Blue light of wavelength 480 nanometers is most strongly reflected off a thin film of oil on a glass slide when viewed near normal incidence. Assuming that the index of refraction of the oil is 1.2 and that of the glass is 1.6, what is the minimum thickness of the oil film (other than zero)?
B. 200 nm
C. 300 nm
D. 400 nm
E. 480 nm
(GR0177 #69)
Solution:
n1 = 1
n2 = 1.2
n3 = 1.6
n1
n2 → Δa = λ/2n2
n3 → Δb = 2t + λ/2The Relative Shift:
Δ = Δb − Δa = 2t + λ/2 − λ/2 = 2t
For Constructive Interference: Δ = mλ
For Destructive Interference: Δ = (m + ½)λ
with m = 0, 1, 2, 3, ...
Blue Light → Constructive interference, Δ = mλ
Δ = mλ = 2t
t = mλ/2
λ = 480 nm
m = 1 for minimum thickness other than 0,
t = 480/2 = 240 nm
Answer: B
Notes: Click HERE for more info on Thin Film.
1 comment :
Since we were in oil, don't forget to use the new wavelength given by wavelength in air divided by the index of refraction of oil. This gives 480/1.2 = 400. Then you can do 400/2 = 200. Still gives B.
Post a Comment