Animated Solution for Physics - Optics: The maximum number of possible interference maxima for slit-separation equal to twice the wavelength in Young's double-slit experiment, is

Imagine standing in a dark room, watching the beautiful, rhythmic pattern of light and dark bands on a screen. The Young's Double Slit Experiment is not just a textbook problem; it is the ultimate proof of the wave nature of light.

In this problem, we are given a very specific and fascinating constraint. The distance between the two slits, $d$, is exactly twice the wavelength of the light being used, $\lambda$.

We need to find out the absolute maximum number of bright fringes (maxima) that can possibly form on the entire infinite screen. Let's dive into the physics and math behind this cosmic dance.

To find a bright fringe on the screen, the light waves from both slits must arrive perfectly in phase. This means their path difference, $\Delta x$, must be an integral multiple of the wavelength.

Geometrically, for a point on the screen at an angle $\theta$ from the central axis, the path difference is given by $d\sin \theta$. Equating these two gives us our master equation:

Here, $n$ is an integer representing the order of the maximum ($n=0$ is the central maximum, $n=\pm 1$ are the first maxima, and so on).

The problem hands us a beautiful simplification: $d=2\lambda$. Let's substitute this directly into our master equation.

Notice how the wavelength $\lambda$ appears on both sides of the equation? This means the specific color of the light doesn't actually matter for the number of fringes, only the ratio of $d$ to $\lambda$. Canceling $\lambda$ from both sides, we get a incredibly simple relation:

Or, rearranging for $\sin \theta$:

Now, we hit a fundamental wall of mathematics. The sine function is a bounded entity. No matter how far out on the screen you look, the angle $\theta$ can only approach $90^{\circ }$ (or $\frac{\pi }{2}$ radians).

Therefore, the value of $\sin \theta$ is strictly trapped between $-1$ and $1$.

This simple mathematical constraint dictates the physical reality of our interference pattern. Let's substitute our expression for $\sin \theta$ into this inequality.

Multiplying the entire inequality by $2$, we find the absolute limits for our integer $n$:

Since $n$ represents the order of the interference maxima, it must be an integer.

What are the integers that lie between $-2$ and $2$, inclusive? Let's list them out:

Counting them up, we have exactly five possible values for $n$.

This means that on the entire screen, from negative infinity to positive infinity, you will only ever see exactly five bright fringes. The correct answer is five!