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Question

A narrow slit S transmitting light of wavelength λ is placed a distance d above a large plane mirror as shown in fig 29.16 (a). The light coming directly from the slit and that after reflection interference coverage at P on the screen placed at a distance D from the slit. What will be the intensity at a point just above O? What will be x for which first maxima occurs?
155737_605e7581d1b642469c850745398d1ada.png

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Solution

a) Since there is a phase difference of π between direct light and reflecting light, the intensity just above the mirror will be zero.

b) Here, 2d= equivalent slit separation
D= Distance between slit and screen.

We know for bright fringe,
Δx=y×2dD=nλ
But as there is a phase reversal of λ/2

y×2dD+λ2=nλ

y×2dD=nλλ2
y=λD4d

282600_155737_ans_e48ec749665640bc9a4a7006dc2e5194.png

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