Physics - Wave Optics Question with Solution | TestHub

PhysicsWave OpticsYDSEHard2 minPYQ_2016
PhysicsHardsingle choice

The intensity at the maximum in a Young's double slit experiment isI0. Distance between two slits isd=5λ, whereλis the wavelength of light used in the experiment. What will be the intensity in front of one of the slits on the screen placed at a distanceD=10d ?NEET 2016 (Phase 1)

Options:

Answer:
D
Solution:

In YDSEImax=I0

Path difference at a point in front of one of shifts is

∆x=dyD=dd2D=d22D ( Here  y= d 2 )

∆x=d2210d=d20=5λ20=λ4

Path difference is

ϕ=2πλΔx=2πλλ4

ϕ=π2

So intensity at that point is

I=Imaxcos2θ2

I=I0cos2π4=I02

Stream:NEETSubject:PhysicsTopic:Wave OpticsSubtopic:YDSE
⏱ 2mℹ️ Source: PYQ_2016

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