Q10 / 107 · NTA Abhyas (JEE Main) 2020 · Physics · Wave Optics

In a Young's double slit experiment using monochromatic light of wavelengths λ, the intensity of light at a point on the screen with a path difference λ is M units and the intensity of light at a point where the path difference is λ3 is nM units. What is the value of n?

Correct answer: 0.25

Solution

Resultant intensity IR=I1+I2+2I1I2cosθ
If path difference, =λ then phase difference =2π ….∵ϕ=2πλ×Δx
∴ IR=I+I+2I×Icos2π …..(∵I1=I2=I)
=4I=M
If path difference =λ3
then phase difference =2πλ×λ3=2π3rad
IR'=I+I+2I×Icos2π3
=2I+2I-12=I=M4=0.25M

Hence, the correct answer is 0.25.

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