JEE Main · 2023 · 25 Jan 2023 · Shift 1 · Wave Optics

In Young's double slits experiment, the position of

5^{th}

bright fringe from the central maximum is 5 cm. The distance between slits and screen is 1 m and wavelength of used monochromatic light is 600 nm. The separation between the slits is:

(A) $60 μ m$ ✓ Correct
(B) $48 μ m$
(C) $12 μ m$
(D) $36 μ m$

Answer: (A)

Solution

The

n

-th bright fringe lies at

y_{n} = \frac{nλ D}{d}

, so

d = \frac{nλ D}{y _{n}}

. With

n = 5

λ = 600 \times 1 0^{- 9} m

D = 1 m

and

y_{5} = 5 \times 1 0^{- 2} m

d = \frac{5 \times 600 \times 1 0 ^{- 9} \times 1}{5 \times 1 0 ^{- 2}} = 60 \times 1 0^{- 6} m

. The separation between the slits is

60 μ m

JEE Main · 2023 · 25 Jan 2023 · Shift 1 · Single correct option

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