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PhysicsWave OpticsYDSEMedium2 minPYQ_2024
PhysicsMediumsingle choice

In Young's double slit experiment, light from two identical sources are superimposing on a screen. The path difference between the two lights reaching at a point on the screen is7λ4. The ratio of intensity of fringe at this point with respect to the maximum intensity of the fringe is:

Options:

Answer:
A
Solution:

Given the path difference is x=7λ4

Hence, the phase difference can be found as follows:

ϕ=2πλx=2πλ×7λ4=7π2

The formula to calculate the intensity is given by

I=Imaxcos2ϕ2   ...1

From equation (1), it follows that

IImax=cos2ϕ2=cos27π2×2=cos27π4=cos22π-π4=cos2π4=12

Stream:JEESubject:PhysicsTopic:Wave OpticsSubtopic:YDSE
2mℹ️ Source: PYQ_2024

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