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In Young's double slit experiment, the intensity at a point where path difference is $\frac{\lambda}{6}$ ( $\lambda$ being the wavelength of light used) is I'. If ' $I_0$ ' denotes the maximum intensity, then $\frac{I}{I_0}$ is equal to
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The correct answer is:
$\frac{3}{4}$
If $\mathrm{I}$ ' is the intensity of each wave, then the resultant intensity at a point is given by
$$
\mathrm{I}=4 \mathrm{I}^{\prime} \cos ^2 \frac{\phi}{2}
$$
For maximum intensity, phase difference $\phi=0$
$$
\therefore \mathrm{I}_0=4 \mathrm{I}^{\prime} \cos ^2 0=4 \mathrm{I}^{\prime}
$$
At a point where path difference is $\frac{\lambda}{6}$, hence phase difference is
$$
\begin{aligned}
& \frac{\pi}{3} . \\
& \therefore \mathrm{I}=4 \mathrm{I}^{\prime} \cos ^2 \frac{\pi}{6}=3 \mathrm{I}^{\prime} \\
& \therefore \frac{\mathrm{I}}{\mathrm{I}_0}=\frac{3}{4}
\end{aligned}
$$
$$
\mathrm{I}=4 \mathrm{I}^{\prime} \cos ^2 \frac{\phi}{2}
$$
For maximum intensity, phase difference $\phi=0$
$$
\therefore \mathrm{I}_0=4 \mathrm{I}^{\prime} \cos ^2 0=4 \mathrm{I}^{\prime}
$$
At a point where path difference is $\frac{\lambda}{6}$, hence phase difference is
$$
\begin{aligned}
& \frac{\pi}{3} . \\
& \therefore \mathrm{I}=4 \mathrm{I}^{\prime} \cos ^2 \frac{\pi}{6}=3 \mathrm{I}^{\prime} \\
& \therefore \frac{\mathrm{I}}{\mathrm{I}_0}=\frac{3}{4}
\end{aligned}
$$
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