Optics Question 1157
Question: In Young’s double slit experiment, the aperture screen distance is 2m. The fringe width is 1 mm. Light of 600 nm is used. If a thin plate of glass (m = 1.5) of thickness 0.06 mm is placed over one of the slits, then there will be a lateral displacement of the fringes by
[BCECE 2005]
Options:
A) 0 cm
B) 5 cm
C) 10 cm
D) 15 cm
Show Answer
Answer:
Correct Answer: B
Solution:
Lateral displacement of fringes = $ \frac{\beta }{\lambda }(\mu -1)t $
$ =\frac{1\times {{10}^{-3}}}{600\times {{10}^{-9}}}(1.5-1)\times 0.06\times {{10}^{-3}} $
$ =\frac{1}{20}m $
$ =5cm. $
