QJEEM24Phfcaa: JEE Main 2024 Physics - Class 12 JEEM24Phfcaa Solution: In a Young's double slit experiment, the intensity at a point…

In a Young's double slit experiment, the intensity at a point can be expressed as a function of the phase difference between the light arriving from the two slits. The intensity at any point on the screen is given by:

$I = I_{\text{max}} \cos^2 \left( \frac{\delta}{2} \right)$

Where:

• $I_{\text{max}}$ is the maximum intensity.
• $\delta$ is the phase difference between the light waves from the two slits.

Given the intensity at a point is $\left(\frac{1}{4}\right)^{\text {th }}$ of the maximum intensity, we can write:

$\frac{I}{I_{\text{max}}} = \frac{1}{4}$

Substituting this into the intensity equation:

$\frac{1}{4} = \cos^2 \left( \frac{\delta}{2} \right)$

Taking the square root of both sides, we get:

$\cos \left( \frac{\delta}{2} \right) = \frac{1}{2}$

The possible solutions for $\delta$ are:

$\frac{\delta}{2} = \frac{\pi}{3}$ or $\frac{\delta}{2} = \left(\pi - \frac{\pi}{3}\right)$ which gives $\delta = \frac{2\pi}{3}$ or $\delta = \frac{4\pi}{3}$ .

Considering the smallest phase difference, $\delta = \frac{2\pi}{3}$ , we use the relation for the phase difference due to path difference:

$\delta = \frac{2 \pi}{\lambda} \cdot \Delta x$

Thus, substituting the value of $\delta$ , we have:

$\frac{2\pi}{\lambda} \cdot \Delta x = \frac{2\pi}{3}$

Solving for $\Delta x$ , we get:

$\Delta x = \frac{\lambda}{3} = \frac{600 \, \mathrm{nm}}{3} = 200 \, \mathrm{nm} = 0.2 \, \mu \mathrm{m}$

The minimum distance of the point from the central maximum on the screen can be found using the interference equation:

$y = \frac{\Delta x \cdot D}{d}$

Substituting the known values:

$y = \frac{0.2 \, \mu \mathrm{m} \cdot 1.0 \, \mathrm{m}}{1.0 \, \mathrm{mm}} = \frac{0.2 \cdot 10^{-6} \, \mathrm{m} \cdot 1.0 \, \mathrm{m}}{1.0 \cdot 10^{-3} \, \mathrm{m}}$

Simplifying this expression:

$y = 0.2 \, \mathrm{mm} = 200 \, \mu \mathrm{m}$

Hence, the minimum distance of the point from the central maximum is:

$200 \, \mu \mathrm{m}$