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JEE Main 2019 (10 Apr, Shift 2) · Superposition, Coherence & Intensity

Superposition, Coherence & Intensity question from JEE Main 2019 (10 Apr, Shift 2)

Q90JEE Main 2019 · 10 APR · SHIFT 2Medium

Superposition, Coherence & Intensity

In a Young's double slit experiment, the ratio of the slit's width is 4: 1. The ratio of the intensity of maxima to minima, close to the central fringe on the screen, will be:

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Answer: (d)

Intensity $\propto$ width: $I_1 : I_2 = 4 : 1$ , amplitudes $2 : 1$ .
$\frac{I_{max}}{I_{min}} = \left(\frac{2+1}{2-1}\right)^2 = 9$
So the answer is (d) $9 : 1$ .

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