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Question

Physics Question on Wave optics

In the Young's double slit experiment using a monochromatic light of wavelength λ\lambda the path difference (in terms of an integer nn) corresponding to any point having half the peak intensity is

A

(2n+1)λ2(2n+1) \frac{\lambda}{2}

B

(2n+1)λ4(2n+1) \frac{\lambda}{4}

C

(2n+1)λ8(2n+1) \frac{\lambda}{8}

D

(2n+1)λ16(2n+1) \frac{\lambda}{16}

Answer

(2n+1)λ4(2n+1) \frac{\lambda}{4}

Explanation

Solution

Imax2=Imcos2(ϕ2)\frac{I_{\max }}{2}=I_{m} \cos ^{2}\left(\frac{\phi}{2}\right)
cos(ϕ2)=12\Rightarrow \cos \left(\frac{\phi}{2}\right)=\frac{1}{\sqrt{2}}
ϕ2=π4\Rightarrow \frac{\phi}{2}=\frac{\pi}{4}
ϕ=π2(2n+1)\Rightarrow \phi=\frac{\pi}{2}(2 n+1)
Δx=λ2πϕ\Rightarrow \Delta x =\frac{\lambda}{2 \pi} \phi
=λ2π×π2(2n+1)=\frac{\lambda}{2 \pi} \times \frac{\pi}{2}(2 n +1)
=λ4(2n+1)=\frac{\lambda}{4}(2 n +1)