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Question

A mica strip and a polystyrene strip are fitted on the two slits of a double slit apparatus. The thickness of the strips is 0.50 mm and the separation between the slits is 0.12 cm. The refractive index of mica and polystyrene are 1.58 and 1.55 respectively for the light of wavelength 590 nm which is used in the experiment. The interference is observed on a screen a distance one meter away. (a) What would be the fringe-width? (b) At what distance from the centre will the first maximum be located?

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Solution


Given: t1=t2=0.5mm=0.5×103m
μm=1.58
μp=1.55
λ=590nm=590×109m
d=0.12m=12×104m
D=1m
(a) Fringe width = Dλd=1×590×10912×104=4.91×104m

(b) When both strips are fitted, the optical path changes by
Δx=(μm1)t1(μp1)t2=(μmμp)t(1.581.55)×0.5×103=0.015×103m

So, Number of fringes shift =0.015×103590××103=25.43

There are 25 fringes and 0.43th of a fringe.
There are 13 bright fringes and 12 dark fringes and 0.43th of a dark fringe. So, position of the first maximum on both sides will be given by
x=0.43×4.91×104=0.021cm
x=(10.43)×4.91×104=0.028cm (since, fringe width=4.91×104m)

849669_833475_ans_7d13b437c33043179ebf4b9b8367cb2f.png

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