JEE Main · 2021 · 31 Aug 2021 · Shift 2 · Wave Optics

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In a Young's double slit experiment, the slits are separated by 0.3 mm and the screen is 1.5 m away from the plane of slits. Distance between fourth bright fringes on both sides of central bright is 2.4 cm. The frequency of light used is ______

\times 1 0^{14} Hz

Answer: 5

Solution

The distance between the

n

-th bright fringe on one side and the

n

-th on the other side is

2 n β

, where

β = \frac{λ D}{d}

is the fringe width. For

n = 4

, the distance is

8 β = 2.4

= 0.024

m.

Thus,

β = \frac{0.024}{8} = 0.003

= 3

mm.

From

β = \frac{λ D}{d}

, we get

λ = \frac{β d}{D} = \frac{0.003 \times 0.3 \times 1 0 ^{- 3}}{1.5} = \frac{0.9 \times 1 0 ^{- 6}}{1.5} = 0.6 \times 1 0^{- 6}

= 600

nm.

Frequency

f = \frac{c}{λ} = \frac{3 \times 1 0 ^{8}}{600 \times 1 0 ^{- 9}} = 5 \times 1 0^{14}

Hz.

JEE Main · 2021 · 31 Aug 2021 · Shift 2 · Integer answer

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