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Home / Doubts and Solutions/NEET/Physics

The maximum intensity og fringes in YDSE is I.If one slit is closed, the.n the intensity at that place is I'.Then

a) I=I'

b)I=2I'

c)I=4I'

d) no relation

Asked by aayush 23rd March 2018, 8:53 PM
Answered by Expert
Answer:
Intensity is proportional to square of amplitude of wave. If amplitude of wave of each source is a, then after superposition intensity is prportional to (2a)2 i.e., 4a2 .   If one source is closed, there is no superposition, hence intensity is proportional to a2. Hence I = 4 I' is the correct answer
Answered by Expert 28th March 2018, 7:15 AM
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