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. $
BETA
