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Question

Physics Question on Wave optics

A glass plate of refractive index 1.51.5 is coated with a thin layer of thickness t and refractive index 1.81.8. Light of wavelength 648nm648 \,nm travelling in air is incident normally on the layer. It is partly reflected at upper and lower surfaces of the layer and the two reflected rays interfere. The least value of tt for which the rays interfere constructively is

A

30 nm

B

60 nm

C

90 nm

D

120nm

Answer

90 nm

Explanation

Solution

Given μ=\mu= refractive index of glass =1.5=1.5
t=t = thickness of coating,
μ=\mu= refractive index of coating =1.8λ=1.8\, \lambda
=648×109m=648 \times 10^{-9} m
For constructive interference 2μt=(2n+1)λ22 \,\mu t=(2 n+1) \frac{\lambda}{2}
2×1.8×t=648×10922 \times 1.8 \times t=\frac{648 \times 10^{-9}}{2}
(tt to be minimum n=0n =0 )
t=6482×2×1.8×109mt=\frac{648}{2 \times 2 \times 1.8} \times 10^{-9} m
t=6487.2×109m=90mt=\frac{648}{7.2} \times 10^{-9} m=90 \,m