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Q.

When a plastic thin film of refractive index 1.45 is placed in the path of one of the interfering waves then the central fringe is displaced through width of 10 fringes. The thickness of the film will be, if the wavelength of light is 5850 A0 .

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a

6.5×10−4 cm

b

7.5×10−4 m

c

13×10−4 cm

d

13×10−5 cm

answer is C.

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Detailed Solution

Refractive index μ=1.45 Number of fringes n=10 Wavelength λ=5850 A0=5850×10−10 m Thickness of film t=? t=nλμ−1 =10×5850×10−101.45−1 =585×10−80.45 =13×10−6m ∴t=13×10−4cm

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