CameraIcon
CameraIcon
SearchIcon
MyQuestionIcon
MyQuestionIcon
1
You visited us 1 times! Enjoying our articles? Unlock Full Access!
Question

The angular width of the central maximum in a single slit diffraction pattern is 60. The width of the slit is 1 μm. The slit is illuminated by monochromatic plane waves. If another slit of the same width is made near it, Young's Fringes can be observed on a screen placed at a distance 50 cm from the slits. If the observed fringe width is 1 cm. What is slit separation distance? (i.e. distance between the centre of each slit)

A
25.00
Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
B
25.0
Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
C
25
Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
Open in App
Solution



In single slit diffraction,

The angular width of central maximum is given by 2θ

Where, θangular position of first minimum

Given that, 2θ=60
θ=30

The condition or first minimum is given by,

asinθ=λ

1×106sin30=λ

λ=0.5×106m=5×107m



In YDSE,

Fringe width, β=λDd

Where d = distance between the slits

1×102=5×107×50×102d

d=25×106 m=25 μm

Why this question?Asked in JEE mains 2018.

flag
Suggest Corrections
thumbs-up
11
similar_icon
Similar questions
View More
Join BYJU'S Learning Program
similar_icon
Related Videos
thumbnail
lock
Diffraction I
PHYSICS
Watch in App
Join BYJU'S Learning Program
CrossIcon