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In the Young's double-slit experiment, the intensity of light at a point on the screen where the path difference is $\lambda$ is $\mathrm{K},(\lambda$ being the wave length of light used). The intensity at a point where the path difference is $\lambda / 4,$ will be :
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Verified Answer
The correct answer is:
$\mathrm{K} / 2$
For path difference $\lambda,$ phase difference $=2 \pi \mathrm{rad}$
For path difference $\frac{\lambda}{4},$ phase difference
$$
=\frac{\pi}{2} \mathrm{rad}
$$
As $\mathrm{K}=4 \mathrm{I}_{0}$ so intensity at given point where
path difference is $\frac{\lambda}{4}$
$\mathrm{K}^{\prime}=4 \mathrm{I}_{0} \cos ^{2}\left(\frac{\pi}{4}\right)\left(\cos \frac{\pi}{4}=\cos 45^{\circ}\right)$
$=2 \mathrm{I}_{0}=\frac{\mathrm{K}}{2}$
For path difference $\frac{\lambda}{4},$ phase difference
$$
=\frac{\pi}{2} \mathrm{rad}
$$
As $\mathrm{K}=4 \mathrm{I}_{0}$ so intensity at given point where
path difference is $\frac{\lambda}{4}$
$\mathrm{K}^{\prime}=4 \mathrm{I}_{0} \cos ^{2}\left(\frac{\pi}{4}\right)\left(\cos \frac{\pi}{4}=\cos 45^{\circ}\right)$
$=2 \mathrm{I}_{0}=\frac{\mathrm{K}}{2}$
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