Animated Solution for Physics - Optics: Consider a Young's double slit experiment as shown in figure. What should be the slit separation $d$ in terms of wavelength $\lambda$ such that the first minima occurs directly in front of the slit ($S_1$)?

The Young's Double Slit Experiment (YDSE) is a beautiful demonstration of the wave nature of light. Usually, we rely on the small-angle approximation where the path difference is simply $d\sin \theta$. However, this problem throws a curveball! The point $P$ is directly in front of one of the slits, and the screen is relatively close. This means we must abandon our approximations and rely on pure, unadulterated geometry.

Let's visualize the physical reality of the setup. We have two coherent slits, $S_1$ and $S_2$, separated by a distance $d$. The screen is placed at a distance $D=2d$ from the plane of the slits.

The point of interest, $P$, lies exactly opposite to the upper slit $S_1$. We are given a crucial piece of information: the first minimum of the interference pattern forms exactly at this point $P$.

What does it mean for a minimum to form? It means the light waves from $S_1$ and $S_2$ arrive at $P$ completely out of phase, destructively interfering with each other.

The general condition for the $n$-th minimum is that the path difference $\Delta x$ must be an odd multiple of half the wavelength:

Since we are dealing with the first minimum, we set $n=1$. This simplifies our master equation to:

To find the path difference, we need the exact distances the light travels from each slit to point $P$.

First, consider the ray from $S_1$. Because $P$ is directly in front of $S_1$, the light travels in a straight horizontal line. The length of this path is simply the distance to the screen:

Next, consider the ray from $S_2$. This ray must travel diagonally to reach $P$. If we look closely, the slits $S_1$, $S_2$, and the point $P$ form a perfect right-angled triangle. The distance between the slits is the vertical leg ($d$), and the distance to the screen is the horizontal leg ($2d$).

Using the Pythagorean theorem, the hypotenuse $S_{2}P$ is:

Now we have both path lengths. The path difference $\Delta x$ is the difference between the longer path and the shorter path:

We can factor out $d$ to make it cleaner:

We know from our master equation that this path difference must equal $\frac{\lambda }{2}$ for the first minimum to occur. Equating the two expressions:

Finally, isolating the slit separation $d$, we get our elegant final answer:

This perfectly matches option (c). By trusting the geometry rather than blindly applying standard formulas, we navigated the trap and arrived at the correct solution!