wiz-icon
MyQuestionIcon
MyQuestionIcon
1
You visited us 1 times! Enjoying our articles? Unlock Full Access!
Question

In Young's double slit experiment the slit separation is 0.5 m from the slits. Distance between the screen and slit is 0.454 mm For a monochromatic light of wavelengths 500 nm, the distance of 3rd maxima from 2nd maxima on the other side is

A
2.75 mm
Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
B
2.5 mm
No worries! We‘ve got your back. Try BYJU‘S free classes today!
C
22.5 mm
No worries! We‘ve got your back. Try BYJU‘S free classes today!
D
2.25 mm
No worries! We‘ve got your back. Try BYJU‘S free classes today!
Open in App
Solution

The correct option is D 2.75 mm
The distance of third maxima from the central maxima=3β=3dλD
The distance of second maxima from central maxima=2β=2dλD
Hence distance between the two maxima=5dλD
=5×0.5×500×1090.454×103m
=2.75mm

flag
Suggest Corrections
thumbs-up
1
Join BYJU'S Learning Program
similar_icon
Related Videos
thumbnail
lock
Intensity in YDSE
PHYSICS
Watch in App
Join BYJU'S Learning Program
CrossIcon