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Question

Consider a plane inclined at an angle 45 with the horizontal has two slits (S1andS2 separated by a distance d = 2 mm The screen is placed at a distance of D = 10 m A parallel monochromatic light beam of wavelength 5000 A is incident on the slits as shown If the fringe width of interference pattern on the screen is k×103 meter then find the value of k
332867_718a9a162a8d400094d675f0292ea594.png

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Solution

Path difference
=S1NMS2=dsin(45°+θ)d2=d[cosθ2+sinθ2]d2, as θ is small
θ=2nλdy=2nλDd
Fringe width=yn+1yn=2Dλd=5×103m
yn+1−yn=2Dλd=5×10−3mHence k=5.

848115_332867_ans_92c8abd3d6d44f678c1928ec914f858c.png

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