In a Young's double slit experiment, the path difference, at a certain point on the screen, between two interfering waves is $1 / 8^{\text {th }}$ of wavelength. The ratio of the intensity at this point to that at the centre of a bright fringe is close to:

JEE Main · 2019 · 11 Jan 2019 · Shift 1 · Wave Optics

In a Young's double slit experiment, the path difference, at a certain point on the screen, between two interfering waves is

1/ 8^{th}

of wavelength. The ratio of the intensity at this point to that at the centre of a bright fringe is close to:

(A)0.74
(B)0.85✓ Correct
(C)0.94
(D)0.80

Answer: (B)

Solution

A path difference of

λ /8

corresponds to a phase difference

Δ ϕ = \frac{2 π}{λ} \cdot \frac{λ}{8} = \frac{π}{4}

. The intensity at phase difference

Δ ϕ

I = I_{m a x} cos^{2} \frac{Δ ϕ}{2}

, where

I_{m a x}

is the central bright-fringe value. So

\frac{I}{I _{m a x}} = cos^{2} \frac{π}{8} = \frac{1 + cos ( π /4 )}{2} = \frac{1 + 0.707}{2} \approx 0.85

JEE Main · 2019 · 11 Jan 2019 · Shift 1 · Single correct option

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