JEE Main 2024 — Wave Optics Question with Solution

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:

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