In Young's double slit experiment, the intensity at a point where path difference is (λ)/(6) ( λ being the wavelength of light used) is ' I '. If ' I₀ ' denotes the maximum intensity, then I/I₀ is equal to (cos 0° = 1, cos 60° = 1/2) (a) √(3)/2 (b) 4/3 (c) 1/√(2) (d) 1/2

Answer: (b). Δ x = λ/6 ⇒ φ = π/3. I/I₀ = cos ²π/6 = (1 + cos 60°)/(2) = 3/4 3/4 is not among the options; option (b) 4/3 is the inverse ratio I₀/I. So I/I₀ = 3/4 (the key marks (b) 4/3).

In Young's double slit experiment, the intensity at a point where… | MHT CET 2021 (23 Sep, Shift 2)

Q81MHT CET 2021 · 23 SEP · SHIFT 2Medium

Superposition, Coherence & Intensity

In Young's double slit experiment, the intensity at a point where path difference is $\frac{\lambda}{6}$ ( $\lambda$ being the wavelength of light used) is ' $I$ '. If ' $I_0$ ' denotes the maximum intensity, then $\frac{I}{I_0}$ is equal to $\left(\cos 0^\circ = 1, \cos 60^\circ = \frac{1}{2}\right)$

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Answer: (b)

$\Delta x = \lambda/6 \Rightarrow \phi = \pi/3$ .
$\dfrac{I}{I_0} = \cos^2\dfrac{\pi}{6} = \dfrac{1 + \cos 60^\circ}{2} = \dfrac{3}{4}$
$\dfrac{3}{4}$ is not among the options; option (b) $\dfrac{4}{3}$ is the inverse ratio $\dfrac{I_0}{I}$ .
So $\dfrac{I}{I_0} = \dfrac{3}{4}$ (the key marks (b) $\dfrac{4}{3}$ ).

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