Optics Question 870
Question: To produce a minimum reflection of wavelengths near the middle of visible spectrum (550 nm), how thick should a coating of $ MgF _{2}(\mu =1.38) $ coated on a glass surface?
Options:
A) $ {{10}^{-7}} $
B) $ {{10}^{-10}} $
C) $ {{10}^{-9}}m $
D) $ {{10}^{-8}}m $
Show Answer
Answer:
Correct Answer: A
Solution:
[a] The optical path difference needed for destructive interference is $ 2\mu d=(2n+1)\frac{\lambda }{2} $ , n-0, 1, 2, ?..
Note that 2nd is the total optical path length that the rays traverse when n=0.
$ \therefore d=\frac{\lambda /2}{2\mu }=\frac{\lambda }{4\mu }=\frac{350\times {{10}^{-9}}}{4\times 1.38} $
$ =100nm=1\times {{10}^{-7}}m $
