NEET Physics Question - Wave Optics - Young's Double Slit Experiment

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

Solution:

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.

Rankers Physics

YDSE Maxima Intensity

Solution:

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