YDSE Maxima Intensity – Rankers Physics

Young's Double Slit Experiment: Practice Problem & Solution

Assertion (A): In Young's double slit experiment, assuming slits to be of equal widths, intensity at interference maxima is four times the intensity due to each slit. Reason (R): Intensity is proportional to the square of amplitude.  
Both (A) & (R) are true and the (R) is the correct explanation of the (A)
Both (A) & (R) are true but the (R) is not the correct explanation of the (A)
(A) is true but (R) is false
Both (A) and (R) are false

Short Solution (Best for NEET):

To solve this problem, we apply the core principles of Young's Double Slit Experiment. Understanding the underlying formula is key to arriving at the correct answer below:

If \(I_0\) is the intensity from each slit, then the amplitude is \(A_0 \propto \sqrt{I_0}\). At maxima, amplitudes add to \(2A_0\), so intensity is \((2A_0)^2 \propto 4A_0^2 = 4I_0\). Intensity is indeed proportional to the square of amplitude, explaining this result. Both A and R are true, and R explains A.

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Similar Practice Questions

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