The ratio of intensities of two points on a screen in Young's double slit experiment when waves from the two…
The ratio of intensities of two points on a screen in Young's double slit experiment when waves from the two slits have a path difference of $\frac{\lambda}{4}$ and $\frac{\lambda}{6}$ is
$\left(\cos 90^{\circ}=0, \cos 60^{\circ}=0.5\right)$
$2: 1$
$2: 3$
$3: 4$
$3: 5$
Solution
The intensity at the point due to interference is given as $I=I_1+I_2+2 \sqrt{I_1 I_2} \cos \phi$
For path difference $\frac{\lambda}{4}$, the phase difference is $\phi_1=\frac{2 \pi}{\lambda} \times \frac{\lambda}{4}=\frac{\pi}{2}$
For path difference $\frac{\lambda}{6}$, the phase difference is $\phi_2=\frac{2 \pi}{\lambda} \times \frac{\lambda}{6}=\frac{\pi}{3}$
Assuming equal intensity of the interfering waves i.e., $\mathrm{I}_1=\mathrm{I}_2=\mathrm{I}_0$
Equation (i) becomes,
$\begin{aligned}
& I=I_0+I_0+2 I_0 \cos \phi \\
& I=2 I_0(1+\cos \phi)
\end{aligned}$
For the given path difference, $I_1=2 I_0\left(1+\cos \frac{\pi}{2}\right)$.
$\begin{aligned}
& \text { and } I_2=2 I_0\left(1+\cos \frac{\pi}{3}\right) \\
& \therefore \quad \frac{I_1}{I_2}=\frac{1+\cos \frac{\pi}{2}}{1+\cos \frac{\pi}{3}}
\end{aligned}$
$\begin{aligned} \frac{\mathrm{I}_1}{\mathrm{I}_2} & =\frac{1+0}{1+0.5} \\ \therefore \quad \frac{\mathrm{I}_1}{\mathrm{I}_2} & =\frac{1}{1.5}=\frac{2}{3}\end{aligned}$