JEE Main · 2025 · 24 Jan 2025 · Shift 1 · Wave Optics

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The Young's double slit interference experiment is performed using light consisting of 480 nm and 600 nm wavelengths to form interference patterns. The least number of the bright fringes of 480 nm light that are required for the first coincidence with the bright fringes formed by 600 nm light is :-

(A)4
(B)8
(C)6
(D)5✓ Correct

Answer: (D)

Solution

The position of the

n^{t h}

bright fringe on the screen is

y_{n} = \frac{nλ D}{d}

. Two fringes coincide when their positions match:

\frac{n _{1} λ _{1} D}{d} = \frac{n _{2} λ _{2} D}{d} \Rightarrow n_{1} λ_{1} = n_{2} λ_{2}

With

λ_{1} = 480

nm and

λ_{2} = 600

nm:

n_{1} \times 480 = n_{2} \times 600 \Rightarrow \frac{n _{1}}{n _{2}} = \frac{600}{480} = \frac{5}{4}

The smallest integers in this ratio are

n_{1} = 5, n_{2} = 4

.

The first coincidence occurs at the 5th bright fringe of the 480 nm pattern.

JEE Main · 2025 · 24 Jan 2025 · Shift 1 · Single correct option

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JEE Main · 2025 · 24 Jan 2025 · Shift 1 · Wave Optics

Solution

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