Animated Solution for Physics - Optics: In Young's double slit arrangement, slits are separated by a gap of $0.5\text{ mm}$, and the screen is placed at a distance of $0.5\text{ m}$ from them. The distance between the first and the third bright fringe formed when the slits are illuminated by a monochromatic light of $5890\text{ A˚}$ is

The YDSE Setup

Imagine you are standing in a dark room, observing a classic Young's Double Slit Experiment (YDSE). We have two incredibly narrow slits separated by a tiny gap $d=0.5\text{ mm}$. A screen is placed at a macroscopic distance $D=0.5\text{ m}$ away from these slits.

When a monochromatic light of wavelength $\lambda =5890\text{ A˚}$ is fired at the slits, the light waves emerge, overlap, and interfere. This interference creates a beautiful, rhythmic pattern of bright and dark bands—called fringes—on the screen. Our goal is to find the exact physical distance between the first and the third bright fringes.

Before we crunch the numbers, let's understand the geometry of the interference pattern. The distance between any two consecutive bright fringes (or dark fringes) is constant and is known as the fringe width, denoted by $\beta$. The master equation for the fringe width is:

The central maximum is located at the center of the screen ($n=0$). The first bright fringe is at a distance of $1\beta$ from the center, the second is at $2\beta$, and the third is at $3\beta$.

The question asks for the distance between the first and the third bright fringe. Let's call this distance $Y$. Mathematically, it is simply the difference in their positions:

Now that we know we need to calculate $2\beta$, let's substitute our known values into the equation. But first, a crucial warning: always convert your units to the standard SI system (meters) to avoid catastrophic order-of-magnitude errors.

- Wavelength: $\lambda =5890\text{ A˚}=5890\times 10^{-10}\text{ m}$ - Slit separation: $d=0.5\text{ mm}=0.5\times 10^{-3}\text{ m}$ - Screen distance: $D=0.5\text{ m}$

Substituting these into our expression for $Y$:

I know this expression looks a bit heavy, but let's take a breath and simplify it step-by-step. Notice how elegantly the $0.5$ in the numerator and the $0.5$ in the denominator cancel each other out!

To match the format of the given options, we can adjust the decimal point by shifting it one place to the left, which increases the power of ten by one:

And there we have it! The distance between the first and the third bright fringe is exactly $1178\times 10^{-6}\text{ m}$. This perfectly matches option (b).

The Way Forward: What if this entire setup was immersed in a liquid, say water? The wavelength of light would decrease (${\lambda }^{\prime }=\lambda /\mu$), which means the fringe width $\beta$ would also decrease. The entire fringe pattern would shrink, bringing the fringes closer together. Always keep these conceptual variations in mind, as they are absolute favorites for JEE!