Optics Question 856

Question: When a plastic thin film of refractive index 1.45 is placed in the path of one of the interfering waves then the central fringe is displaced through width of five fringes. The thickness of the film, if the wavelength of light is $ 5890\overset{o}{\mathop{A}} $ , will be

Options:

A) $ 6.544\times {{10}^{-4}}cm $

B) $ 6.544\times {{10}^{-4}}m $

C) $ 6.54\times {{10}^{-4}}cm $

D) $ 6.5\times {{10}^{-4}}cm $

Show Answer

Answer:

Correct Answer: A

Solution:

[a]

$ \therefore X _{0}=\frac{\beta }{\lambda }(\mu -1)t\Rightarrow 5\beta =\frac{\beta (0.45)t}{5890\times {{10}^{-10}}} $

$ \therefore t=\frac{5\times 5890\times {{10}^{-10}}}{0.45}=6.544\times {{10}^{-4}}cm $



NCERT Chapter Video Solution

Dual Pane