Q129 / 150 · MHT CET 2021 (22 Sep Shift 2) · Physics
In Young's double slit experiment using monochromatic light of wavelength ' $\lambda$ ', the maximum intensity of light at a point on the screen is $\mathrm{K}$ units. The intensity of light at point where the path difference is $\frac{\lambda}{3}$
- Correct answerA$\frac{\mathrm{K}}{4}$
- B$\frac{3 \mathrm{~K}}{4}$
- C$\mathrm{K}$
- D$\frac{\mathrm{K}}{2}$
Solution
The intensity is given by
$$
\mathrm{I}=4 \mathrm{I}_0 \cos ^2 \frac{\phi}{2}
$$
Maximum intensity $\mathrm{K}=4 \mathrm{I}_0$ when $\phi=0$
When path difference is $\frac{\lambda}{3}, \phi=\frac{2 \pi}{3}$
$$
\therefore \mathrm{I}=\mathrm{K} \cos ^2 \frac{\pi}{3}=\mathrm{K}\left(\frac{1}{2}\right)^2=\frac{\mathrm{K}}{4}
$$
Hence, the correct answer is option: 'A'.