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

In a double slit experiment, the separation between the slits is d and distance of the screen from slits is D. If the wavelength of light used is λ and I is the intensity of central bright fringe, then intensity at distance x from central maximum is

A
Icos2(π2xdλD)
No worries! We‘ve got your back. Try BYJU‘S free classes today!
B
I2sin2(πxd2λD)
No worries! We‘ve got your back. Try BYJU‘S free classes today!
C
Icos2(πxdλD)
Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
D
Isin2(πxdλD)
No worries! We‘ve got your back. Try BYJU‘S free classes today!
Open in App
Solution

The correct option is C Icos2(πxdλD)
For a path difference Δ, the phase difference (in radians) is
δ=2πλΔ
We know,
Path difference, Δ=xndD
where xn is the distance from the central maxima
d is the separation between the slits
D is the distance of the screen from slits
Here,
Given xn=x
Path difference, Δ=xdD

δ=2πλxdD=2πxdDλ

Intensity due to interference is given by
I=I1+I2+2I1I2Cosδ

At central maximum, δ=0,
Given, Icenter=I
Let the intensity of light from each slit be I1=I2=Is
I=Is+Is+2IsIsCos0
I=4Is

Now, intensity at a distance x is given by
Ix=Is+Is+2IsIsCosδ
Ix=2Is+2IsCosδ
Ix=I2+I2Cos(2πxdDλ)
Ix=I2(1+Cos(2πxdDλ))
Using
Cos2θ=2Cos2θ1
1+Cos2θ2=Cos2θ
We have,
Ix=ICos2(πxdDλ)

Hence, the correct answer is OPTION C.

flag
Suggest Corrections
thumbs-up
0
Join BYJU'S Learning Program
similar_icon
Related Videos
thumbnail
lock
Intensity in YDSE
PHYSICS
Watch in App
Join BYJU'S Learning Program
CrossIcon