Optics Question 157

Question: A flake of glass (refractive index 1.5) is placed over one of the openings of a double slit apparatus. The interference pattern displaces itself through seven successive maxima towards the side where the flake is placed. if wavelength of the diffracted light is $ \lambda =600nm $ , then the thickness of the flake is

Options:

A) 2100 nm

B) 4200 nm

C) 8400 nm

D) None of these

Show Answer

Answer:

Correct Answer: C

Solution:

Shift $ =\frac{\beta }{\lambda }(\mu -1)t $

$ \Rightarrow 7\beta =\frac{\beta }{\lambda }(\mu -1)t\Rightarrow t=\frac{7\lambda }{(\mu -1)} $

$ =\frac{7\times 600}{(1.5-1)}=8400nm. $



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