In a double slit experiment, at a certain point on the screen the path difference between the two interfering waves is 1/8 th of a wavelength. The ratio of the intensity of light at that point to that at the centre of a bright fringe is: (a) 0.853 (b) 0.568 (c) 0.672 (d) 0.760

Answer: (a). Δ x = λ/8 ⇒ φ = π/4. IImax = cos ²π/8 = (1 + cos 45°)/(2) = (1 + 0.707)/(2) ≈ 0.853 So the answer is (a) 0.853.

In a double slit experiment, at a certain point on the screen the… | JEE Main 2020 (8 Jan, Shift 2)

Q85JEE Main 2020 · 8 JAN · SHIFT 2Medium

Superposition, Coherence & Intensity

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

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Answer: (a)

$\Delta x = \lambda/8 \Rightarrow \phi = \pi/4$ .
$\dfrac{I}{I_{max}} = \cos^2\dfrac{\pi}{8} = \dfrac{1 + \cos 45^\circ}{2} = \dfrac{1 + 0.707}{2} \approx 0.853$
So the answer is (a) $0.853$ .

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Superposition, Coherence & Intensity question from JEE Main 2020 (8 Jan, Shift 2)

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