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JEE Main 2025 (23 Jan, Shift 2) · Superposition, Coherence & Intensity

Superposition, Coherence & Intensity question from JEE Main 2025 (23 Jan, Shift 2)

Q28JEE Main 2025 · 23 JAN · SHIFT 2Medium

Superposition, Coherence & Intensity

The width of one of the two slits in Young's double slit experiment is d while that of the other slit is $x\,d$ . If the ratio of the maximum to the minimum intensity in the interference pattern on the screen is $9 : 4$ then what is the value of $x$ ? (Assume that the field strength varies according to the slit width.)

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Answer: (d)

Field amplitude $\propto$ slit width, so $a_1 : a_2 = x : 1$ .
$\frac{x+1}{x-1} = \sqrt{\frac{9}{4}} = \frac{3}{2}$
$2x + 2 = 3x - 3 \Rightarrow x = 5$
So the answer is (d) 5.

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