In a Young's double slit experiment, the intensity at a point is $\left(\frac{1}{4}\right)^{\text {th }}$ of the maximum intensity, the minimum distance of the point from the central maximum is _________ $\mu \mathrm{m}$. (Given : $$\lambda=600 \mathrm{~nm}, \mathrm{~d}=1.0 \mathrm{~mm}, \mathrm{D}=1.0 \mathrm{~m}$$)

<p>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:</p> <p> <p>$I = I_{\text{max}} \cos^2 \left( \frac{\delta}{2} \right)$</p> </p> <p>Where: <ul> <li>$I_{\text{max}}$ is the maximum intensity.</li> <li>$\delta$ is the phase difference between the light waves from the two slits.</li> </ul> </p> <p>Given the intensity at a point is $\left(\frac{1}{4}\right)^{\text {th }}$ of the maximum intensity, we can write:</p> <p> <p>$\frac{I}{I_{\text{max}}} = \frac{1}{4}$</p> </p> <p>Substituting this into the intensity equation:</p> <p> <p>$\frac{1}{4} = \cos^2 \left( \frac{\delta}{2} \right)$</p> </p> <p>Taking the square root of both sides, we get:</p> <p> <p>$\cos \left( \frac{\delta}{2} \right) = \frac{1}{2}$</p> </p> <p>The possible solutions for $\delta$ are:</p> <p> <p>$\frac{\delta}{2} = \frac{\pi}{3}$ or $\frac{\delta}{2} = \left(\pi - \frac{\pi}{3}\right)$ which gives $\delta = \frac{2\pi}{3}$ or $\delta = \frac{4\pi}{3}$.</p> </p> <p>Considering the smallest phase difference, $\delta = \frac{2\pi}{3}$, we use the relation for the phase difference due to path difference:</p> <p> <p>$\delta = \frac{2 \pi}{\lambda} \cdot \Delta x$</p> </p> <p>Thus, substituting the value of $\delta$, we have:</p> <p> <p>$\frac{2\pi}{\lambda} \cdot \Delta x = \frac{2\pi}{3}$</p> </p> <p>Solving for $\Delta x$, we get:</p> <p> <p>$$\Delta x = \frac{\lambda}{3} = \frac{600 \, \mathrm{nm}}{3} = 200 \, \mathrm{nm} = 0.2 \, \mu \mathrm{m}$$</p> </p> <p>The minimum distance of the point from the central maximum on the screen can be found using the interference equation:</p> <p> <p>$y = \frac{\Delta x \cdot D}{d}$</p> </p> <p>Substituting the known values:</p> <p> <p>$$y = \frac{0.2 \, \mu \mathrm{m} \cdot 1.0 \, \mathrm{m}}{1.0 \, \mathrm{mm}} = \frac{0.2 \cdot 10^{-6} \, \mathrm{m} \cdot 1.0 \, \mathrm{m}}{1.0 \cdot 10^{-3} \, \mathrm{m}}$$</p> </p> <p>Simplifying this expression:</p> <p> <p>$y = 0.2 \, \mathrm{mm} = 200 \, \mu \mathrm{m}$</p> </p> <p>Hence, the minimum distance of the point from the central maximum is:</p> <p> <p>$200 \, \mu \mathrm{m}$</p> </p>