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 $
