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Wave Optics MCQs for NEET — Physics Questions with Answers

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In the Young’s double slit experiment with sodium light. The slits are 0.589 m a part. The angular separation of the third maximum from the central maximum will be (given λ = 589 nm)

$\sin^{- 1} (0 .33 \times 10^{8})$

$\sin^{- 1} (0 .33 \times 10^{- 6})$

$\sin^{- 1} (3 \times 10^{- 8})$

$\sin^{- 1} (3 \times 10^{- 6})$

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Explanation

Using relation, $d \sin θ = n λ \Rightarrow \sin θ = \frac{n λ}{d}$

For n = 3, $\sin θ = \frac{3 λ}{d} = \frac{3 \times 589 \times 10^{- 9}}{0 .589}$

= 3 × 10^–6 or $θ = \sin^{- 1} (3 \times 10^{- 6})$

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If the sodium light in Young’s double slit experiment is replaced by red light, the fringe width will

Decrease

Increase

Remain unaffected

First increase, then decrease

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Explanation

$B \propto λ$

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In Young’s double slit experiment the wavelength of light was changed from 7000 Å to 3500 Å. While doubling the separation between the slits which of the following is not true for this experiment

The width of the fringes changes

The colour of bright fringes changes

The separation between successive bright fringes changes

The separation between successive dark fringes remains unchanged

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Explanation

$β \propto \frac{λ}{d}$

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In Young’s double-slit experiment, an interference pattern is obtained on a screen by a light of wavelength 6000 Å, coming from the coherent sources S₁ and S₂. At certain point P on the screen third dark fringe is formed. Then the path difference S₁P – S₂P in microns is

0.75

1.5

3.0

4.5

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Explanation

For dark fringe at P

$S_{1} P - S_{2} P = Δ = (2 n - 1) λ / 2$

Here n =3 and λ = 6000

So, $Δ = \frac{5 λ}{2} = 5 \times \frac{6000}{2} = 15000 Å = 1 .5 m i c r o n$

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In Young’s double-slit experiment the fringe width is β. If entire arrangement is placed in a liquid of refractive index n, the fringe width becomes

$\frac{β}{n + 1}$

nβ

$\frac{β}{n}$

$\frac{β}{n - 1}$

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Explanation

$β = \frac{λ D}{d}$ and $λ \propto \frac{1}{μ}$

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In Young’s double slit experiment, distance between two sources is 0.1 mm. The distance of screen from the sources is 20 cm. Wavelength of light used is 5460 Å. Then angular position of the first dark fringe is

0.08°

0.16°

0.20°

0.313°

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Explanation

Angular position of first dark fringe

$θ = \frac{λ}{d} = \frac{5460 \times 10^{- 10}}{0 .1 \times 10^{- 3}} \times \frac{180}{π}$ (in degree)

= 0.313°

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In a Young’s double slit experiment, the slit separation is 0.2 cm, the distance between the screen and slit is 1m. Wavelength of the light used is 5000 Å. The distance between two consecutive dark fringes (in mm) is

0.25

0.26

0.27

0.28

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Explanation

Distance between two consecutive dark fringes $β = \frac{λ D}{d} = \frac{5000 \times 10^{- 10} \times 1}{0 .2 \times 10^{- 2}} = 0 .25$ mm.

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

3250 Å

6.5 × 10^–4 mm

1.24 microns

2.6 × 10^–4 cm

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Explanation

For first minima $θ = \frac{λ}{a}$ or $a = \frac{λ}{θ}$

$∴ a = \frac{6500 \times 10^{- 8} \times 6}{π}$ (As 30^o = $\frac{π}{6}$ radian)

$= 1 .24 \times 10^{- 4} c m = 1 .24$ microns

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The radius of central zone of the circular zone plate is 2.3 mm. The wavelength of incident light is $5893 Å .$ Source is at a distance of 6m. Then the distance of the first image will be

9 m

12 m

24 m

36 m

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Explanation

$f_{1} = \frac{r^{2}}{λ} = \frac{{(2 .3 \times 10^{- 3})}^{2}}{5893 \times 10^{- 10}} = 9 m .$

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What will be the angular width of central maxima in Fraunhoffer diffraction when light of wavelength $6000 Å$ is used and slit width is 12×10^–5 cm

2 rad

3 rad

1 rad

8 rad

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Explanation

Angular width $= \frac{2 λ}{d} = \frac{2 \times 6000 \times 10^{- 10}}{12 \times 10^{- 5} \times 10^{- 2}} = 1 r a d .$

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