Q146 / 150 · MHT CET 2024 (03 May Shift 2) · Physics
In Young's double slit experiment using monochromatic light of wavelength ' $\lambda$ ', the intensity of light at a point on the screen where path difference ' $\lambda$ ' is K units. The intensity of light at a point where the path difference is $\frac{\lambda}{6}$ is $\left[\cos \frac{\pi}{6}=\sin \frac{\pi}{3}=\frac{\sqrt{3}}{2}\right]$
- AK
- Correct answerB$\frac{3 \mathrm{~K}}{4}$
- C$\frac{\mathrm{K}}{2}$
- D$\frac{\mathrm{K}}{4}$
Solution
The intensity is given by
$I=4 I_0 \cos ^2 \frac{\phi}{2}$
Maximum intensity $\mathrm{K}=4 \mathrm{I}_0$ when $\phi=0$ when path difference is $\frac{\lambda}{6}$, $\begin{aligned} & \phi=\frac{2 \pi}{\lambda} \times \text { path difference }=\frac{\pi}{3} \\ \therefore \quad & I=K \cos ^2\left(\frac{\pi}{6}\right)=K\left(\frac{1}{4}\right)=\frac{3 K}{4} \end{aligned}$ ...[From(i) and (ii)]
Maximum intensity $\mathrm{K}=4 \mathrm{I}_0$ when $\phi=0$ when path difference is $\frac{\lambda}{6}$, $\begin{aligned} & \phi=\frac{2 \pi}{\lambda} \times \text { path difference }=\frac{\pi}{3} \\ \therefore \quad & I=K \cos ^2\left(\frac{\pi}{6}\right)=K\left(\frac{1}{4}\right)=\frac{3 K}{4} \end{aligned}$ ...[From(i) and (ii)]
Hence, the correct answer is option: 'B'.