JEE Main · 2023 · 08 Apr 2023 · Shift 2 · Wave Optics

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The width of fringe is 2 mm on the screen in a double slits experiment for the light of wavelength of 400 nm . The width of the fringe for the light of wavelength 600 nm will be:

(A)4 mm
(B)1.33 mm
(C)3 mm✓ Correct
(D)2 mm

Answer: (C)

Solution

In Young's double-slit experiment the fringe width is

β = \frac{D λ}{d}

, so with the screen distance

D

and slit separation

d

unchanged,

β \propto λ

. Hence

\frac{β _{2}}{β _{1}} = \frac{λ _{2}}{λ _{1}} = \frac{600}{400} = \frac{3}{2}

. Taking

β_{1} = 2 mm

β_{2} = \frac{3}{2} \times 2

. The fringe width for the

600 nm

light is

3 mm

JEE Main · 2023 · 08 Apr 2023 · Shift 2 · Single correct option

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