Calculate Intensity at Fringe Opposite to Slit in Young Double Slit Experiment

In a Young's double slit experiment, let the two slits be $S_1$ and $S_2$, separated by a distance $d$. The screen is placed at a distance $D$ from the plane of the slits.

A point $P$ on the screen that is directly opposite to one of the slits (say $S_1$) lies at a distance $y$ from the central line on the screen, given by: $y = \frac{d}{2}$

The path difference $\Delta x$ between the waves arriving at point $P$ from slits $S_1$ and $S_2$ is given by the standard approximation: $\Delta x = \frac{y d}{D}$

Substituting $y = \frac{d}{2}$ into the formula for path difference: $\Delta x = \frac{\left(\frac{d}{2}\right) d}{D} = \frac{d^2}{2D}$

Given that $d = 5\lambda$ and $D = 10d$: $\Delta x = \frac{d^2}{2(10d)} = \frac{d}{20} = \frac{5\lambda}{20} = \frac{\lambda}{4}$

The corresponding phase difference $\phi$ at point $P$ is given by: $\phi = \frac{2\pi}{\lambda} \Delta x = \frac{2\pi}{\lambda} \times \frac{\lambda}{4} = \frac{\pi}{2}$

The resultant intensity $I$ at any point on the screen with phase difference $\phi$ is given by: $I = I_0 \cos^2\left(\frac{\phi}{2}\right)$

Substituting $\phi = \frac{\pi}{2}$: $I = I_0 \cos^2\left(\frac{\pi}{4}\right) = I_0 \left(\frac{1}{\sqrt{2}}\right)^2 = \frac{I_0}{2}$

Hence, the intensity of the fringe opposite to one of the slits is $\frac{I_0}{2}$.

Correct Option: B ($\frac{I_0}{2}$)