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 (μ = 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
see full answer
Your Exam Success, Personally Taken Care Of
1:1 expert mentors customize learning to your strength and weaknesses – so you score higher in school , IIT JEE and NEET entrance exams.
An Intiative by Sri Chaitanya
a
0 cm
b
5 cm
c
10 cm
d
15 cm
answer is B.
(Unlock A.I Detailed Solution for FREE)
Best Courses for You
JEE
NEET
Foundation JEE
Foundation NEET
CBSE
Detailed Solution
Lateral displacement of fringes = βλ(μ−1)t=1×10−3600×10−9(1.5−1)×0.06×10−3=120m=5cm