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In young's double-slit experiment, the intensity of light at a point on the screen where the path difference is $\lambda$ is $\mathrm{I}, \lambda$ being the wavelength of light used. The intensity at a point where the path difference is $\frac{\lambda}{4}$ will be
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Verified Answer
The correct answer is:
$\frac{\mathrm{I}}{2}$
For path difference $\lambda$, phase
$$
\begin{array}{l}
\text { difference }=2 \pi\left(\mathrm{Q}=\frac{2 \pi}{\lambda} \mathrm{x}=\frac{2 \pi}{\lambda} \lambda=2 \pi\right) \\
\Rightarrow \mathrm{I}=\mathrm{I}_{0}+\mathrm{I}_{0}+2 \mathrm{I}_{0} \cos 2 \pi \\
\Rightarrow \mathrm{I}=4 \mathrm{I}_{0}(\therefore \cos 2 \pi=1)
\end{array}
$$
For $x=\frac{\lambda}{4}$, phase difference $=\frac{\pi}{2}$
$$
\therefore I^{\prime}=I_{1}+I_{2}+2 \sqrt{I_{1}} \sqrt{I_{2}} \cos \frac{\pi}{2}
$$
If $I_{1}=I_{2}=I_{0}$ then $I^{\prime}=2 I_{0}=2 \cdot \frac{I}{4}=\frac{I}{2}$
$$
\begin{array}{l}
\text { difference }=2 \pi\left(\mathrm{Q}=\frac{2 \pi}{\lambda} \mathrm{x}=\frac{2 \pi}{\lambda} \lambda=2 \pi\right) \\
\Rightarrow \mathrm{I}=\mathrm{I}_{0}+\mathrm{I}_{0}+2 \mathrm{I}_{0} \cos 2 \pi \\
\Rightarrow \mathrm{I}=4 \mathrm{I}_{0}(\therefore \cos 2 \pi=1)
\end{array}
$$
For $x=\frac{\lambda}{4}$, phase difference $=\frac{\pi}{2}$
$$
\therefore I^{\prime}=I_{1}+I_{2}+2 \sqrt{I_{1}} \sqrt{I_{2}} \cos \frac{\pi}{2}
$$
If $I_{1}=I_{2}=I_{0}$ then $I^{\prime}=2 I_{0}=2 \cdot \frac{I}{4}=\frac{I}{2}$
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