Young’s Double Slit Experiment – Fringe Intensity – Rankers Physics

Young's Double Slit Experiment: Practice Problem & Solution

Assertion (A): When a width of one of the slits of Young's double slit experiment is double that of the other than brighter fringes are nine times brighter than the dark fringes. Reason (R): The amplitude of the wave is proportional to the width of the slit.  
(1) Both (A) & (R) are true and the (R) is the correct explanation of the (A)
(2) Both (A) & (R) are true but the (R) is not the correct explanation of the (A)
(3) (A) is true but (R) is false
(4) 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 Young's double slit experiment, intensity is proportional to the square of amplitude \(I \propto A^2\). The amplitude \(A\) is proportional to the slit width \(w\). If \(w_1 = 2w_2\), then \(A_1 = 2A_2\). The ratio of maximum to minimum intensity is \(I_{max}/I_{min} = ((A_1+A_2)/(A_1-A_2))^2\). Substituting \(A_1 = 2A_2\), we get \(I_{max}/I_{min} = ((2A_2+A_2)/(2A_2-A_2))^2 = (3A_2/A_2)^2 = 3^2 = 9\). So, (A) is true. (R) is also true, as amplitude is proportional to slit width. And (R) correctly explains (A).

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