## Optics

## Diffraction And Polarisation Of Light Multiple Choice Questions

**Question 1. A light bea is i incident from ait to a glass slab at Brewster angle, A polaroid placed in the path of the refracted beam is being rotated about an axis passing through the center and perpendicular to the plane of the polaroid**

- For a particular orientation, there shall be darkness as observed through the polaroid
- The intensity of light as seen through the polaroid shall be independent of the rotation
- The intensity of light as seen through the polaroid shall go through a maximum but not zero for two orientations of the polaroid
- The intensity of light as seen through the polaroid shall go through a minimum but not zero for two orientations of the polaroid

**Answer:** 3

**Question 2. Consider the diffraction pattern of a small pinhole. As the size of the hole Is Increased the changes that occur In the pattern**

- The size decreases
- The intensity increases
- The size Increases
- The intensity decreases

**Answer:** 1 And 2

**Question 3. Geometrical optics is valid when the dimensions of the aperture are**

- Of the same order as the wavelength of light
- Much smaller than the wavelength of light
- Most the larger than the wavelength of light
- Of the order of 1 A

**Answer:** 3. Much of the larger than the wavelength of light

**Read and Learn More Class 12 Physics Multiple Choice Questions**

**Question 4. Light appears to propagate in straight lines because**

- It is reflected by the upper part of the atmosphere
- It is not absorbed in the atmosphere
- Its speed is very large
- Its wavelength is very small

**Answer:** 4. Its wavelength is very small

**Question 5. Which of the following does not support the corpuscular nature of light?**

- Photoelectric effect
- Compton effect
- Diffraction
- Blackbody radiation

**Answer:** 2. Compton effect

**Question 6. Maximum diffraction takes place in the case of**

- Ultraviolet rays
- Radio waves
- γ-rays
- Infrared waves

**Answer:** 4. Infrared waves

**Question 7. Among the following conditions which one is an essential condition for Fresnel diffraction?**

- The source and screen will remain at infinite distance from the slit
- Either source or screen will remain at infinite distance from the slit
- Neither source nor screen will remain at infinite distance from the slit
- None of the above

**Answer:** 3. Neither source nor screen will remain at infinite distance from the slit

**Question 8. In single slit diffraction of light of wavelength λ, the angular width of the central maximum (considering slit width d) will be**

- \(\frac{d}{\lambda}\)
- \(\frac{\lambda}{d}\)
- \(\frac{2 \lambda}{d}\)
- \(\frac{2 d}{\lambda}\)

**Answer:** 3. \(\frac{2 \lambda}{d}\)

**Question 9. In the diffraction of light of wavelength λ at a single slit of width a, the angle θ between the central maximum and first minimum on either side is**

- \(\frac{d}{\lambda}\)
- \(\frac{\lambda}{2 a}\)
- \(\frac{\lambda}{4 a}\)
- \(\frac{\pi}{2}\)

**Answer:** 1. \(\frac{d}{\lambda}\)

**Question 10. The width of single-slit diffraction fringes varies**

- Directly with the distance between the slit and the fringes
- Inversely with the wavelength of light
- Directly with the width of the slit
- None of above

**Answer:** 2. Inversely with the wavelength of light

**Question 11. If the width of the slit is increased, what changes would be there in the linear width of the central maximum?**

- Increases
- Decreases
- Remains unchanged
- None of the above

**Answer:** 2. Decreases

**Question 12. In a single slit (width = a) diffraction of light wavelength λ, fringes are produced with a diffraction angle θ. The condition of formation of the first minimum is**

- λsin θ = a
- a cos θ = λ
- a sin θ = λλcos θ= a

**Answer:** 3. a sin θ = λ

**Question 13. In a single slit experiment, the angular width of the first minima obtained by using the light of wavelength G980 A is 2°. The width of the slit is**

- 2 × 10
^{-5}mm - 0.02 mm
- 0.2mm
- 2 mm

**Answer:** 2. 0.02mm

**Question 14. In a single slit experiment, the angular width of the first minima obtained by using the light of wavelength 6980 A° is 2°. The width of the slit is**

- 2.4 mm
- 4.8 mm
- 9.6 mm
- None of these

**Answer:** 4. None of these

**Question 15. The width of the diffraction band is**

- Inversely proportional to the wavelength of the incident light
- Inversely proportional to the size of the source which illuminates the slit.
- Directly proportional to the distance between the slit and the screen
- Directly proportional to the width of the slit

**Answer:** 3. Directly proportional to the distance between the slit and the screen

**Question 16. In a single slit Fraunhofer diffraction experiment, the wavelength of the light used is 400 nm and the first minimum is obtained for a diffraction angle of 30°. The magnitude of 8 for first secondary maximum is**

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

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

**Question 17. In a single slit experiment, a beam of parallel monochro¬ matic light falls on a slit perpendicularly. A diffraction fringe is produced on a screen kept perpendicular to the direction of propagation of light. What would be the phase difference of the two rays coming from two ends of the slit at the first minimum of fringe?**

- 0
- \(\frac{\pi}{2}\)
- π
- 2π

**Answer:** 4.

**Question 18. In a single-slit experiment, the tire width of the slit is reduced. The linear width of the central maximum**

- Will Increase but the intensity will be reduced
- Will decrease but Intensity will be increased more
- Will increase but the Intensity will be Increased more
- Will decrease but the intensity will be reduced

**Answer:** 1. The linear width of the central maximum

**Question 19. In a single-slit experiment, the width of the slit is reduced to half. To keep the width of the central maximum the same, what is to be done of the following?**

- The distance between the slit and the screen is to be reduced to half
- The distance between the slit and the screen is to be reduced to one-fourth
- The distance between the slit and the screen is to be doubled
- Nothing to be done, as the width of central maximum does not depend upon the width of the slit

**Answer:** 1. The distance between the slit and the screen is to be reduced to half

**Question 20. A narrow slit of 2 mm width is illuminated by a monochromatic light of wavelength 500 nm. What would be the intermediate distance between two first minima on either side of a screen kept away?**

- 5 mm
- 0.5 mm
- 1mm
- 10mm

**Answer:** 2. 0.5 mm

**Question 21. The resolving power of a microscope is**

- Inversely proportional to the numerical aperture of the lens of the objective
- Directly proportional to the wavelength of the light
- Directly proportional to the square of the wavelength of the light
- Directly proportional to the numerical aperture of the lens of the objective

**Answer:** 4. Directly proportional to the numerical aperture of the lens of the objective

**Question 22. For the minimum angular distance between two stars, which a telescope can analyze, which of the following statements is true (here aperture means the diameter of the objective of the telescope)?**

- Angular distance decreases with the increase in the aperture of the telescope
- Angular distance does not depend on the aperture of the telescope.
- Angular distance increases linearly with the aperture of the telescope.
- Angular distance increases with the second power of the aperture of the telescope

**Answer:** 1. Angular distance decreases with the increase in the aperture of the telescope

**Question 23. An unpolarised light of intensity I _{o} falls on a pair of Nicol prism. The angle between the two prisms is 60°. the intensity of the light emitted from the prism will be**

- I
_{o} - \(\frac{I_0}{2}\)
- \(\frac{I_0}{4}\)
- \(\frac{I_0}{8}\)

**Answer:** 4. \(\frac{I_0}{8}\)

**Question 24. A polaroid is inclined at an angle of 45° with an incident light intensity of 70°. After polarization, the intensity of the light emitted from the polaroid is**

- I
_{o} - \(\frac{I_0}{2}\)
- \(\frac{I_0}{4}\)
- 0

**Answer:** 2. \(\frac{I_0}{2}\)

**Question 25. Amplitude of unpolarised light on a polariser is a. Amplitude of polarised light which passes through the polariser is**

- \(\frac{a}{2}\)
- \(\frac{a}{\sqrt{2}}\)
- \(\frac{\sqrt{3} a}{2}\)
- \(\frac{3 a}{4}\)

**Answer:** 2. \(\frac{a}{\sqrt{2}}\)

**Question 26. When a light ray in the air is incident on glass at 57°, the reflected rays are completely polarised. The same ray is incident on the water at an angle 6, also the reflected ray is completely polarised. Then**

- θ >57°
- θ <57°
- θ = 57°
- θ = 90°

**Answer:** 2. θ <57°

**Question 27. The Brewster’s law polarisation of fight is**

- μ sin i
_{p}= 1 - μ cos i
_{p}= 1 - μ tan i
_{p}= 1 - μ cot i
_{p }= 1

**Answer:** 4. μ cot i_{p }= 1

**Question 28. From Brewster’s law, it follows that the polarising angle depends on**

- Wavelength of fight
- Frequency
- Plane of polarisation
- Plane of vibration

**Answer:** 1. Wavelength of fight

**Question 29. The tangent of the polarising angle is equal to the refractive indices of the reflecting medium. This is called**

- Brewster’s law
- Law of Malus
- Braggs law
- Grimaldi’s law

**Answer:** 1. Brewster’s law

**Question 30. A ray of light is incident on a glass. The reflected ray gets totally polarized. The magnitude of the incidence angle is (refractive index of glass =p)**

- \(\sin ^{-1}(\mu)\)
- \(\sin ^{-1}\left(\frac{1}{\mu}\right)\)
- \(\tan ^{-1}\left(\frac{1}{\mu}\right)\)
- \(\tan ^{-1}(\mu)\)

**Answer:** 4. \(\tan ^{-1}(\mu)\)

**Question 31. The angle of polarization for a medium is 60°. Critical angle**

- \(\sin ^{-1} \sqrt{3}\)
- \(\tan ^{-1} \sqrt{3}\)
- \(\cos ^{-1} \sqrt{3}\)
- \(\sin ^{-1} \sqrt{\frac{1}{3}}\)

**Answer:** 4. \(\sin ^{-1} \sqrt{\frac{1}{3}}\)

**Question 32. An unpolarized ray of light is an Incident on the water surface The incident angle for which the reflected and refracted rays become perpendicular to each other is \(\mu_w=\frac{4}{3}\)–**

- \(\sin ^{-1} \frac{4}{3}\)
- \(\tan ^{-1} \frac{3}{4}\)
- \(\tan ^{-1} \frac{4}{3}\)
- \(\sin ^{-1} \frac{1}{3}\)

**Answer: **3. \(\tan ^{-1} \frac{4}{3}\)

**Question 33. The critical angle of a medium is sin-1(0.6). The polarising angle for that medium is**

- tan
^{-1}(1.5) - sin
^{-1}(0.8) - tan
^{-1}(1-6667) - tan
^{-1 }(0.6667)

**Answer:** 3. tan^{-1}(1-6667)

**Question 34. Brewster’s angle at the glass-air interface is 54.74°. If a ray of light is incident on glass from the air at an angle of 45°, then the magnitude of the refracted angle will be (given tan54.74° = \(\sqrt{2}\))**

- 60°
- 30°
- 25°
- 54.74°

**Question 35. A bird sitting on the top of a light post, which is at the side of the lake. It observed stop should of at height be completely posted, the angle which of polarised theis at indent light with the plane of the surface of the water? (Refractive index of water = 1.33 )**

- 30°
- 35°
- 37°
- 38°

**Answer:** 3. 37°

**Hint:** Here, μ = tan i_{p} or, tan i_{p} = 1.33

i_{p} = tan^{-1}(1.33) = 53°

The angle between the incident ray and the surface of the water,

θ = 90°- i_{p} = 90° -53° = 37°

**Question 36. If μ _{O} and μ_{E} are the refractive indices of the crystal for O-ray and E-ray respectively, then which of the following relations is correct for a negative crystal**

- μ
_{E}= μ_{O} - μ
_{E}>μ_{O} - μ
_{E}<μ_{O} - μ
_{E}≥μ_{O}

**Answer:** 3.μ_{E}<μ_{O}

**Question 37. In double refraction we get two refracted rays i.e., O-ray and E-ray. Which one of the following statements is correct?**

- Only O-ray is polarised
- Only E-ray is polarised
- Both O-ray and E-ray are polarised
- None of the O-ray and E-ray is polarised

**Answer:** 3. Both O-ray and E-ray are polarised

**Question 38. When a polaroid, is placed In the path of Unlit, I* rotated, the Intensity of light appears to vary but never reduce* to two. The Light Is**

- Unpolarlsed
- Plane polarised
- Partially plane polarised
- No conclusion can be drawn

**Answer:** 3. Partially plane polarised

**Question 39. When a polaroid, placed in the path of light, Is rotated, there is no change intensity of light. The Incident light –**

- Totally polarised
- partially plane polarised
- Unpolarised
- None of above

**Answer:** 3. Unpolarised

**Question 40. Which of the following undergoes larger diffraction?**

- Ultraviolet light
- Radio waves
- γ-rays
- Sound waves.

**Answer:** 3 And 4

**Question 41. The width of single-slit diffraction fringes varies**

- Directly as the distance between the slit and the screen
- Directly as the width of the slit
- Inversely as the width of the slit
- Inversely as the wavelength of light

**Answer:** 1 And 3

**Question 42. Diffraction patterns can be observed with**

- Two narrow sifts
- A large number of narrow slits
- One narrow slit
- One wide slit

**Answer:** 1, 2, And 3

**Question 43. Interference differs from diffraction in that**

- It cannot be observed with white light
- Unlike diffraction, the interference fringes are of varying intensity
- Interference minima are perfectly dark and those of diffraction may not be dark
- Interference fringes may or may not be of the same width but diffraction fringes are never of the same width

**Answer:** 3 And 4

**Question 44. Ordinary light falling at a polarising angle on a glass slab placed in ai r is partly reflected in air and partly refracted in the slab. Then:**

- The reflected light is completely polarised
- The reflected light is partially polarised
- The refracted light is completely polarised
- The refracted light is partially polarised

**Answer:** 1 And 4

**Question 45. A slit of width a is illuminated by red light of wavelength 650 nm. The first diffraction minimum is observed at an angle θ _{1} = 5.2° from the direction of the incident beam**

**1. The widt****h of the slit and in mm**

- 5.12
- 7.17
- 3.21
- 4.25

**Answer:** 2 . 7.17

**2. The angle θ _{2} at which the second minimum Is observed is**

- 10.4
- 8.6
- 7.5
- 9.5

**Answer:** 1. 10.4

**Question 46. Light of wavelength 0500 A passes through a slit 0.1 cm wide and forms a diffraction pattern on a screen 1.8 m away.**

**1. The width of the central maximum in mm is**

- 2.5
- 1.32
- 2.34
- 1.72

**Answer:** 3. 2.34

**2. The width of the central maximum in mm when the apparatus is immersed in water of refractive index is ____**

- 2.12
- 1.53
- 2.54
- 1.755

**Answer:** 4.1.755

**Question 47. If a light of wavelength λ falling on a single slit of width diffracts an angle θ, the conditions of first minima will be **

- λ sin θ = a
- a cos θ= λ
- a sin θ= λ
- λ cos θ= a

**Answer:** 3. a sin θ= λ

In this case, the condition of the first minima will be a sin θ= λ

**Question 48. Which one does not change in the polarisation of light?**

- Intensity
- Phase
- Frequency
- None of these

**Answer:** 3. Frequency

**Question 49. Find the right condition(s) for Fraunhofer diffraction due to a single slit.**

- The source is at an infinite distance and the incident beam has converged at the slit
- The source is near to the slit and the incident beam is parallel.
- The source is at infinity and the incident beam is parallel.
- The source is converged at the slit.

**Answer: ** 2 And 3

The beam of light incident on the slit has to be parallel—here the incident wavefront would become parallel wavefront f.’ or this the source could be kept at infinity or the source could be placed near the slit and the parallel beam of light could be made to fall on the slit by using the proper lens.

**Question 50. Two booms, A and R, 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 30° makes the two beams appear equally bright. If the initial intensities of the two beams arc IA and IB respectively, then l _{A}/ I_{B} equals**

- \(\frac{1}{3}\)
- 3
- \(\frac{3}{2}\)
- 1

**Answer:** 1. \(\frac{1}{3}\)

According to Malus’ law, I = I_{o}cos ² θ

I_{A’ }= I_{A} cos ² θ_{1}

= I_{A’} cos ² 30°

= \(\frac{3}{4}\) I_{A}

Since θ_{1}= 30°

Again ,I_{B’ }= I_{B} cos ² θ_{2}

= I_{B} cos² 60°

⇒ \(\frac{1}{4}\) I_{B}

Since θ_{2}= 90 °- θ_{1}= 60°

According to question I_{A’} = I_{B’}

Or, \(\frac{1}{4}\)I_{A} = \(\frac{1}{4}\)I_{B}

Or, \(\frac{I_A}{I_B}\)IB = \(\frac{1}{3}\)

**Question 51. Assuming the human pupil to have a radius of 0.25 cm comfortable viewing distance of 25 cm, the minimum separation between two objects that the human eye can resolve at 500 nm wavelength Is**

- 1 μm
- 30 μm
- 100 μm
- 300 μm

**Answer:** 2. 30 μm

Diameter, d = 0.25 × 2 = 0.5 cm

= 0.005 m

D = 25 cm = 0.25 m

λ = 500 nm = 5 ×10^{-7}m

Angular resolution, A = \(\Delta \theta=\frac{1.22 \lambda}{d}\)

If the required minimum separation is x

= \(\Delta \theta=\frac{\Delta x}{D}\)

= \(\Delta x=D \Delta \theta=0.25 \times \frac{1.22 \times\left(5 \times 10^{-7}\right)}{0.005}\)

= 30.5 ×10^{-6}m = 30.5μm

= 30.5μm ≈ = 30 μm

**Question 52. The Box of n pinhole camera of length L _{1 }has a hole of radius a. It is assumed that when the hole Is Illuminated by a parallel beam of light of wavelength A, 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 when**

- a = \(\frac{\lambda^2}{L} \text { and } b_{\min }=\frac{2 \lambda^2}{L}\)
- a = \(\sqrt{\lambda L} \text { and } b_{\min }=\frac{2 \lambda^2}{L}\)
- a = \(\sqrt{\lambda L} \text { and } b_{\min }=\sqrt{4 \lambda L}\)
- a = \(\frac{\lambda^2}{L} \text { and } b_{\min }=\sqrt{4 \lambda L}\)

**Answer:** 3. a= \(\sqrt{\lambda L} \text { and } b_{\min }=\sqrt{4 \lambda L}\)

sin θ = \(\frac{\lambda}{a} \)

tan θ = \(\frac{C E}{A E}\)

Since θ is very small, so tan θ ≈ sin θ

∴ sin = \(\frac{C E}{A E}\)

Or, = \(\frac{\lambda}{a}=\frac{C E}{L}\)

Or, CE = \(\frac{L \lambda}{a}\)

Hence diameter of the spot,

B = EF + CE + FD = 2a +\(\frac{2 L \lambda}{a}\)

The diameter of the spot will be minimal when

⇒ \(\frac{d B}{d a}\) = 0

or, 2 – \(\frac{2 L \lambda}{a}\) = 0

Or, \(\frac{L \lambda}{a^2}\) = 1

Or, a = \(\sqrt{L \lambda}\)

B_{min } = 2\(2 \sqrt{L \lambda}+2 \sqrt{L \lambda}=4 \sqrt{L \lambda}\)

Hence radius of the spot

b_{min }= \(\frac{1}{2} \times 4 \sqrt{L \lambda}=2 \sqrt{L \lambda}=\sqrt{4 L \lambda}\)

**Question 53. Unpolarlsed of Intensify I passes through an ideal polariser A, Another identical polariser B is placed behind A. The Intensity flight beyond B is found to be \(\frac{I}{2}\) Now another Identical polarizer C fa placed between A and B. The Intensity beyond I Is now found to be \(\frac{I}{8}\). The angle between polarisers A and C is**

- 45°
- 60°
- 0°
- 30°

**Answer:** 1. 45°

It’s The Intensity of unpolarised light after patting through polariser A=\(\frac{I}{2}\). Since the intensity of light does not change after passing through B, the optic axes of polarizers A and B arc parallel,

Let the angle between the optic axes of polarisers A and C be θ. Then the angle between the optic axes polarisers B and C is also θ.

Applying Malus law

⇒ \(\frac{I}{8}=\left(\frac{I}{2} \times \cos ^2 \theta\right) \times \cos ^2 \theta \quad \text { or, } \cos ^2 \theta=\frac{1}{2}\)

Or, = 45°

**Question 54. The angular width of the central maximum in a single slit diffraction pattern is 60°. The width of the slit is 1μm. 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., the distance between the centers of each slit)
Answer: **

The condition for the formation of first minima at points O_{1}, and O_{2
}

d sin θ = λ

d = width of the slit, θ = angle of diffraction)

Or, \(\lambda=\frac{d}{2}\left[\text { since } \theta=30^{\circ}\right]\)

Or, \(\lambda=\frac{1 \times 10^{-6}}{2} \mathrm{~m}\)

Or, λ = 5000 A°

The fringe within Young’s double slit experiment

= \(\frac{\lambda D}{d^{\prime}}\)

d’ = \(\frac{5000 \times 10^{-10} \times 0.5}{10^{-2}}\)

= \(2.5 \times 10^{-5}\)m

= 25 μm

**Question 55. A beam of light of = 600nm from a distant source falls on a single slit 1mm wide and the resulting diffraction pattern is observed on a screen 2m away. The distance between the first dark fringes on either side of the central bright fringe is**

- 1.2cmλ
- 1.2mm
- 2.4 cm
- 2.4mm

**Answer:** 1. 1.2cm

The distance of the n -th minima from the central maxima,

x_{n} = \(\frac{n f \lambda}{a}\) [ a = slit width ]

x_{1} = \(\frac{1 \times 200 \times 600 \times 10^{-7}}{0.1}=\frac{12 \times 10^{-3}}{0.1}\)

= 0. 12 cm

Distance of the first minima from the central maxima

= 2x_{1} = 2 × 1.2 = 2.4 mm

**Question 56. For a parallel beam of monochromatic light of wavelength A, diffraction is produced by a single slit whose width a is of the order 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**

- \(\frac{2 D \lambda}{a}\)
- \(\frac{D \lambda}{a}\)
- \(\frac{D a}{\lambda}\)
- \(\frac{2 D a}{\lambda}\)

**Answer: **1. \(\frac{2 D \lambda}{a}\)

The width ofthe central maximum will be \(\frac{2 D \lambda}{a}\)

**Question 57. In a double-slit experiment, the two slits are 1 mm apart and the tire screen is placed 1 m away. A monochromatic light of wavelength 500 nm is used. What will be the width of each slit for obtaining ten maxima of double slit within the central maxima of single slit pattern?**

- 0.2 mm
- 0.1mm
- 0.5mm
- 0.02mm

**Answer:** 1. 0.2 mm

Width of the interference fringe due to the double slit,

y =\(\frac{D}{d} \lambda\)

D = lm,rf= 1 mm and A = 500 nm

Now, the width of the central bright diffraction band due to the single slit,

y ‘= \(\frac{2 D}{a} \lambda\) a = width of the slit

According to the question

⇒ \(\frac{2 D}{a} \lambda=10 \frac{D}{d} \lambda\)

Or, \(a\frac{2}{10} d=\frac{2}{10} \times 1 \mathrm{~mm}\)

= 0.2mm

**Question 58. A parallel beam of light of wavelength A incident normally on a single slit of width d. Diffraction bands are obtained on a screen placed at a distance D from the slit. The second dark band from the central bright band will be at a distance given by**

- \(\frac{2 \lambda D}{d}\)
- λdD
- \(\frac{\lambda D}{2 d}\)
- \(\frac{2 \lambda d}{D}\)

**Answer:** 1. \(\frac{2 \lambda D}{d}\)

The distance of n -th dark band from the central bright band = \(\frac{n \lambda D}{d}\)

For n = 2 distance = \(\frac{2 \lambda D}{d}\)