Interference fringes were produced in Young's double slit experiment using light of wavelength 5000 A. When a film of material 2.5 x 1O-3 cm thick was placed over one of the slits, the fringe pattern shifted by a distance equal to 20 fringe widths. The reactive index of the material of the film is
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a
1.25
b
1.33
c
1.4
d
1.5
answer is C.
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Detailed Solution
S=(μ−1)t(D/2d) 2d is taken as distance between the slits. But β=λD2d or D2d=βλ ∴ S=(μ−1)t(β/λ) or 20β=(μ−1)2⋅5×10−3β/5000×10−8 or (μ−1)=20×5000×10−82.5×10−3=0.4 or μ=1⋅4