A parallel beam of white light is incident normally on a water film 1.0 × 10−4 cm thick. Find the wavelengths in the visible range (400 nm − 700 nm) which are strongly transmitted by the film. Refractive index of water = 1.33.
Answers
Given,
Wavelength of light used λ=400×10⁻⁹m to 700×10⁻⁹ nm
Refractive index of water μ=1.33
The thickness of film, t=10⁻⁴ cm=10⁻⁶ m
The condition for strong transmission:
2μt=nλ,
where n is an integer.
⇒ λ=2μt/n
⇒ λ=2×1.33×10⁻⁶/n =2660×10⁻⁹/n m
Putting n = 4, we get, λ1 = 665 nm.
Putting n = 5, we get, λ2 = 532 nm.
Putting n = 6, we get, λ3 = 443 nm.
Therefore, the wavelength (in visible region) which are strongly transmitted by the film are 665 nm, 532nm and 443 nm.
Answer:
For Maximum Transmission
Optical path should be integral multiple of λ
μ2d=nλ (Constructive Interference)
where μ refractive index of film
λ wavelength of light
d thickness of film
λ=
n
2μd
λ=
n
2×1.33×1.0×10
−4
cm
λ=
n
2.66×10
−4
cm
when
n=4,λ
1
=665nm
n=5,λ
2
=532nm
n=6,λ
3
=443nm