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 0.5mm and the screen is 0.5m away from the slit. For a monochromatic light of wavelength 500nm, the distance of 3rd maxima from the 2nd minima on the other side of central maxima is

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

The correct option is B 2.25mm
Distance between third maxima and second minima on the other side,
x=3λDd+(2×21)λD2d=9λD2d
x=9×500×109×0.52×0.5×103=2250×106=2.25mm

flag
Suggest Corrections
thumbs-up
2
Join BYJU'S Learning Program
similar_icon
Related Videos
thumbnail
lock
Superposition Recap
PHYSICS
Watch in App
Join BYJU'S Learning Program
CrossIcon