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Question

In a Young's double slit experiment, the slits are placed 0.320 mm apart. Light of wavelength λ=500nm is incident on the slits. The total number of bright fringes that are observed in the angular range 30oθ30o is :

A
320
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B
641
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C
321
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D
640
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Solution

The correct option is B 641
Pam difference
dsinθ=nλ
where d = seperation of slits
λ = wave length
n = no. of maximas
0.32×103sin30=n×500×109
n=320
Hence total no. of maximas observed in angular range 30oθ30o is
maximas = 320 + 1 + 320 = 641


1142123_1329045_ans_db47bed1b8214e24a56eca2c9ab14220.png

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