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 $



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