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) 0.94 (b) 0.85 (c) 0.74 (d) 0.80

Answer: (b). Δ x = λ/8 ⇒ φ = π/4. IImax = cos ²π/8 = (1 + cos 45°)/(2) ≈ 0.85 So the answer is (b) 0.85.

In a Young's double slit experiment, the path difference, at a… | JEE Main 2019 (11 Jan, Shift 1)

Q92JEE Main 2019 · 11 JAN · SHIFT 1Medium

Superposition, Coherence & Intensity

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

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Answer: (b)

$\Delta x = \lambda/8 \Rightarrow \phi = \pi/4$ .
$\dfrac{I}{I_{max}} = \cos^2\dfrac{\pi}{8} = \dfrac{1 + \cos 45^\circ}{2} \approx 0.85$
So the answer is (b) $0.85$ .

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