In Young's double slit experiment, the intensity at a point where path difference is $\frac{\lambda}{6}$ (…
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
$\frac{\sqrt{3}}{2}$
$\frac{3}{4}$
$\frac{1}{\sqrt{2}}$
$\frac{1}{2}$
Solution
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
$\begin{aligned}
& \frac{\pi}{3} . \\
& \therefore \mathrm{I}=4 \mathrm{I}^{\prime} \cos ^2 \frac{\pi}{6}=3 \mathrm{I}^{\prime} \\
& \therefore \frac{\mathrm{I}}{\mathrm{I}_0}=\frac{3}{4}
\end{aligned}$