JEE Main · 2023 · 06 Apr 2023 · Shift 2 · Wave Optics

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A beam of light consisting of two wavelengths 7000 Å and 5500 Å is used to obtain interference pattern in Young's double slit experiment. The distance between the slits is 2.5 mm and the distance between the plane of slits and the screen is 150 cm. The least distance from the central fringe, where the bright fringes due to both the wavelengths coincide, is

n \times 1 0^{- 5}

m. The value of n is ____.

Answer: 462

Solution

The distance of a bright fringe from the centre is

y = n \frac{λ D}{d}

. Bright fringes of the two wavelengths coincide when

n_{1} λ_{1} = n_{2} λ_{2}

, that is

\frac{n _{1}}{n _{2}} = \frac{λ _{2}}{λ _{1}} = \frac{5500}{7000} = \frac{11}{14}

. The least distance corresponds to the smallest integers,

n_{1} = 11

for

7000

Å and

n_{2} = 14

for

5500

Å. Then

y = 11 \times \frac{7000 \times 1 0 ^{- 10} \times 1.5}{2.5 \times 1 0 ^{- 3}} = 462 \times 1 0^{- 5}

m. The value of

n

462

JEE Main · 2023 · 06 Apr 2023 · Shift 2 · Integer answer

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JEE Main · 2023 · 06 Apr 2023 · Shift 2 · Wave Optics

Solution

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