The ratio of intensities of two points on a screen in Young's double… | MHT CET 2023 (12 May, Shift 2)

Q60MHT CET 2023 · 12 MAY · SHIFT 2Medium

Superposition, Coherence & Intensity

The ratio of intensities of two points on a screen in Young's double slit experiment when waves from the two slits have a path difference of $\frac{\lambda}{4}$ and $\frac{\lambda}{6}$ is

$(\cos 90^\circ = 0, \cos 60^\circ = 0.5)$

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Answer: (b)

$I = I_{max}\cos^2\dfrac{\phi}{2}$ , $\phi = \dfrac{2\pi}{\lambda}\Delta x$ .
$\lambda/4$ : $\phi = \pi/2$ , $I_1 = \dfrac{I_{max}}{2}$ (using $\cos^2\theta = \frac{1+\cos 2\theta}{2}$ with $\cos 90^\circ = 0$ ).
$\lambda/6$ : $\phi = \pi/3$ , $I_2 = I_{max}\dfrac{1 + \cos 60^\circ}{2} = \dfrac{3I_{max}}{4}$ .
$I_1 : I_2 = \dfrac{1}{2} : \dfrac{3}{4} = 2 : 3$ . So the answer is (b) $2 : 3$ .

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