A glass surface is coated by an oil film of uniform thickness 1.00 × 10−4 cm. The index of refraction of the oil is 1.25 and that of the glass is 1.50. Find the wavelengths of light in the visible region (400 nm − 750 nm) which are completely transmitted by the oil film under normal incidence
Answers
Given:
Wavelength of light used λ=400×10⁻⁹ to 750×10⁻⁹ m
Refractive index of oil, μoil, is 1.25 and that of glass, μg, is 1.50.
The thickness of the oil film,
d=1×10⁻⁴cm=10⁻⁶m,
The condition for the wavelengths which can be completely transmitted through the oil film is given by
λ=2μd/(n+1/2) =2×10⁻⁶×1.25×2/2n+1
=5×10⁻⁶/ 2n+1 m
⇒ λ=5000/2n+1 nm
Where n is an integer.
For wavelength to be in visible region i.e (400 nm to 750 nm)
When n = 3, we get,
λ=5000/2×3+1 =5000/7=714.3 nm
When, n = 4, we get,
λ=5000/2×4+1 =5000/9=555.6 nm
When, n = 5, we get,
λ=5000/2×5+1 =5000/11=454.5 nm
Thus the wavelengths of light in the visible region (400 nm − 750 nm) which are completely transmitted by the oil film under normal incidence are 714 nm, 556 nm, 455 nm.
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