If a transparent medium of refractive index μ=1.5 and thickness t=2.5×10−5 m is inserted in front of one of the slits of Young's double slit experiment, how much will be the shift in the interference pattern? The distance between the slits is 0.5 mm and that between slits and screen is 100 cm

a) 5 cm

b) 2.5 cm

c) 0.25 cm

d) 0.1 cm

Answer: b) 2.5 cm

Solution:

Shift in the fringe pattern x = (μ – 1)t.D/d

= [(1.5 – 1) x 2.5 x 10-5 x 100 x 10-2]/0.5 x 10-3

= 2.5 cm

Was this answer helpful?

 
   

3.5 (1)

(0)
(0)

Choose An Option That Best Describes Your Problem

Thank you. Your Feedback will Help us Serve you better.

Leave a Comment

Your Mobile number and Email id will not be published. Required fields are marked *

*

*

BOOK

Free Class

Ask
Question