Q136 / 180 · KCET 2010 · Physics
The wavelength of the light used in Young's double slit experiment is $\lambda$. The intensity at a point on the screen is I, where the path difference is $\frac{\lambda}{6}$. If $\mathrm{I}_{0}$ denotes the maximum intensity, then the ratio of I and $\mathrm{I}_{0}$ is
- A$0.866$
- B$0.5$
- C$0.707$
- Correct answerD$0.75$
Solution
Phase difference, $\phi=\frac{2 \pi}{\lambda} \times$ path difference
$$
\phi=\frac{2 \pi}{\lambda} \times \frac{\lambda}{6}=\frac{\pi}{3}=60^{\circ}
$$
Intensity, $\mathrm{I}=\mathrm{I}_{0} \cos ^{2}\left(\frac{\phi}{2}\right)$
$$
\frac{\mathrm{I}}{\mathrm{I}_{0}}=\cos ^{2}\left(30^{\circ}\right)=\left(\frac{\sqrt{3}}{2}\right)^{2}=0.75
$$
Hence, the correct answer is option: 'D'.