Animated Solution for Physics - Optics: A parallel monochromatic beam of light is incident normally on a narrow slit. A diffraction pattern is formed on a screen placed perpendicular to the direction of the incident beam. At the first minimum of the diffraction pattern, the phase difference between the rays coming from the two edges of the slit is

Have you ever wondered why light doesn't just travel in perfectly straight lines? When light encounters an obstacle or a narrow opening, it bends and spreads out. This beautiful phenomenon is known as diffraction.

Imagine a parallel beam of monochromatic light—light of a single, pure color—striking a narrow slit. According to classical ray optics, we should just see a sharp, bright line on the screen behind the slit. But nature is far more poetic. Instead, we see a central bright band flanked by alternating dark and bright fringes. This is the diffraction pattern, a direct consequence of the wave nature of light.

The Single Slit

Let's set up our mental laboratory. We have a slit of width $b$. A screen is placed at a distance $D$ away, where $D$ is much larger than $b$. A plane wavefront of light with wavelength $\lambda$ hits the slit normally.

According to Huygens' Principle, every point on the wavefront within the slit acts as a source of secondary spherical wavelets. These wavelets travel forward and interfere with each other on the screen. The central point on the screen, directly opposite the center of the slit, is where all these wavelets arrive in phase. They have traveled the exact same distance, so they add up constructively, creating the intense central maximum.

But what happens as we move away from the center? Let's look at a point $P$ on the screen at an angle $\theta$ from the central axis. The wavelets from different parts of the slit now have to travel different distances to reach $P$.

To find the first minimum—the first dark fringe—we use a clever trick. We conceptually divide the slit into two equal halves: a top half and a bottom half. For every point source in the top half, there is a corresponding point source in the bottom half exactly $b/2$ distance away.

If the path difference between these paired sources is exactly $\lambda /2$, they will arrive at point $P$ completely out of phase and cancel each other out. Since this pairing works for every single point in the slit, the entire slit's light cancels out, resulting in a dark spot!

Let's translate this geometry into math. If the path difference between sources separated by $b/2$ is $\lambda /2$, then the path difference between the extreme edges of the slit (separated by the full width $b$) must be twice as much.

Therefore, the path difference $\Delta x$ between a ray from the very top edge of the slit and a ray from the very bottom edge is:

From the geometry of the setup, if we drop a perpendicular from the top edge to the ray originating from the bottom edge, we can see that this path difference is also given by $b\sin \theta$. This gives us the famous condition for the first minimum:

The Final Leap

Now, we must be careful. The question doesn't ask for the path difference; it asks for the phase difference.

Path difference is a physical distance, measured in meters. Phase difference is an angle, measured in radians, representing how far out of sync the wave cycles are. The bridge between them is the wave number, $k=\frac{2\pi }{\lambda }$. The relationship is:

We already established that for the first minimum, the path difference between the extreme edges is $\Delta x=\lambda$. Let's substitute this into our phase equation:

The wavelengths cancel out perfectly, leaving us with:

And there we have it! The phase difference between the rays coming from the two edges of the slit at the first minimum is exactly $2\pi$. This elegant result reminds us that understanding the physical geometry of interference is the key to unlocking the secrets of wave optics.