In double slit experiment, instead of taking slits of equal widths, one slit is made twice as wide as the other. Then in interference pattern (a) the intensity of the maxima decreases and the minima has zero intensity. (b) the intensity of maxima decreases and that of the minima increases. (c) the intensity of the maxima increases and the minima has zero intensity. (d) the intensities of both the maxima and the minima increase.

Answer: (d). Equal slits ( I, I): Imax = 4I, Imin = 0. One slit twice as wide gives I and 2I: Imax = (1+√(2))² I ≈ 5.83I. Imin = (√(2)-1)² I ≈ 0.17I > 0 Both maxima and minima become brighter. So the answer is (d).

In double slit experiment, instead of taking slits of equal widths… | MHT CET 2024 (15 May, Shift 1)

Q42MHT CET 2024 · 15 MAY · SHIFT 1Medium

Superposition, Coherence & Intensity

In double slit experiment, instead of taking slits of equal widths, one slit is made twice as wide as the other. Then in interference pattern

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Answer: (d)

Equal slits ( $I, I$ ): $I_{max} = 4I$ , $I_{min} = 0$ .
One slit twice as wide gives $I$ and $2I$ : $I_{max} = (1+\sqrt{2})^2 I \approx 5.83I$ .
$I_{min} = (\sqrt{2}-1)^2 I \approx 0.17I > 0$
Both maxima and minima become brighter. So the answer is (d).

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