Q110 / 150 · MHT CET 2024 (09 May Shift 2) · Physics
In the Young's double slit experiment, the intensity at a point on the screen, where the path difference is $\lambda(\lambda=$ wavelength $)$ is $\beta$. The intensity at a point where the path difference is $\lambda / 3$, will be $\left.\cos \frac{\pi}{3}=1 / 2\right]$
- A$\beta$
- B$\beta / 2$
- Correct answerC$\frac{\beta}{4}$
- D$\beta / 8$
Solution
The intensity is given by
$\mathrm{I}=4 \mathrm{I}_0 \cos ^2 \frac{\phi}{2}$
Maximum intensity $\beta=4 \mathrm{I}_0$ when $\phi=0$ when path difference is $\frac{\lambda}{3}, \phi=\frac{2 \pi}{3}$ $\therefore \quad I=\beta \cos ^2 \frac{2 \pi}{3}=\beta\left(-\frac{1}{2}\right)^2=\frac{\beta}{4}$
Maximum intensity $\beta=4 \mathrm{I}_0$ when $\phi=0$ when path difference is $\frac{\lambda}{3}, \phi=\frac{2 \pi}{3}$ $\therefore \quad I=\beta \cos ^2 \frac{2 \pi}{3}=\beta\left(-\frac{1}{2}\right)^2=\frac{\beta}{4}$
Hence, the correct answer is option: 'C'.