YDSE Intensity and Energy Conservation – Rankers Physics

Young's Double Slit Experiment: Practice Problem & Solution

Assertion (A): In Young's double slit experiment if intensity of each source is \(I_0\) then minimum and maximum intensity is zero and \(4I_0\) respectively. Reason (R): In Young's double slit experiment energy conservation is not followed.  
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:

In YDSE with coherent sources of intensity (I_0) each, \(I_{\text{min}} = (\sqrt{I_0} - \sqrt{I_0})^2 = 0\) and \(I_{\text{max}} = (\sqrt{I_0} + \sqrt{I_0})^2 = 4I_0\). Thus, A is true. Energy is conserved in interference; it's redistributed, not destroyed. Thus, R is false.

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