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In Young's double slit experiment, the intensity of light at a point on the screen is ${ }^{\prime} \mathrm{K}^{\prime}$ unit for path difference ${ }^{\prime} \lambda^{\prime}$. What would be the intensity at a point if path
difference is $\frac{\lambda^{\prime}}{4}$ ?
Options:
difference is $\frac{\lambda^{\prime}}{4}$ ?
Solution:
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Verified Answer
The correct answer is:
$\frac{K}{2}$
When path difference is $\lambda$, phase difference $\phi$ is $2 \pi$ and when path difference is $\frac{\lambda}{4}$, the phase difference is $\frac{\pi}{2}$. If $\mathrm{I}_{0}$ is the intensity of the two waves then the resultant intensity is given by
$$
\begin{aligned}
I &=4 I_{0} \cos ^{2} \frac{\phi}{2} \\
\therefore K &=4 I_{0} \cos ^{2} \frac{2 \pi}{2}=4 I \cos ^{2} \pi=4 I_{0} \quad \because \cos \pi=-1
\end{aligned}
$$
$$
\begin{array}{l}
\text { When } \phi=\frac{\pi}{2} \\
\qquad \begin{aligned}
\mathrm{I}=4 \mathrm{I}_{0} \cos ^{2} \cdot \frac{\pi}{4} & \\
&=4 \mathrm{I}_{0} \frac{1}{2}=\frac{\mathrm{K}}{2} \quad \because \cos \frac{\pi}{4}=\frac{1}{\sqrt{2}}
\end{aligned}
\end{array}
$$
$$
\begin{aligned}
I &=4 I_{0} \cos ^{2} \frac{\phi}{2} \\
\therefore K &=4 I_{0} \cos ^{2} \frac{2 \pi}{2}=4 I \cos ^{2} \pi=4 I_{0} \quad \because \cos \pi=-1
\end{aligned}
$$
$$
\begin{array}{l}
\text { When } \phi=\frac{\pi}{2} \\
\qquad \begin{aligned}
\mathrm{I}=4 \mathrm{I}_{0} \cos ^{2} \cdot \frac{\pi}{4} & \\
&=4 \mathrm{I}_{0} \frac{1}{2}=\frac{\mathrm{K}}{2} \quad \because \cos \frac{\pi}{4}=\frac{1}{\sqrt{2}}
\end{aligned}
\end{array}
$$
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