Optics Question 866
Question: A wedged shaped air film having an angle of 40 second is illuminated by a monochromatic light and the fringes are observed vertically down through a microscope. The fringe separation between two consecutive bright fringes is 0.12 cm. The wavelength of light is:
Options:
A) $ ~5545\overset{o}{\mathop{A}} $
B) $ 6025\overset{o}{\mathop{A}} $
C) $ 4925\overset{o}{\mathop{A}} $
D) $ ~4655\overset{o}{\mathop{A}} $
Show Answer
Answer:
Correct Answer: D
Solution:
[d] $ \beta =\frac{\lambda }{2\mu \tan \alpha }\simeq \frac{\lambda }{2\mu \alpha } $
$ \therefore \lambda =\beta \times 2\mu \alpha $
$ =0.12\times {{10}^{-2}}\times 2\times 1\times ( \frac{40}{60\times 60}\times \frac{\pi }{180} ) $
$ =4655\overset{o}{\mathop{A}} $ .
BETA
