Optics Question 847
Question: Interference fringes were produced in Young’s double slit experiment using light of wave length $ 5000\overset{o}{\mathop{A}} $ . When a film of material $ 2.5\times {{10}^{-3}}cm $ thick was placed over one of the slits, the fringe pattern shifted by a distance equal to 20 fringe width. The refractive index of the material of the film is
Options:
A) 1.25
B) 1.33
C) 1.4
D) 1.513
Show Answer
Answer:
Correct Answer: C
Solution:
[c] $ n=\frac{(\mu -1)tD}{d} $ but $ \beta =\frac{\lambda D}{d}\Rightarrow \frac{D}{d}=\frac{\beta }{\lambda } $
$ n=(\mu -1)t\beta /\lambda $
$ 20\beta =(\mu -1)2.5\times {{10}^{-3}}(\beta /5000\times {{10}^{-8}}) $
$ \mu -1=\frac{20\times 5000\times {{10}^{-8}}}{2.5\times {{10}^{-3}}}\Rightarrow \mu =1.4 $
BETA
