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}} $ .



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