Question
Physics Question on Wave optics
A glass plate of refractive index 1.5 is coated with a thin layer of thickness t and refractive index 1.8. Light of wavelength 648nm 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 t 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= refractive index of glass =1.5
t= thickness of coating,
μ= refractive index of coating =1.8λ
=648×10−9m
For constructive interference 2μt=(2n+1)2λ
2×1.8×t=2648×10−9
(t to be minimum n=0 )
t=2×2×1.8648×10−9m
t=7.2648×10−9m=90m