Animated Solution for Physics - Optics: A narrow slit of width $1\text{ mm}$ is illuminated by monochromatic light of wavelength $600\text{ nm}$. The distance between the first minima on either side of a screen at a distance of $2\text{ m}$ is

Imagine you are standing in a dark room, holding a laser pointer. You shine it through a tiny, narrow slit. What do you expect to see on the wall opposite to you? A single, sharp line of light? Surprisingly, nature has a beautiful trick up its sleeve. Instead of a single line, you see a central bright band of light, flanked by alternating dark and bright bands fading away into the darkness. This mesmerizing phenomenon is called diffraction.

In our problem, we are given a narrow slit of width $d=1\text{ mm}$. We are illuminating it with a monochromatic light—meaning light of a single, pure color—with a wavelength $\lambda =600\text{ nm}$. The screen where we observe this beautiful pattern is placed at a distance $D=2\text{ m}$ away. Our goal is to find the distance between the first dark fringes (minima) on either side of the central bright band.

To solve this, we need to dive into the mathematics of single-slit diffraction. The condition for finding the dark fringes (minima) is given by the elegant equation:

Here, $n$ represents the order of the minimum. For the first minimum, $n=1$, so our equation simplifies to:

Now, let's look at the geometry of our setup. The angle $\theta$ is the angle from the central axis to the first minimum. Because the screen is quite far away ($D=2\text{ m}$) compared to the slit width ($d=1\text{ mm}$), the angle $\theta$ is incredibly small. In the realm of small angles, we can use a handy mathematical approximation: $\sin \theta \approx \tan \theta$.

From our geometry, $\tan \theta$ is simply the opposite side over the adjacent side, which is $y/D$, where $y$ is the distance from the center of the pattern to the first minimum. Substituting this into our equation, we get:

Rearranging this equation to solve for $y$, we find:

This $y$ gives us the distance from the very center of the bright fringe to the first dark fringe on one side. But wait! The question asks for the distance between the first minima on either side. This means we need the total width of the central maximum, which spans from the first minimum on the left to the first minimum on the right. Therefore, the total distance is $2y$:

Now comes the fun part—plugging in the numbers! To avoid any silly mistakes, it is always a great practice to convert all our measurements into a single, consistent unit. Let's use millimeters ($\text{mm}$).

- The wavelength $\lambda =600\text{ nm}=600\times 10^{-6}\text{ mm}$. - The distance to the screen $D=2\text{ m}=2\times 10^3\text{ mm}$. - The slit width $d=1\text{ mm}$.

Let's substitute these into our master equation:

Let's crunch the numbers. Combining the powers of ten, $10^{-6}\times 10^3=10^{-3}$. Multiplying the constants, $2\times 600\times 2=2400$.

And there we have it! The distance between the first minima on either side is exactly $2.4\text{ mm}$. This perfectly matches option (d).

Changing the Medium

What if we took this entire experiment and submerged it underwater? It’s a classic twist! When light enters a denser medium like water, its speed decreases, and consequently, its wavelength $\lambda$ decreases. Looking at our formula $2y=\frac{2\lambda D}{d}$, we can see that the spread of the central maximum is directly proportional to the wavelength. Therefore, if the wavelength decreases, the entire diffraction pattern shrinks, and the distance between the minima would become smaller. Physics is all about understanding these interconnected relationships!