitude of the electric field is much larger as compared to the magnitude of the magnetic field. We generally prefer to describe light as electric field oscillations.

Unpolarised light: In ordinary light (light from the sun, bulb, etc.) the electric field vectors are distributed in all directions in a light that is called unpolarised light. The oscillation of propagation of light wave resolved into horizontal and vertical components.

Polarised light: The phenomenon of limiting the vibrating of electric field vector in one direction in a plane perpendicular to the direction of propagation of the light wave is called polarization of light

The plane in which oscillation occurs in the polarised light is the led plane of oscillation.

The plane perpendicular to the plane of oscillation is the led plane of polarisation.

Light can be polarised by transmitting through certain crystals such as tourmaline or polaroids.

Maluslaw: This law states that the intensity of the polarised light transmitted through analyzers varies as the square of the cosine of the angle between the plane of transmission of the analyzers and the plane of the polariser.

Ι = Ι₀cos²θ and A² = A²₀cos²θ ⇒ A = A₀cos θ

If θ = 0º, Ι = Ι0 , A = A0. If θ = 90º , Ι = 0 , A = 0

If Ι1 = intensity of unpolarised light.

So Ι0= \(\frac{\mathrm{I}_1}{2}\) Ιi.e. if an unpolarised light is converted into plane polarised light (say by passing through though a Polaroid or a Nicol prism, it intensity becomes half. and I = \(\frac{\mathrm{I}_1}{2}\)cos²θ

Solved Miscellaneous Problems

Problem 1. Consider interference between waves from two sources of intensities Ι and 4Ι. Find intensities at points where the phase difference is π.
Solution:

I = R² = a_1² + a_2² + 2a₁a₂cos δ = Ι + 4Ι + 4Ι cos π

Ι = 5Ι – 4Ι = Ι

Problem 2. The width of one of the slots sits in Young’s double slits experiment is double of the other slit. Assuming that the amplitude of the light coming from a proportion abortion to slit-width. Find the ratio of the maximum to the minimum intensity in the interference pattern.
Solution:

Suppose the amplitude of the light wave coming from the narrow slit is A and that coming from the wider slit is 2A. The maximum intensity occurs at a place where constructive interference takes place. Then the resultant amplitude is the sum of the individual amplitudes. Thus, Amax = 2A + A = 3A

The minimum intensity occurs at a place where destructive interference takes place. The resultant amplitude is then the difference of the individual amplitudes.

Thus, Amin = 2A – A = A.

∴\(\frac{I_{\max }}{I_{\min }}=\frac{\left(A_{\max }\right)^2}{\left(A_{\min }\right)^2}=\frac{(3 A)^2}{(A)^2}=9\)

Problem 3. n Young’s experiment, the separation between the slits is 0.10mm, the wavelength of light used is 600nm and the interference pattern is observed on a screen 1.0 m away. Find the separation between the successive bright fringes.
Solution:

The separation between the successive bright fringes is –

⇒ \(\beta=\frac{D \lambda}{d}=\frac{1 \times 600 \times 10^{-9}}{.1 \times 10^{-3}}\)

β = 6.0 mm

Problem 4. Two waves originating from a rom source having zero phase difference and common wavelength λ will show completely destructive interference at a point P if (S₁ P – S₂ P) is

1. 0
2. 5λ
3. 11λ/4
4. 11λ/2

Solution:

For destructive interference :

Path difference = S₁ P – S₂ P = (2n^-1) λ/2

For n = 1, S₁ P – S₂ P = (2 × 1 ^-1)λ/2 = λ/2

n = 2, S₁ P – S₂ P = (2 × 2 ^-1)λ/2 = 3λ/2

n = 3, S₁ P – S₂ P = (2 × 3 ^-1)λ/2 = 5λ/2

n = 4, S₁ P – S₂ P = (2 × 4 ^-1)λ/2 = 7λ/2

n = 5, S₁ P – S₂ P = (2 × 5 ^-1)λ/2 = 9λ/2

n = 6, S₁ P – S₂ P = (2 × 6 ^-1)λ/2 = 11λ/2

So, a destructive pattern is possible only for path difference = 11λ/2.

Problem 5. In Young’s experiment, the wavelength of red light is 7.5×10–5 cm. and that of blue light is 5.0 × 10^-5 cm. The value of n for which (n +1)th the blue bright band coincides with nth red band.
Solution:

n1λ1 = n₂λ₂for bright fringe

n(7.5 × 10^-5) = (n + 1) (5 × 10^-1)

⇒ \(n=\frac{5.0 \times 10^{-5}}{2.5 \times 10^{-5}}=2\)

Problem 6. In Young’s slit experiment, carried out with lights of wavelength λ = 5000 Aº, the distance between the slit is 0.2 mm and the screen is at0 cm from the slits. The central maximum is at x = 0. The third maximum will be at x equal to.
Solution:

⇒ \(X_n=\frac{n \lambda D}{d} \quad \text { or } \quad X_3=\frac{3 \lambda D}{d}\)

⇒ \(x_3=\frac{3 x\left(5000 \times 10^{-8}\right) \times 200}{0.02}=1.5 \mathrm{~cm}\)

Problem 7. Two slits separated by a distance of 1mm are illuminated with red light of wavelength 6.5 × 10^-7 m. The interference fringes are observed on a screen placed 1m from the slits. The distance between the third dark fringe and the fifth bright fringe.
Solution:

β= \(\beta=\frac{\lambda D}{d}=\frac{6.5 \times 10^{-7} \times 1}{10^{-3}}\)

β = .65 ×10–3 m = .65 mm

The distance between the fifth bright fringe from third dark fringe = 5β – 2.5 β ⇒ 2.5 β = 2.5 × .65 = 1.63 mm

Problem 8. In an experiment, the two slits are 0.5 mm apart and the fringes are observed to be 100 cm from the plane of the slits. The distance of the 11th bright fringe from the Ist bright fringe is 9.72 mm. Calculate the wavelength.
Solution:

Given d = .5 mm = 5× 10^-2 cm,

D = 100 cm

X_n= X₁₁ – X₁= 9.72 mm

∴\(X_n=\frac{n \lambda D}{d}\) n = 11 – 1 = 10

⇒ λ = \(\frac{X_n d}{n D}=\frac{.972 \times 5 \times 10^{-2}}{10 \times 100}\)

⇒ λ = 4.86 × 10^-5 cm

Problem 9.  In Young’s experiment, two coherent sources are placed 0.90 mm apart and the fringes are observed one meter away. If it produces the second dark fringe at a distance of 1mm from the central fringe, the wavelength of monochromatic light hissed.
Solution:

D =1m, d = .90 mm = .9 × 10^-3 m

The distance of the second dark ring from the center = 10^-3 m

∴ \(\mathrm{X}_{\mathrm{n}}=(2 \mathrm{n}-1) \frac{\lambda}{2} \frac{\mathrm{D}}{\mathrm{d}}\)

for n = 2,

⇒\(\frac{3 \lambda}{2} \frac{\mathrm{D}}{\mathrm{d}} \Rightarrow \lambda=\frac{2 \mathrm{X}_{\mathrm{n}} \mathrm{d}}{3 \mathrm{D}}=\frac{2 \times 10^{-3} \times .9 \times 10^{-3}}{3}\)

λ = 6 × 10^-7 m ⇒ λ = 6 ×10^-5 cm

Problem 10. A beam of light consisting of two wavelengths 6500Aº and 5200Aº is used to obtain interference fringesYoung’soung’s double slit experiment. The distance between the slits is 2.0 mm and the distance between the plane of the slits and the screen is 120 cm. What is the least distance from the central maximum where the bright fringes due to both wavelength lengths coincide?
Solution:

Suppose the mth bright fringe of 6500Aº coincides with the nth bright fringe of 5200Aº.

⇒ \(x_n=\frac{m \lambda_1 D}{d}=\frac{n \lambda_2 D}{d}\)

⇒ \(\frac{m \times 6500 \times D}{d}=\frac{n \times 5200 \times D}{d} \Rightarrow \frac{m}{n}=\frac{5200}{6500}=\frac{4}{5}\)

∴ distance y is \(y=\frac{m \lambda_1 D}{d}\)⇒ y = 0.156 cm .

Problem 11. Interference fringes were producYoung’soung’s double slit experiment using lig wavelength length 5000 Aº. When a film of material 2.5 × 10^-3 cm thick was placed over one of the slits, the fringe pattern shifted by a distance equal to 20 fringe width. The refractive index of the material of the film is
Solution:

n = \(n=\frac{(\mu-1) t D}{d} \quad \text { but } \beta=\frac{\lambda D}{d} \Rightarrow \frac{D}{d}=\frac{\beta}{\lambda}\)

n =(μ−1)β/λ

20β = (μ – 1) 2.5 × 10–3(β/5000 ×10^-8)

⇒ \(\mu-1=\frac{20 \times 5000 \times \times 10^{-8}}{2.5 \times 10^{-3}} \quad \Rightarrow \quad \mu=1.4\)

Problem 12. The path difference between two interfering waves at a point on the screen is 171.5 times the wavelength. If the path difference is 0.01029 cm. Find the wavelength.
Solution:

Path difference = 171.5 λ

⇒ \(\frac{343}{2}\)λ= odd multiple of half wavelength It means a dark fringe is observed

According to question .0.01029 \(=\frac{343}{2} \lambda \quad \Rightarrow \quad \lambda=\frac{0.01029 \times 2}{343}=6 \times 10^{-5} \mathrm{~cm}\) λ = 6000Aº.

Problem 13. Find the minimum thickness of a film that witheringly reflects the light of wavelength 589 nm. The refractive index of the material of the film is 1.25
Solution:

For strong reflection, the least optical path difference introduced by the film should beλ/2. The optical path difference between the waves reflected from the surfaces of the film is 2μd. Thus, for strong reflection,

2μd = λ/2

⇒ \(\mathrm{d}=\frac{\lambda}{4 \mu}=\frac{589}{4 \times 1.25}=118 \mathrm{~nm}\)

Key Concept

Wavefronts :

Rays are perpendicular to wavefronts.

The time taken by light to travel from one wavefront to another is the same along any ray.

Figure: Wavefronts and the corresponding rays in case of plane wave.

Huygens’ Principle :

All points on a wavefront serve as point sources of spherical secondary wavelets. After a time t, the new position of the wavefront will be that of a surface tangent to these secondary wavelets.

Interference of waves of intensity Ι₁and Ι₂:

resultant intensity,

Ι = Ι₁+ Ι₂+2\(\sqrt{I_1 I_2}\) cos (Δφ) where, Δφ = phase difference.

For Constructive Interference :

Ι_max= \(\left(\sqrt{I_1}+\sqrt{I_2}\right)^2\)

For Destructive interference :

Ι_min = \(\left(\sqrt{I_1}-\sqrt{I_2}\right)^2\)

If sources are incoherent

Ι = Ι₁+ Ι₂, at each point.

YDSE :

Path difference, Δp = S₂P – S₁P = d sin θ

if d < < D = \(\frac{d y}{D}\)if y << D

for maxima,

Δp = nλ ⇒ y = nβ n = 0, ±1, ±2 …….

For minima

Δp = Δp = \(\begin{cases}(2 n-1) \frac{\lambda}{2} & n=1,2,3 \ldots \ldots \ldots \ldots \\ (2 n+1) \frac{\lambda}{2} & n=-1,-2,-3 \ldots \ldots \ldots\end{cases}\)

⇒ y = \(\begin{cases}(2 n-1) \frac{\beta}{2} & n=1,2,3 \ldots \ldots \ldots \ldots \\ (2 n+1) \frac{\beta}{2} & n=-1,-2,-3 \ldots \ldots .\end{cases}\)

where, fringe width β = \(\frac{\lambda \mathrm{D}}{\mathrm{d}}\)

Here, λ = wavelength in medium.

Highest order maxima : nmax= \(\left[\frac{\mathrm{d}}{\lambda}\right]\)

total number of maxima = 2nmax+ 1

Highest order minima : nmax= \(=\left[\frac{d}{\lambda}+\frac{1}{2}\right]\)

total number of minima = 2nmax.

Intensity on screen :

Ι = Ι₁+ Ι₂+ \(\sqrt{I_1 I_2}\) Ιcos (Δφ) where, Δφ = \(\frac{2 \pi}{\lambda} \Delta p\)

If Ι₁= Ι₂, Ι = 4Ι1cos2 \(\left(\frac{\Delta \phi}{2}\right)\)

YDSE with two wavelengths λ₁ and λ₂:

The nearest point to central maxima where the bright fringes coincide:

y = n₁β₁= n₂β₂= Lcm of β₁and β₂

The nearest point to central maxima where the two dark fringes coincide,

y = \(\left.\left(n_1-\frac{1}{2}\right) \beta_1=n_2-\frac{1}{2}\right) \beta_2\)

Optical path difference ΔP_opt = μΔp

Phase difference Δφ = \(\frac{2 \pi}{\lambda} \quad \Delta p=\frac{2 \pi}{\lambda_{\text {vacuum }}} \Delta \mathrm{p}_{\text {opt. }}\)

Displacement of fringe pattern on the introduction glass slabs-slab in the path of one of the slabs:

Δ = (μ – 1) t. \(\frac{D}{d}=(\mu-1) t \frac{B}{\lambda}\)

This shift is in the direction of the slit before which the glass slab is placed. If the glass slab is placed before the upper slit, the fringe pattern gets shifted upwards and if the glass slab is placed before the lower slit the fringe pattern gets shifted downwards.

YDSE with Oblique Incidence :

In YDSE, the ray is incident on the slit at an inclination of θ0 to the axis of symmetry of the experimental set-up

We obtain central maxima at a point where Δp = 0.

or θ₂= θ0.

This corresponds to the point O’ in the diagram. Hence we have or path difference.

⇒ Δp = \(\left\{\begin{array}{l}d\left(\sin \theta_0+\sin \theta\right)-\text { for points above } O \\d\left(\sin \theta_0-\sin \theta\right)-\text { for points between } O \&O^{\prime} \ldots \text { (8.1) } \\d\left(\sin \theta-\sin \theta_0\right)-\text { for points below } O^{\prime}\end{array}\right.\)

Thin-Film Interference :

Case – 1: Medium on the two sides of the film are either denser or rarer n for destructive interference

for interference in reflected light 2μd = \(\begin{cases}\mathrm({n} \lambda) \text { for destructive interference } \\ \left(\mathrm{n}+\frac{1}{2}\right) \lambda \text { for constructive interference }\end{cases}\)

for interference in transmitted light 2μd = \(\begin{cases}\mathrm({n} \lambda) \text { for destructive interference } \\ \left(\mathrm{n}+\frac{1}{2}\right) \lambda \text { for constructive interference }\end{cases}\)

Case – 2: Medium on one side of the film is denser and that on the other side is rarer. Here condition for interference in reflected light is the same as the condition for interference in transmitted light of case 1, and vice versa.

Fresnel’s Birpism Experiment :

• Itan is an optical device to obtains two coherent sources by refraction of lights.
• The angle of biprism is 179º and the refracting angle is α = 1/2º.
• Distance between source and screen D = a + b.

Distance between two coherent source = d = 2a (μ –1)α

Where a = distance between source and Biprism

b = distance between screen and Biprism

μ = refractive index of the material of the prism.

λ =\(\frac{d \beta}{D}=\frac{2 a(\mu-1) \alpha \beta}{(a+b)}\)

Diffraction

Diffraction DefinItion: The phenomenon of bending of light waves around the sharp edges of opaque obstacles or aperture and their encroachment in the geometrical shadow of obstacle or aperture is defined as the diffraction of light.

Necessary Conditions of Diffraction of Waves

The size of the obstacle (a) must be of the order of the wavelength of the waves (λ).

⇒ \(\frac{\mathrm{a}}{\lambda} \approx 1\)

Fraunhofer diffraction for single slit :

In this diffraction pattern, central maxima areas are bright on both sides, and  maxima and minima occur symmetrically

For Diffraction Maxima : a sin θ = (2n + 1) λ/2

For Diffraction Minim: a sin θ = nλ

The maxima or minima areas are observed due to the superposition of waves emerging from infinite secondary sources between the A and B points of the slit.

Fringe width :

The distance between two secondary minima formed on two sides of the central maximum is known as the width of the central maximum

⇒ \(W=\frac{2 f \lambda}{a}\)

f = focal distance of convex lens a = width of slit

Angular width = \(\mathrm{W}_\theta=\frac{2 \lambda}{\mathrm{a}}\)

Polarisation of Light :

Unpolarised light: In ordinary light (light from the sun, but b, etc.) the electric field vectors are distributed in all directions in a light that is called unpolarised light.

Polarised light: The phenomenon of limiting the vibrating of electric field vector in one direction in a plane perpendicular to the direction of propagation of light wave is called polarization of light

The plane in which oscillation occurs in the polarised light is called the plane of oscillation.

The plane perpendicular to the plane of oscillation is canceled there of polarization.

Light can be polarised by transmitting through certain crystals such as tourmaline or polaroids.

Brewster’s Low: For light incident at the Brewster angle θB, The reflected and refracted rays are perpendicular to each other. The refracted light has only perpendicular components. The reflected light is then fully polarized perpendicular to the plane of incidence.

Brewster angle = \(\theta_{\mathrm{B}}=\tan ^{-1} \frac{\mathrm{n}_2}{\mathrm{n}_1}\)

Wave Optics Exercise – 1

Section (1): Principle Of Superposition, Path Difference, Wavefronts, And Coherence

Question 1. The ratio of intensities of two light waves is given by 4 :1. The ratio of the amplitudes of the waves is :

1. 2: 1
2. 1:2
3. 4 :1
4.  :4

Solution: 1. 2 :1

Question 2. Two coherent monochromatic light beams of intensities I and 4I are superposed; the maximum and minimum possible intensities in the resulting beam are :

1. 5Ι and Ι
2. 5Ι and 3Ι
3. 9Ι and Ι
4. 9Ι and 3Ι

Solution: 3. 9Ι and Ι

Question 3. Two sources of waves are called coherent if:-

1. Both have the same amplitude of vibrations
2. Both produce waves of the same wavelength
3. Both produce waves of the same wavelength having constant phase difference
4. Both produce waves having the same velocity

Solution: 3. Both produce waves of the same wavelength having a  constant phase difference

Question 4. The meaning of wavefront is:-

1. All the particles on its surface vibrate in the  same phase
2. All the particles on its surface vibrate in opposite phase
3. Some particles vibrate in the  same phase and some in the  opposite phase
4. None of the above

Solution: 1. All the particles on its surface vibrate in the  same phase

Question 5. Two coherent monochromatic light beams of intensities I and 4I are superposed. The maximum and minimum possible intensities in the resulting beam are

1. 5I and I
2. 5I and 3I
3. 9I and I
4. 9I and 3I

Solution: 3. 9I and I

Question 6. Two beams of light having intensities I and 4I interfere to produce a fringe pattern on a screen. The phase difference between the beams is π/2 at point A and π at point B. Then the difference between the 2 resultant intensities at A and B is

1. 2I
2. 4I
3. 5I
4. 7I

Solution: 2. 4I

Question 7. The phase difference between the  incident wave and reflected wave is 180° when light ray :

1. Incident on glass from air
2. Incident on air from glass
3. Incident on glass from diamond
4. Incident on water from glass

Solution: 1. Incident on glass from air

Question 8. Which one of the following statements is true?

1. Both light and sound waves in air are transverse
2. The sound waves in air are longitudinal while the light waves are transverse
3. Both light and sound waves in air are longitudinal
4. Both light and sound waves can travel in a vacuum

Solution: 2. The sound waves in air are longitudinal while the light waves are transverse

Question 9. Two waves have equations : y₁= a sin ( ωt + φ₁ );y₂ asin( ωt + φ₂ ).= If the amplitude of the resultant wave is equal to the amplitude of each of the superimposing waves, then what will be the phase differences between them?

1. \(\frac{2 \pi}{3}\)
2. \(\frac{\pi}{3}\)
3. \(\frac{\pi}{4}\)
4. \(\frac{\pi}{2}\)

Solution: 1. \(\frac{2 \pi}{3}\)

Question 10. Two periodic waves of intensities I₁₁ and I₁₂ pass through a region at the same time in the same direction. The sum of the maximum and minimum intensities is

1. Ι₁ + Ι₂
2. \(\left(\sqrt{I_1}+\sqrt{I_2}\right)^2\)
3. \(\left(\sqrt{I_1}-\sqrt{I_2}\right)^2\)
4. 2(I₁+ I₂)

Solution: 4. 2(I₁+ I₂)

Question 11. Two points are located at a distance of 10 m and 15 m from the source of oscillation. The period of oscillation is 0.05 s and the velocity of the wave is 300 m/s. What is the phase difference between the oscillations of two points?

1. \(\frac{\pi}{3}\)
2. \(\frac{2\pi}{3}\)
3. π
4. \(\frac{\pi}{6}\)

Solution: 2. \(\frac{2\pi}{3}\)

Question 12. Electromagnetic waves are transverse as evidenced by

1. Polarization
2. Interference
3. Reflection
4. Diffraction

Solution: 1. polarization

Question 13. To demonstrate the phenomenon of interference we require two sources that emit radiation of

1. Nearly the same frequency
2. The same frequency
3. Different wavelength
4. The same frequency and having a definite phase relationship

Solution: 4. The same frequency and having a definite phase relationship

Section (2): YDSE With Monochromatic Light

Question 1. The contrast in the fringes in any interference pattern depends on :

1. Fringe width
2. Wavelength
3. Intensity ratio of the sources
4. Distance between the sources

Solution: 3. Intensity ratio of the sources

Question 2. Yellow light emitted by sodium lamp in Young’s double slit experiment is replaced by monochromatic blue light of the same intensity :

1. Fringe width will decrease.
2. Fringe width will increase.
3. Fringe width will remain unchanged.
4. Fringes will become less intense.

Solution: 1. Fringe width will decrease.

Question 3. Young’s experiment is performed in air and then performed in water, the fringe width:-

1. Will remain same
2. Will decrease
3. Will increase
4. All the above types of waves

Solution: 3. Will increase

Question 4. In Young’s double slit experiment, the phase difference between the light waves reaching the  third bright fringe from the central fringe will be [λ = 6000 A0]:-

1. Zero
2. 2π
3. 4π
4. 6π

Solution: 1. Zero

Question 5. Interference was observed in the the  interference chamber when air was present, now the chamber is evacuated and if the same light is used, a careful observer will see

1. No interference
2. Interference with bright bands
3. Interference with dark bands
4. Interference in which the width of the fringe will be slightly increased

Solution: 4. Interference in which the width of the fringe will be slightly increased

Question 6. In the double slits experiment, for the light of which color the fringe width will be minimum:-

1. Violet
2. Red
3. Green
4. Yellow

Solution: 2. Red

Question 7. If two line slits are illuminated by a wavelength 5 x 10^-7 m and the distance between two bright fringes is 0.005 m on a screen 1 m away, then the distance between the slits is:-

1. 10 cm
2. 1 cm
3. 10^-1cm
4. 10^-2cm

Solution: 4. 10^-2cm

Question 8. The fringe width in the Young’s double slit experiment is 2 x 10-4 m. If the distance between the slits is halved and the slit screen distance is double, then the new fringe width will be:-

1. 2 x 10-4m
2. 1 x 10-4m
3. 0.5 x 10-4m
4. 8 x 10-4m

Solution: 4. 8 x 10-4m

Question 9. The fringe width observed in Young’s double slit experiment is β. If the frequency of the source is doubled, the fringe width will :

1. become 2 β
2. become \(\frac{3 \beta}{2}\)
3. remain as β
4. become \(\frac{\beta}{2}\)

Solution: 4. become \(\frac{\beta}{2}\)

Question 10. The distance between two slits in the double-slit experiment is 1 mm. The distance between the slits and the screen is 1 m. If the distance of the  10th fringe from the central fringe is 5 mm, then the wavelength of light is:-

1. 5000 A⁰
2. 6000 A⁰
3. 7000 A⁰
4. 8000 A⁰

Solution: 1. 5000 A⁰

Question 11. If the slit distance in Young’s double slit experiment is reduced to 1/3 rd, the fringe width becomes times. The value of n is:

1. 3
2. 1/3
3. 9
4. 1/9

Solution: 1. 3

Question 12. In Young’s double slit experiment, if the slit widths are in the ratio 1: 9, then the ratio of the intensity at minima to that at maxima will be

1. 1
2. 1/9
3. 1/4
4. 1/3

Solution: 3. 1/4

Question 13. The monochromatic green light of wavelength 5 × 10^-7 m illuminates a pair of slits 1 mm apart. The separation of bright lines on the interference pattern formed on a screen 2 m away is

1. 0.25 mm
2. 0.1 mm
3. 1.0 mm
4. 0.01 mm

Solution: 3. 1.0 mm

Question 14. The figure shows that double-slit experiments P and Q are the slits. The path lengths PX and QX are nλ and (n + 2)λ respectively, where n is a whole number and λ is the wavelength. Taking the central fringe as zero, what is formed at X

1. First bright
2. First dark
3. Second bright
4. Second dark

Solution: 3. Second bright

Question 15. In Young’s double slit experiment, a glass plate is placed before a slit which absorbs half the intensity of light. Under this case

1. The brightness of fringes decreases
2. The fringe width decreases
3. No fringes will be observed
4. The bright fringes become fainter and the dark fringes have finite light intensity

Solution: 4. The bright fringes become fainter and the dark fringes have finite light intensity

Question 16. In the double slit experiment, the angular width of the fringes is 0.20 for the sodium light (λ =5890 Å).To increase the angular width of the fringes by 10%, the necessary change in the wavelength is

1. Increase of 589 Å
2. Decrease of 589 Å
3. Increase of 6479 Å
4. Zero

Solution: 1. Increase of 589 Å

Question 17. In Young’s double slit experiment, 62 fringes are seen in the visible region for sodium light of wavelength 5893 Å. If violet light of wavelength 4358 Å is used in place of sodium light, then the  number of fringes seen will be

1. 54
2. 64
3. 74
4. 84

Solution: 4. 84

Question 18. If in Young’s double slit experiment, the slit distance is 3 cm, the separation between slits and screen is 70cm and the  wavelength of light is 1000 Å, then the  fringe width will be

1. 2 × 10^-5 m
2. 2 × 10^-3 m
3. 0.2 × 10^-3 m
4. None of these

Solution: 4. None of these

Question 19. The separation between slits is halved and between screen and slits is doubled. Final fringe width if the original is w :

1. w
2. 9w
3. 4w
4. 2w

Solution: 3. 4W

Question 20. The maximum number of possible interference maxima first–separations equal to twice the wavelength in Young’ double slit experiment is

1. Infinite
2. Five
3. Three
4. Zero

Solution: 2. Five

Question 21. To demonstrate the phenomenon of interference we require two sources that emit radiations of

1. Nearly the same frequency
2. The same frequency
3. Differentwavelengthh
4. The same frequency and having a definite phase relationship

Solution: 4. The same frequency and having a definite phase relationship

Question 22. Monochromatic light of frequency 5 × 10¹⁴ Hz traveling in a vacuum enters a medium of refractive index 1.5. Its wavelength in the medium is

1. 4000 Å
2. 5000 Å
3. 6000 Å
4. 5500 Å

Solution: 1. 4000 Å

Question 23. In Young’s double slit experiment when the wavelength used 6000 and the screen is 40 cm from the slits, the fringes are 0.012 cm apart. What is the distance between the slits:–

1. 0.024 cm
2. 2.4 cm
3. 0.24 cm
4. 0.2 cm

Solution: 4. 0.2 cm

Question 24. The maximum number of possible interference maxim slit separation-ration equal to twice the wavelength in Young’s double-slit experiment is

1. Infinite
2. Five
3. Three
4. Zero

Solution: 2. Five

Question Young’soung’s double slit experiment uses a monochromatic source. The shape of the interference fringes formed on a screen.

1. Straight line
2. Parabola
3. Hyperbola
4. Circle

Solution: 3. Hyperbola

Question 26. In a Young’s double slit experiment the intensity at a point where the path difference is λ/6(λ being the wavelength of the light used) is I. If I₀ denotes the maximum intensity, I/I₀is equal to:

1. \(\frac{1}{\sqrt{2}}\)
2. \(\frac{\sqrt{3}}{2}\)
3. \(\frac{1}{2}\)
4. \(\frac{3}{4}\)

Solution: 4. \(\frac{3}{4}\)

Question 27. In Young’s double-slit experiment, an electron beam is used to form a fringe pattern instead of light. If the speed of the electrons is increased then the fringe width will :

1. Increase
2. Decrease
3. Remains same
4. No fringe pattern will be formed

Solution: 2. Decrease

Section (3): Ydse With Polychromatic Light

Question 1. What happens by the use of white light in Young’s double slit experiment:-

1. Bright fringes are obtained
2. Only bright and dark fringes are obtained
3. Central fringe shares bright and two or three colored and dark fringes are observed
4. None of the above

Solution: 3. Central fringes are bright and two or three colored and dark fringes are observed

Question 2. The length of the optical path of two media in contact of lengths d1and d2of refractive indices μ₁and μ₂ respectively is

1. \(\mu_1 d_1+\mu_2 d_2\)
2. \(\mu_1 d_2+\mu_2 d_1\)
3. \(\frac{d_1 d_2}{\mu_1 \mu_2}\)
4. \(\frac{d_1+d_2}{\mu_1 \mu_2}\)

Solution: 1. \(\mu_1 d_1+\mu_2 d_2\)

Section (4): Ydse With Glass Slab, Optical Path, Thin Film Interference

Question 1. In Young’s experiment, light wavelength 4000 A0 is used, and fringes are formed at a 2-metre distance and have a fringe width of 0.6 mm. If the whole of the experiment is performed in a liquid of refractive index 1.5, then the width of the fringe will be:-

1. 0.2 mm
2. 0.3 mm
3. 0.4 mm
4. 1.2 mm

Solution: 3. 0.4 mm

Question 2. The slits in Young’s double-slit experiment have equal widths and the source is placed symmetrically relative to the slits. The intensity at the central fringe is Ι₀. If one of the slits is closed, the intensity at this point will be:

1. Ι₀
2. Ι₀/4
3. Ι₀/2
4. 4Ι₀

Solution: 2. Ι₀/4

Question 3. A thin mica sheet of thickness 2 × 10^-6 m and refractive index (μ = 1.5) is introduced in the path of the first wave. The wavelength of the wave used is 5000Å. The central bright maximum will shift

1. 2 fringes upward
2. 2 fringes downward
3. 10 fringes upward
4. None of these

Solution: 1. 2 fringes upward

Question 4. Colors of thin films result from:-

1. Dispersion of light
2. Interference of light
3. Absorption of light
4. Scattering of light

Solution: 2. Interference of light

Question 5. White light may be considered to be a mixture of waves with λ ranging between 3900 Å and 7800 Å. An oil film of thickness 10,000 Å is examined normally by reflected light. If μ = 1.4, then the film appears bright for

1. 4308 Å, 5091 Å, 6222 Å
2. 4000 Å, 5091 Å, 5600 Å
3. 4667 Å, 6222 Å, 7000 Å
4. 4000 Å, 4667 Å, 5600 Å, 7000Å

Solution: 1. 4308 Å, 5091 Å, 6222 Å

Question 6. If a mica sheet of thickness t and refractive index μ is placed in the path of one of the interfering beams in a double slit experiment then the displacement of fringes will be :

1. \(\frac{D}{d} \mu t\)
2. \(\frac{\mathrm{D}}{\mathrm{d}}(\mu-1) \mathrm{t}\)
3. \(\frac{\mathrm{D}}{\mathrm{d}}(\mu+1) \mathrm{t}\)
4. \(\frac{\mathrm{D}}{\mathrm{d}}\left(\mu^2-1\right) \mathrm{t}\)

Solution: 2.\(\frac{\mathrm{D}}{\mathrm{d}}(\mu-1) \mathrm{t}\)

Question 7. If a thin mica sheet of thickness t and refractive index μ = [5/3] is placed in the path of one of the interfering beams as shown in fig. then the distance placement of the fringe system is-

1. \(\frac{D t}{3 \mathrm{~d}}\)
2. \(\frac{\mathrm{Dt}}{5 \mathrm{~d}}\)
3. \(\frac{D t}{4 d}\)
4. \(\frac{2 D t}{5 d}\)

Solution: 1. \(\frac{D t}{3 \mathrm{~d}}\)

Question 8. A double slit experiment is performed with light of wavelength 500 nm. A thin film of a thickness of 2 mm and a refractive index of 1.5 is introduced in the path of the upper beam. The location of the central maximum will

1. Remain unshifted
2. Shift downward by nearly two fringes
3. Shift upward by nearly two fringes
4. Shift downward by 10 fringes

Solution: 3. Shift upward by nearly two fringes

Section (5) : Fresnal Biprism And Diffraction Of Light

Question 1. In Fresnel’s biprism (μ = 1.5) experiment the distance between the source and biprism is 0.3 m and between biprism and screen is 0.7m and the angle of the prism is 1°. The fringe width with light of wavelength 6000 Å will be

1. 3 cm
2. 0.011 cm
3. 2 cm
4. 4 cm

Solution: 2. 0.011 cm

Question 2. In a Fresnel biprism experiment, the two positions of the lens give separation between the slits as 16 cm and 9 cm respectively. What is the actual distance between the slits?

1. 10.5 cm
2. 12 cm
3. 13 cm
4. 14 cm

Solution: 2. 12 cm

Question 3. What is the effect on the fresnel biprism experiment when the white light is used:-

1. Fringe are affected
2. The diffraction pattern is spread more
3. Central fringes are white and all others are colored
4. None of these

Solution: 3. Central fringes are white and all others are colored

Question 4. A slit of width is illuminated by light. For red light (λ = 6500Å), the first minima is obtained at θ = 30° Then the value of a will be

1. 3250 Å
2. 6.5×10^-4 mm
3. 1.3 µm
4. 2.6×10^-4 cm

Solution: 4. 2.6×10^-4cm

Question 5. The light of wavelength 6328 Å is incident on a slit of width 0.2 mm perpendicularly, the angular width of central maxima will be

1. 0.36
2. 0.18°
3. 0.72°
4. 0.09°

Solution: 1. 0.36

Question 6. A slit of size 0.15 cm is placed at 2.1 m from a screen. On illuminated it by a light of wavelength 5 × 10^-5 cm. The width of the central maxima will be.

1. 70 mm
2. 0.14 mm
3. 1.4 mm
4. 0.14 cm

Solution: 3. 1.4 mm

Question 7. A diffraction is obtained by using a beam of red light. What will happen if the red light is replaced by the blue light?

1. Bands will narrower and crowd full together
2. Bands become broader and farther apart
3. No change will take place
4. Bands disappear

Solution: 1. Bands will narrower and crowd full together

Question 8. What will be the angle of diffraction for the first minimum due to Fraunhofer diffraction with sources of light of wavelength 550 nm and slit of width 0.55 mm?

1. 0.001 rad
2. 0.01 rad
3. 1 rad
4. 0.1 rad

Solution: 1. 0.001 rad

Question 9. The angular width (β)of the central maximum of a diffraction pattern on a single slit does not depend upon

1. Distance between slit and source
2. The wavelength of light used
3. Width of the slit
4. Frequency of light used

Solution: 1. Distance between slit and source

Question 10. A single slit of width 0.20 mm is illuminated with light of wavelength 500nm. The observing screen is placed 80 cm from the slit. The width of the central bright fringe will be

1. 1mm
2. 2mm
3. 4mm
4. 5mm

Solution: 3. 4mm

Question 11. A plane wavefront (λ = 6×10^-7m) falls on a slit 0.4 mm wide. A convex lens of focal length 0.8m placed behind the slit focuses the light on a screen. What is the linear diameter of the second maximum

1. 6mm
2. 12mm
3. 3mm
4. 9mm

Solution: 1. 6mm

Question 12. Yellow light is used in a single slit diffraction experiment with a slit width of 0.6 mm. If yellow light is replaced by X-rays then the pattern will reveal

1. The central maxima are narrower
2. No diffraction pattern
3. More number of fringes
4. Less number of fringes

Solution: 2. No diffraction pattern

Question 13. A beam of light of wavelength 600 nm from a distant source falls on a single slit 1 mm wide and the resulting diffraction pattern is observed on a screen 2 m away. The distance between the first dark fringes on either side of the central bright fringe is

1. 1.2 mm
2. 1.2 cm
3. 2.4 cm
4. 2.4 mm

Solution: 4. 2.4 mm

Question 14. In the far-field diffraction pattern of a single slit under polychromatic illumination, the first minimum with the wavelength λ₁ is found to be coincident with the third maximum at λ₂. So

1. 3λ₁ = 0.3λ₂
2. 3λ₁ = λ₂
3. λ₁ = 3.5λ₂
4. 0.3λ₁ = 3λ₂

Solution: 3. λ₁ = 3.5λ₂

Question 15. If in Fraunhofer diffraction due to a single slit, the slit width is increased, the width of the central maximum will

1. Increase
2. Decrease
3. Not change
4. Change depending upon the wavelength of light used

Solution: 2. Decrease

Question 16. A slit of size 0.15 cm is placed at 2.1 m from a screen. On illuminated it by a light of wavelength 5 x 10^-5 cm. The width of the diffraction pattern will be:-

1. 70 mm
2. 0.14 mm
3. 1.4 cm
4. 0.14 cm

Solution: 4. 0.14 cm

Question 17. What is the effect on Fresnel’s biprism experiment when the use of white light is made

1. Fringe are affected
2. The diffraction pattern is spread more
3. The central fringe is white and all are coloured
4. None of these

Solution: 3. The Central fringe is white and all are colored

Question 18. A slit of size 0.15 cm is placed at 2.1 m from a screen. On illuminated it by a light of wavelength 5 × 10^-5 cm. The width of the central maxima will be

1. 70 mm
2. 0.14 mm
3. 1.4 mm
4. 0.14 cm

Solution: 3. 1.4 mm

Question 19. The diameter of the human eye lens is 2 mm, which is the minimum distance between two points to resolve them, which are situated at a distance of 50 meters from the eye. The wavelength of light is 5000 Å:

1. 2.32 m
2. 4.28 mm
3. 1.25 cm
4. 12.48 cm

Solution: 3. 1.25 cm

Question 20. If Ι₀ is the intensity of the principal maximum in the single slit diffraction pattern, then what will be its intensity when the slit width is doubled?

1. Ι₀
2. Ι₀ /2
3. 2Ι₀
4. 4Ι₀

Solution: 4. 4Ι₀

Section (6) : Polarisation

Question 1. The angle of polarisation for any medium is 60° which will be a critical angle for this

1. sin–1 √3
2. tan–1 √3
3. cos–1 √3
4. sin–1 √1/3

Solution: 4. sin–1 1/3

Question 2. The angle of incidence at which reflected light is polarized for reflection from air to glass (refraction index)

1. sin^-1(n)
2. sin^-1\(\left(\frac{1}{n}\right)\)
3. tan^-1 \(\left(\frac{1}{n}\right)\)
4. tan^-1(n)

Solution: 4. tan^-1(n)

Question 3. A polaroid is placed at 45° to an incoming light of intensity I₀. Now the intensity of light passing through the polaroid after polarisation would be

1. I₀
2. I₀/2
3. I₀/4
4. Zero

Solution: 1.I₀

Question 4. Plane polarised light is passed through a polaroid. On viewing through the polaroid we find that when the polaroid is given one complete rotation about the direction of the light one of the following is observed

1. The intensity of light gradually decreases to zero and remains at zero
2. The intensity of light gradually increases to a maximum and remains at a maximum
3. There is no change in intensity
4. The intensity of light is twice the maximum and twice zero

Solution: 4. The intensity of light is twice the maximum and twice zero

Question 5. A ray of light is incident on the surface of a glass plate at an angle of incidence equal to Brewster’s angle φ. If μ represents the refractive index of glass concerning air then the angle between reflected and refracted rays is

1. 90 + φ
2. sin^-1 (μcos φ)
3. 90°
4. 90°– sin^-1 (sin φ /μ)

Solution: 3. 90°

Question 6. The figure represents a glass plate placed vertically on a horizontal table with a beam of unpolarised light falling on its surface at the polarising angle of 57° with the normal. The electric vector in the reflected light on screen S will vibrate concerning the plane of incidence in a

1. Vertical plane
2. Horizontal plane
3. The plane makes an angle of 45° with the vertical
4. The plane makes an angle of 57° with the horizontal

Solution: 1. Vertical plane

Question 7. A beam of light AO is incident on a glass slab (μ= 1.54) in a direction as shown in the figure. The reflected ray OB is passed through a Nicol prism. On viewing through a Nicole prism we find on rotating the prism that

1. The intensity is reduced to zero and remains zero
2. The intensity is reduced somewhat and rises again
3. There is no change in intensity
4. The intensity gradually reduces to zero and then again increases

Solution: 4. The intensity gradually reduces to zero and then again increases

Question 8. In the propagation of electromagnetic waves, the angle between the direction of propagation and the plane of polarization is

1. 0°
2. 45°
3. 90°
4. 180°

Solution: 1. 0°

Question 9. Unpolarized light falls on two polarizing sheets placed one on top of the other. What must be the angle between the characteristic directions of the sheets if the intensity of the final transmitted light is one-third the maximum intensity of the first transmitted beam

1. 75°
2. 55°
3. 35°
4. 15°

Solution: 2. 55°

Question 10. Unpolarized light of intensity 32Wm^-2 passes through three polarizers such that the transmission axes of the first and second polarizer make an angle of 30° with each other and the transmission axis of the last polarizer is crossed with that of the first. The intensity of the final emerging light will be

1. 32 Wm^-2
2. 3 Wm^-2
3. 8 Wm^-2
4. 4 Wm^-2

Solution: 2. 3 Wm^-2

Question 11. Two polaroids are placed in the path of an unpolarized beam of intensity Ι₀ such that no light is emitted from the second polaroid. If a third polaroid whose polarization axis makes an angle θ with the polarization axis of the first polaroid, is placed between these polaroids then the intensity of light emerging from the last polaroid will be

1. \(\left(\frac{I_0}{8}\right) \sin ^2 2 \theta\)
2. \(\left(\frac{I_0}{4}\right) \sin ^2 2 \theta\)
3. \(\left(\frac{I_0}{2}\right) \cos ^4 \theta\)
4. \(I_0 \cos ^4 \theta\)

Solution: 1. \(\left(\frac{I_0}{8}\right) \sin ^2 2 \theta\)

Question 12. A beam of natural light falls on a system of 6 polaroids, which are arranged in succession such that each polaroid is turned through 30° concerning the preceding one. The percentage of incident intensity that passes through the system will be

1. 100%
2. 50%
3. 30%
4. 12%

Solution: 4. 12%

Question 13. When an unpolarized light of intensity Ι0 is incident on a polarizing sheet, the intensity of the light that does not get transmitted is

1. zero
2. Ι0
3. \(\frac{1}{2} I_0\)
4. \(\frac{1}{4} I_0\)

Solution: 2. Ι0

Wave Optics Exercise – 2

Question 1. The distance between two successive atomic planes of a calcite crystal is 0.3 nm. The minimum angle for Brag scattering of 0.3 Å X-rays will be

1. 1.43°
2. 1.56°
3. 2.86°
4. 30°

Solution: 3. 2.86°

Question 2. Two coherent sources S₁ and having the same phase, emit light of wavelength λ. The separation between S₁ and S₂ 2λ. The light collected on a screen placed at a distance D > > λ from slit S₁ is shown in the figure. Find the minimum distance so that the intensity at P is equal to the intensity at O.

1. \(\frac{D}{\sqrt{2}}\)
2. \(\sqrt{2} \mathrm{D}\)
3. \(\frac{D}{\sqrt{3}}\)
4. \(\mathrm{D} \sqrt{3}\)

Solution: 4. \(\mathrm{D} \sqrt{3}\)

Question 3. What happens to the fringe pattern when Young’s double slit experiment is performed in water instead or air then fringe width

1. Shrinks
2. Disappear
3. Unchanged
4. Enlarged

Solution: 1. Shrinks

Question 4. Due to the effect of interference, the floating oil layer in water is visible in colored, for observation of this event the thickness of the oil layer should be :

1. 100 nm
2. 1000 nm
3. 1 mm
4. 10 mm

Solution: 1. 100 nm

Question 5. A ray of light of intensity I is incident on a parallel glass slab at point A as shown in Fig. It undergoes partial reflection and refraction. At each reflection, 25% of incident energy is reflected. The rays AB and A’B’ undergo interference. The ratio Ι_max/Ι_min is

1. 4: 1
2. 8: 1
3. 7: 1
4. 49: 1

Solution: 4. 49: 1

Question 6. In Young’s double slit experiment, the intensity on the screen at a point where the path difference is λ is K. What will be the intensity at the point where the path difference is λ/4

1. \(\frac{\mathrm{K}}{4}\)
2. \(\frac{\mathrm{K}}{2}\)
3. K
4. Zero

Solution: 2. \(\frac{\mathrm{K}}{2}\)

Question 7. When one of the slits of Young’s experiment is covered with a transparent sheet of thickness 4.8 mm, the central fringe shifts to a position originally occupied by the 30th bright fringe. What should be the thickness of the sheet if the central fringe has to shift to the position occupied by the 20th bright fringe?

1. 3.8 mm
2. 1.6 mm
3. 7.6 mm
4. 3.2 mm

Solution: 4. 3.2 mm

Question 8. In Young’s double slit experiment, how many maxims can be obtained on a screen (including the central maximum) on both sides of the central fringe if λ = 2000 Å λ = 2000Åand d = 7000 Å

1. 12
2. 7
3. 18
4. 4

Solution: 2. 7

Question 9. In a single slit diffraction of light of wavelength λ by a slit of width e, the size of the central maximum on a screen at a distance b is

1. 2b e λ+(2)
2. \(\frac{2 \mathrm{~b} \lambda}{\mathrm{e}}\)
3. \(\frac{2 b \lambda}{e}+e\)
4. \(\frac{2 b \lambda}{e}-e\)

Solution: 3. \(\frac{2 b \lambda}{e}+e\)

Question 10. Among the two interfering monochromatic sources A and B; A is ahead of B in phase by 66°. If the observation is taken from point P, such that PB – PA = λ/4. Then the phase difference between the waves from A and B reaching P is

1. 156°
2. 140°
3. 136°
4. 126°

Solution: 1. 156°

Question 11. In an experiment, the two slits are 0.5 mm apart and the fringes are observed to be 100 cm from the plane of the slits. The distance of the 11th bright fringe from the Ist bright fringe is 9.72 mm. Calculate the wavelength-

1. 4.86 × 10^-5 cm
2. 4.86 × 10^-8 cm
3. 4.86 × 10^-6 cm
4. 4.86 × 10^-7 cm

Solution: 1. 4.86 × 10^-5 cm

Question 12. In Young’s double slit experiment, a mica slip of thickness t and refractive index μ is introduced in the ray from the first source S₁. By how much distance, the fringe pattern will be displaced-[d= separation between slits]

1. \(\frac{d}{D}(\mu-1) \mathrm{t}\)
2. \(\frac{D}{d}(\mu-1) t\)
3. \(\frac{d}{(\mu-1) D}\)
4. \(\frac{D}{d}(\mu-1)\)

Solution: 2. \(\frac{D}{d}(\mu-1) t\)

Question 13. Young’s double slit experiment is performed with blue and with green light of wavelengths 4360 Å and 5460 Å respectively. If X is the distance of the 4th maximum from the central one, then :

1. X(blue) = X(green)
2. X(blue) > X(green)
3. X(blue) < X(green)
4. X(blue) 5460 = X(green)

Solution: 3. X(blue) < X(green)

Question 14. The two slits at a distance of 1 mm are illuminated by the light of wavelength 6.5 x 10 m. The interference fringes are observed on a screen placed at a distance of 1 m. The distance between the third dark fringe and the fifth bright fringe will be:-

1. 0.65 mm
2. 1.63 mm
3. 3.25 mm
4. 4.88 mm

Solution: 2. 1.63 mm

Question 15. A two-slit Young’s interference experiment is done with monochromatic light of wavelength 6000 Å. The slits are 2 mm apart. The fringes are observed on a screen placed 10 cm away from the slits. Now a transparent plate of thickness 0.5 mm is placed in front of one of the slits and it is found that the interference pattern shifts by 5 mm. The refractive index of the transparent plate is :

1. 1.2
2. 0.6
3. 2.4
4. 1.5

Solution: 1. 1.2

Question 16. In Young’s experiment, using light of λ = 5893 A₀, 62 fringes are observed in the field of view. How many fringes will be formed in the field of view for wavelength of 5461 A₀:-

1. 62
2. 66
3. 67
4. 68

Solution: 3. 67

Question 17. In two separate set-ups of Young’s double-slit experiment, fringes of equal width are observed when lights of wavelengths in the ratio 1: 2 are used. If the ratio of the slit separation in the two cases is 2: 1, the ratio of the distances between the plane of the slits and the screen in the two set-ups is:-

1. 4: 1
2. 1: 1
3. 1: 4
4. 2: 1

Solution: 1. 4: 1

Question 18. White light is incident normally on a glass plate (in air) of thickness 500 nm and refractive index of 1.5. The wavelength (in nm) in the visible region (400 nm – 700nm) that is strongly reflected by the plate is:

1. 450
2. 600
3. 400
4. 500

Solution: 2. 600

Question 19. Which of the following phenomena exhibits the particle’s nature of light?

1. Interference
2. Diffraction
3. Polarisation
4. Photoelectric effect

Solution: 4. Photoelectric effect

Question 20. In a double-slit experiment, instead of taking slits of equal widths, one slit is made twice as wide as the other. Then, in the interference pattern

1. The intensities of both the maxima and the minima increase
2. The intensity of the maxima increases and the minima have zero intensity
3. The intensity of the maxima decreases and that of the minima increases
4. The intensity of the maxima decreases and the minima has zero intensity.

Solution: 1. The intensities of both the maxima and the minima increase

Question 21. In the ideal double-slit experiment, when a glass plate (refractive index 1.5) of thickness t is introduced in the path of one of the interfering beams (wavelength λ), the intensity at the position where the central maximum occurred previously remains unchanged. The minimum thickness of the glass plate is:

1. 2 λ
2. 2 λ/3
3. λ/3
4. λ

Solution: 1. 2 λ

Question 22. A parallel beam of light of wavelength λ is incident on a plane mirror at an angle θ as shown in the figure. With maximum intensity at point P, which of the following relation is correct?

1. cos θ − sec θ = \(\frac{\lambda}{4d}\)
2. cos θ = \(\frac{\lambda}{4d}\)
3. cos θ − sin θ = \(\frac{\lambda}{d}\)
4. cos θ = \(\frac{\lambda}{2d}\)

Solution: 2. cos θ = \(\frac{\lambda}{4d}\)

Question 23. In a YDSE arrangement composite lights of different wavelengths λ₁= 560 nm and λ₂= 400 nm are used. If D = 1m, d = 0.1 mm. Then the distance between two completely dark regions is :

1. 4 mn
2. 5.6 mm
3. 14 mm
4. 28 mm

Solution: 4. 28 mm

Question 24. In Young’s double-slit experiment, an electron beam is used to form a fringe pattern instead of light. If the speed of the electrons is increased then the fringe width will :

1. Increase
2. Decrease
3. Remains same
4. No fringe pattern will be formed

Solution: 2. Decrease

Question 25. In Young’s double slit experiment, the maximum intensity is Ι than the angular position where the intensity becomes 1/4 Ιis :

1. sin^-1 \(\left(\frac{\lambda}{d}\right)\)
2. sin^-1 \(\left(\frac{\lambda}{3d}\right)\)
3. sin^-1 \(\left(\frac{\lambda}{2d}\right)\)
4. sin^-1 \(\left(\frac{\lambda}{4d}\right)\)

Solution: 2. sin–1 \(\left(\frac{\lambda}{3d}\right)\)

Question 26. Young’s double slit experiment is carried out by using green, red, and blue light, one color at a time. The fringe widths recorded are β_G, β_R, and β_B, respectively. Then

1. β_G> β_B> β_R
2. β_B> β_G> β_R
3. β_R> β_B> β_G
4. β_R> β_G> β_B

Solution: 4. β_R> β_G> β_B

Question 27. A light source, which emits two wavelengths λ₁= 400 nm and λ₂ = 600 nm, is used in Young’s double slit experiment. If recorded fringe widths for λ₁and λ₂are β₁and β₂and the number of fringes for them within a distance y on one side of the central maximum are m₁and m₂, respectively, then the incorrect statement is

1. β₂> β₁
2. m₁> m₂
3. From the central maximum, 3rd maximum of λ₂overlaps with 5th minimum of λ₁
4. The angular separation of fringes for λ₁is greater than λ₂

Solution: 4. The angular separation of fringes for λ₁is greater than λ₂

Wave Optics Exercise – 3

Question 1. In Young’s double slit experiment, the slits are 2 mm apart and are illuminated by photons of two wavelengths λ1= 12000 Å and λ2= 10000 Å. At what minimum distance from the common central bright fringe on the screen, 2 m from the slit will a bright fringe from one interference pattern coincide with a bright fringe from the other?

1. 6 mm
2. 4 mm
3. 3 mm
4. 8 mm

Solution: 1. 6 mm

Question 2. A parallel beam of fast-moving electrons is incident normally on a narrow slit. A fluorescent screen is placed at a large distance from the slit. If the speed of the electrons is increased, which of the following statements is correct?

1. The angular width of the central maximum of the diffraction pattern will increase.
2. The angular width of the central maximum will decrease.
3. The angular width of the central maximum will be unaffected.
4. The diffraction pattern is not observed on the screen in the case of electrons.

Solution: 2. The angular width of the central maximum will decrease.

Question 3. In a double-slit experiment, the two slits are 1mm apart and the screen is placed 1 m away. A monochromatic light wavelength of 500 nm is used. What will be the width of each slit for obtaining ten maxima of double slit within the central maxima of a single slit pattern?

1. 0.1 mm
2. 0.5 mm
3. 0.02 mm
4. 0.2 mm

Solution: 4. 0.2 mm

Question 4. For a parallel beam of monochromatic light of wavelength ‘λ’, diffraction is produced by a single slit whose width ‘a is of the wavelength of the light. If ‘D’ is the distance of the screen from the slit, the width of the central maxima will be

1. \(\frac{\mathrm{D} \lambda}{\mathrm{a}}\)
2. \(\frac{D \lambda}{a}\)
3. \(\frac{2 \mathrm{Da}}{\lambda}\)
4. \(\frac{2 D \lambda}{a}\)

Solution: 4. \(\frac{2 D \lambda}{a}\)

Question 5. In a diffraction pattern due to a single slit of width ‘a’ the first minimum is observed at an angle 30° when light of wavelength 5000 Å is incident on the slit. The first secondary maximum is observed at an angle of :

1. sin^-1 \(\left(\frac{3}{4}\right)\)
2. sin^-1 \(\left(\frac{1}{4}\right)\)
3. sin^-1 \(\left(\frac{2}{3}\right)\)
4. sin^-1 3 \(\left(\frac{2}{3}\right)\)

Solution: 1. sin^-1 \(\left(\frac{3}{4}\right)\)

Question 6. A linear aperture whose width is 0.02 cm is placed immediately in front of a lens of focal length 60 cm. The aperture is illuminated normally by a parallel beam of wavelength 5 × 10^-5 cm. The distance of the first dark band of the diffraction pattern from the center of the screen is :

1. 0.15 cm
2. 0.10 cm
3. 0.25 cm
4. 0.20 cm

Solution: 1. 0.15 cm

Question 7. The ratio of resolving powers of an optical microscope for two wavelengths λ₁ = 4000 Å and λ₂ = 6000 Å is :

1. 8: 27
2. 9: 4
3. 3: 2
4. 16: 81

Solution: 3. 3: 2

Question 8. Young’s double slit experiment is first performed in air and then in a medium other than air. It is found that the 8th bright fringe in the medium lies where the 5th dark fringe lies in the air. The refractive index of the medium is nearly :

1. 1.25
2. 1.59
3. 1.69
4. 1.78

Solution: 4. 1.78

Question 9. Unpolarised light is incident from the air on a plane surface of a material of refractive index ‘μ’. At a particular angle of incidence ‘i’, it is found that the reflected and refracted rays are perpendicular to each other. Which of the following options is correct for this situation?

1. Reflected light is polarised with its electric vector parallel to the plane of incidence
2. \(\mathrm{i}=\tan ^{-1}\left(\frac{1}{\mu}\right)\)
3. \(\mathrm{i}=\sin ^{-1}\left(\frac{1}{\mu}\right)\)
4. Reflected light is polarised with its electric vector perpendicular to the plane of incidence

Solution: 2. \(\mathrm{i}=\tan ^{-1}\left(\frac{1}{\mu}\right)\)

Question 10. In Young’s double slit experiment the separation d between the slits is 2 mm, the wavelength λ of the light used is 5896 Å, and the distance D between the screen and slits is 100 cm. It is found that the angular width of the fringes is 0.20º. To increase the fringe angular width to 0.21º (with the same λ and D) the separation between the slits needs to be changed to :

1. 1.8 mm
2. 1.7 mm
3. 2.1 mm
4. 1.9 mm

Solution: 1. 1.8 mm

Question 11. In a double slit experiment, when light of wavelength 400 nm was used, the angular width of the first minima formed on a screen placed 1 m away, was found to be 0.2º. What will be the angular width of the first minima, if the entire experimental apparatus is immersed in water? (µwater = 4/3)

1. 0.1º
2. 0.266º
3. 0.15º
4. 0.05º

Solution: 3. 0.15º

Question 12. In Young’s double-slit experiment, if there is no initial phase difference between the light from the two slits, a point on the screen corresponding to the fifth minimum has a path difference

1. \(5 \frac{\lambda}{2}\)
2. \(10 \frac{\lambda}{2}\)
3. \(9 \frac{\lambda}{2}\)
4. \(11 \frac{\lambda}{2}\)

Solution: 3. \(10 \frac{\lambda}{2}\)

Question 13. The angular width of the central maximum in the Fraunhofer diffraction for λ = 6000 Å is θ0. When the same slit is illuminated by another monochromatic light, the angular width decreases by 30%. The wavelength of this light is :

1. 1800 Å
2. 4200 Å
3. 6000 Å
4. 420 Å

Solution: 2. 4200 Å

Question 14. At two points P and Q on a screen in Young’s double slit experiment, waves from slits S₁ and S₂ have a path difference of 0 and λ/4 respectively. The ratio of intensities at P and Q will be :

1. 2: 1
2. 2 :1
3. 4: 1
4. 3: 2

Solution: 1. 2: 1

Question 15. In Young’s double slit experiment, the two slits act as coherent sources of waves of equal amplitude A and wavelength λ. In another experiment with the same arrangement, the two slits are made to act as incoherent sources of waves of the same amplitude and wavelength. If the intensity at the middle point of Ι₁ the screen in the first case is Ι₁ and in the second case is Ι₂, then the ratio\(\frac{\mathrm{I}_1}{\mathrm{I}_2}\) is :

1. 2
2. 1
3. 0.5
4. 4

Solution: 1. 2

Question 16.

Statement – 1: On viewing the clear blue portion of the sky through a Calcite Crystal, the intensity of transmitted light varies as the crystal is rotated.

Statement 2: The light coming from the sky is polarized due to the scattering of sunlight by particles in the atmosphere. The scattering is largest for blue light

1. Statement-1 is true, and statement-2 is false.
2. Statement-1 is true, statement-2 is true, statement-2 is the correct explanation of statment-1
3. Statement-1 is true, statement-2 is true, statement-2 is not the correct explanation of statement-1
4. Statement-1 is false, and statement-2 is true.

Solution: 4. Statement-1 is false, statement-2 is true.

Question 17. Direction :

The question has a paragraph followed by two statements, Statement –1 and Statement –2. Of the given four alternatives after the statements, choose the one that describes the statements. A thin air film is formed by putting the convex surface of a plane-convex lens over a plane glass plate. With monochromatic light, this film gives an interference pattern due to light reflected from the top (convex) surface and the bottom (glass plate) surface of the film.

Statement –1: When light reflects from the air-glass plate interface, the reflected wave suffers a phase change of π

Statement 2: The center of the interference pattern is dark.

1. Statement –1 is true, statement –2 is false.
2. Statement –1 is true, Statement –2 is true, Statement–2 is the correct explanation of Statement–1
3. Statement –1 is true, Statement –2 is true, Statement–2 is not the correct explanation of Statement–1
4. Statement–1 is false, Statement –2 is true

Solution: 4. Statement–1 is false, Statement –2 is true

Question 18. Two coherent point sources S₁ and S₂ are separated by a small distance ‘d’ as shown. The fringes obtained on the screen will be :

1. Points
2. Straight lines
3. Semi-circles
4. Concentric circles

Solution: 4. Concentric circles

Question 19. Tow beams, A and B, of plane-polarized light with mutually perpendicular planes of polarization are seen through a polaroid. From the position when beam A has maximum intensity (and beam B has zero intensity), a rotation of polaroid through 30º makes the two beams appear equally bright. If the initial intensities of the two beams are and IB respectively, then \(\frac{I_A}{I_B}\) equals

1. 3
2. 3/2
3. 1
4. 1/3

Solution: 4. 1/3

Question 20. On a hot summer night, the refractive index of air is smallest near the ground and increases with height from the ground. When a light beam is directed horizontally, the Huygens’ principle leads us to conclude that as it travels, the light beam :

1. Becomes narrower
2. Goes horizontally without any deflection
3. Bends downwards
4. Bends upwards

Solution: 4. Bends upwards

Question 21. Assuming the human pupil to have a radius of 0.25 cm and a comfortable viewing distance of 25 cm, the minimum separation between two objects that the human eye can resolve at 500 nm wavelength is :

1. 1 μm
2. 30 μm
3. 100 μm
4. 300 μm

Solution: 2. 30 μm

Question 22. The box of a pinhole camera, of length L, has a hole of radius a. It is assumed that when the hole is illuminated by a parallel beam of light of wavelength λ the spread of the spot (obtained on the opposite wall of the camera) is the sum of its geometrical spread and the spread due to diffraction. The spot would then have its minimum size (say bmin) when :

1. \(a=\sqrt{\lambda L} and b_{\min }=\left(\frac{2 \lambda^2}{L}\right)\)
2. \(a=\sqrt{\lambda L} and b_{\min }=\sqrt{4 \lambda L}\)
3. \(\mathrm{a}=\frac{\lambda^2}{\mathrm{~L}} and \mathrm{b}_{\min }=\sqrt{4 \lambda \mathrm{L}}\)
4. \(\mathrm{a}=\frac{\lambda^2}{\mathrm{~L}} and \mathrm{b}_{\text {min }}=\left(\frac{2 \lambda^2}{\mathrm{~L}}\right)\)

Solution: 2. \(a=\sqrt{\lambda L} and b_{\min }=\sqrt{4 \lambda L}\)

Question 23. Unpolarized light of intensity I pass through an ideal polarizer A. Another identical polarizer B is placed behind A. The intensity of light beyond B is found to be I/2. Now another identical polarizer C is placed between A and B. The intensity beyond B is now found to be I/8. The angle between polarizers A and C is :

1. 45°
2. 60°
3. 0°
4. 30°

Solution: 1. 45°

Question 24. The angular width of the central maximum in a single slit diffraction pattern is 60°. The width of the slit is 1 μm. The slit is illuminated by monochromatic plane waves. If another slit of the same width is made near it, Young’s fringes can be observed on a screen placed at a distance of 50 cm from the slits. If the observed fringe width is 1 cm, what is the slit separation distance?
(i.e. distance between the centers of each slit.)

1. 75 μm
2. 100 μm
3. 25 μm
4. 50 μm

Solution: 3. 25 μm

Question 25. Two coherent sources produce waves of different intensities that interfere. After interference, the ratio of the maximum intensity to the minimum intensity is 16. The intensity of the waves is in the ratio :

1. 4: 1
2. 16: 9
3. 5 : 3
4. 25: 9

Solution: 4. 25: 9

Question 26. In Young’s double-slit experiment, the slits are placed 0.320 mm apart. Light of wavelength λ = 500 nm is incident on the slits. The total number of bright fringes that are observed in the angular range – 30º ≤ θ ≤ 30º is :

1. 320
2. 641
3. 640
4. 321

Solution: 2. 641

Question 27. Consider Young’s double-slit experiment as shown in the figure. What should be the slit separation d in terms of wavelength λ such that the first minima occurs directly in front of the slit (S₁)

1. \(\frac{\lambda}{(\sqrt{5}-2)}\)
2. \(\frac{\lambda}{2(5-\sqrt{2})}\)
3. \(\frac{\lambda}{2(\sqrt{5}-2)}\)
4. \(\frac{\lambda}{(5-\sqrt{2})}\)

Solution: 2. \(\frac{\lambda}{2(5-\sqrt{2})}\)

Question 28. In a double-slit experiment, green light (5303Å) falls on a double slit having a separation of 19.44 μm The number of bright fringes between the first and second diffraction minima is :

1. 09
2. 05
3. 04
4. 10

Solution: 3. 04

Question 29. White light is used to illuminate the two silts in Young’s double-slit experiment. The separation between the slits is b and the screen is at a distance d (> > b) from the slits. At a point on the screen directly in front of one of the slits, certain wavelengths are missing. Some of these missing wavelengths are :

1. \(\lambda=\frac{b^2}{2 d}\)
2. \(\lambda=\frac{2 b^2}{d}\)
3. \(\lambda=\frac{b^2}{3 d}\)
4. \(\lambda=\frac{2 b^2}{3 d}\)

Solution: 3.\(\lambda=\frac{b^2}{3 d}\)

Question 30. In a double-slit experiment, instead of taking slits of equal widths, one slit is made twice as wide as the other. Then, in the interference pattern

1. The intensities of both the maxima and the minima increase
2. The intensity of the maxima increases and the minima have zero intensity
3. The intensity of the maxima decreases and that of the minima increases
4. The intensity of the maxima decreases and the minima has zero intensity.

Solution: 1. The intensities of both the maxima and the minima increase

Question 31. In the ideal double-slit experiment, when a glass plate (refractive index 1.5) of thickness t is introduced in the path of one of the interfering beams (wavelength λ), the intensity at the position where the central maximum occurred previously remains unchanged. The minimum thickness of the glass plate is:

1. 2 λ
2. 2 λ/3
3. λ/3
4. λ

Solution: 1. 2 λ

Question 32. A parallel beam of light of wavelength λ is incident on a plane mirror at an angle θ as shown in the figure. With maximum intensity at point P, which of the following relation is correct?

1. cos θ − sec θ =\(\frac{\lambda}{4d}\)
2. cos θ = \(\frac{\lambda}{4d}\)
3. cos θ − sin θ =\(\frac{\lambda}{d}\)
4. cos θ = \(\frac{\lambda}{2d}\)

Solution: 2. cos θ = \(\frac{\lambda}{4d}\)

Question 33. In a YDSE arrangement composite lights of different wavelengths λ₁= 560 nm and λ₂= 400 nm are used. If D = 1m, d = 0.1 mm. Then the distance between two completely dark regions is

1. 4 mn
2. 5.6 mm
3. 14 mm
4. 28 mm

Solution: 4. 28 mm

Question 34. In Young’s double slit experiment, the maximum intensity is Ι then the angular position where the intensity becomes Ι/4 is :

1. sin^-1
2. sin^-1
3. sin^-12d
4. sin^-1

Solution: 2. sin^-1