JEE Main · 2023 · 31 Jan 2023 · Shift 2 · Wave Optics

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Two light waves of wavelengths 800 and 600 nm are used in Young's double slit experiment to obtain interference fringes on a screen placed 7 m away from plane of slits. If the two slits are separated by 0.35 mm, then shortest distance from the central bright maximum to the point where the bright fringes of the two wavelength coincide will be ______ mm.

Answer: 48

Solution

Bright fringes of the two colours fall on the same point when

n_{1} λ_{1} = n_{2} λ_{2}

, that is

\frac{n _{1}}{n _{2}} = \frac{λ _{2}}{λ _{1}} = \frac{600}{800} = \frac{3}{4}

. The nearest such point to the centre uses the smallest whole numbers,

n_{1} = 3

and

n_{2} = 4

. Its distance from the central maximum is

y = \frac{n _{1} λ _{1} D}{d} = \frac{3 \times ( 800 \times 1 0 ^{- 9} ) \times 7}{0.35 \times 1 0 ^{- 3}} = 0.048 m

, which is

48 mm

JEE Main · 2023 · 31 Jan 2023 · Shift 2 · Integer answer

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JEE Main · 2023 · 31 Jan 2023 · Shift 2 · Wave Optics

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