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Wave Optics MCQs for NEET — Physics Questions with Answers

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In a biprism experiment, by using light of wavelength 5000 Å, 5 mm wide fringes are obtained on a screen 1.0 m away from the coherent sources. The separation between the two coherent sources is

1.0 mm

0.1 mm

0.05 mm

0.01 mm

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Explanation

$d = \frac{D λ}{β} = \frac{1 \times 5 \times 10^{- 7}}{5 \times 10^{- 3}} = 10^{- 4} m = 0 .1 m m$ .

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The slits in a Young's double-slit experiment have equal widths and the source is placed symmetrically relative to the slits. The intensity at the central fringes is I₀. If one of the slits is closed, the intensity at this point will be

I₀

I₀/ 4

I₀/ 2

4I₀

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Explanation

If intensity of each wave is I, then initially at central position I₀ = 4I. when one of the slit is covered then intensity at central position will be I only i.e., $\frac{I_{0}}{4}$ .

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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 5000 Å. The central bright maximum will shift

2 fringes upward

2 fringes downward

10 fringes upward

None of these

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Explanation

Shift $= \frac{β}{λ} (μ - 1) t = \frac{β}{(5000 \times 10^{- 10})} (1 .5) \times 2 \times 10^{- 6} = 2 β$

i.e., 2 fringes upwards.

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

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Explanation

$n_{1} λ_{1} = n_{2} λ_{2}$

$\Rightarrow 62 \times 5893 = n_{2} \times 4358$

$\Rightarrow n_{2} = 84 .$

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In Young's double slit experiment, angular width of fringes is 0.20^o for sodium light of wavelength 5890 Å. If complete system is dipped in water, then angular width of fringes becomes

0.11^o

0.15^o

0.22^o

0.30^o

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Explanation

Angular fringe width $θ = \frac{λ}{d} \Rightarrow θ \propto λ$

$λ_{w} = \frac{λ_{a}}{μ_{w}}$

So $θ_{w} = \frac{θ_{air}}{μ_{w}}$ $= \frac{0 .20}{\frac{4}{3}} = 0 .15 °$

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In Young's double slit experiment, the distance between the two slits is 0.1 mm and the wavelength of light used is 4×10^–7 m. If the width of the fringe on the screen is 4 mm, the distance between screen and slit is

0.1 mm

1 cm

0.1 cm

1 m

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Explanation

$β = \frac{λ D}{d} \Rightarrow (4 \times 10^{- 3}) = \frac{4 \times 10^{- 7} \times D}{0 .1 \times 10^{- 3}} \Rightarrow D = 1 m$

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In Young's double slit experiment using sodium light (λ = 5898 Å), 92 fringes are seen. If given colour (λ = 5461 Å) is used, how many fringes will be seen

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Explanation

$(n_{1} λ_{1} = n_{2} λ_{2})$

$\frac{n_{1}}{n_{2}} = \frac{λ_{2}}{λ_{1}} \Rightarrow \frac{n_{1}}{92} = \frac{5898}{5461} \Rightarrow n_{1} = 99$

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In Young’s experiment, the distance between slits is 0.28 mm and distance between slits and screen is 1.4 m. Distance between central bright fringe and third bright fringe is 0.9 cm. What is the wavelength of used light

5000 Å

6000 Å

7000 Å

9000 Å

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Explanation

Position of 3^rd bright fringe $x_{3} = \frac{3 D λ}{d}$

$\Rightarrow λ = \frac{x_{3} d}{3 D} = \frac{(0 .9 \times 10^{- 2}) \times (0 .28 \times 10^{- 3})}{3 \times 1 .4} = 6000 Å$

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If a transparent medium of refractive index μ = 1.5 and thickness t = 2.5 × 10^–5 m is inserted in front of one of the slits of Young’s Double Slit experiment, how much will be the shift in the interference pattern? The distance between the slits is 0.5 mm and that between slits and screen is 100 cm

5 cm

2.5 cm

0.25 cm

0.1 cm

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Explanation

Shift in the fringe pattern $x = \frac{(μ - 1) t . D}{d}$

$= \frac{(1 .5 - 1) \times 2 .5 \times 10^{- 5} \times 100 \times 10^{- 2}}{0 .5 \times 10^{- 3}} = 2 .5 c m .$

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Two slits, 4 mm apart, are illuminated by light of wavelength 6000 Å. What will be the fringe width on a screen placed 2m from the slits

0.12 mm

0.3 mm

3.0 mm

4.0 mm

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Explanation

$β = \frac{λ D}{d} = \frac{6000 \times 10^{- 10} \times 2}{4 \times 10^{- 3}} = 3 \times 10^{4} m = 0 .3 m m$

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