Animated Solution for Physics - Optics: A mixture of light, consisting of wavelength $590\text{ nm}$ and an unknown wavelength, illuminates Young's double slit and gives rise to two overlapping interference patterns on the screen. The central maximum of both lights coincide. Further, it is observed that the third bright fringe of known light coincides with the $4^{\text{th}}$ bright fringe of the unknown light. From this data, the wavelength of the unknown light is

Imagine you are standing in a dark room, observing a classic Young's Double Slit Experiment (YDSE). But instead of a single monochromatic laser, you are shining a mixture of two different lights through the slits. One light has a known wavelength, ${\lambda }_1=590\text{ nm}$, and the other is a mystery, ${\lambda }_2$.

What happens on the screen? Each wavelength creates its own independent interference pattern. Because they have different wavelengths, their fringe widths will be different. The central maxima for both will perfectly align at the center (where the path difference is zero), but as you move away from the center, the fringes will start to drift apart. However, at certain specific distances, a bright fringe from the first light will perfectly overlap with a bright fringe from the second light. This is the phenomenon of coinciding fringes.

The problem gives us a beautiful piece of forensic evidence: the $3^{\text{rd}}$ bright fringe of the known light perfectly coincides with the $4^{\text{th}}$ bright fringe of the unknown light.

What does "coincide" mean mathematically? It means that if you measure the distance from the central maximum to the $3^{\text{rd}}$ bright fringe of ${\lambda }_1$, and then measure the distance to the $4^{\text{th}}$ bright fringe of ${\lambda }_2$, those two distances are exactly the same. Let's call this common distance $y$.

To solve this, we need our trusty tool: the formula for the position of the $n^{\text{th}}$ bright fringe in a YDSE setup. The distance $y_n$ from the central maximum is given by:

Here, $D$ is the distance to the screen, and $d$ is the separation between the slits. Since both lights are passing through the exact same physical apparatus, $D$ and $d$ are identical for both.

Now, we translate our physical observation into a mathematical equation. We equate the position of the $3^{\text{rd}}$ bright fringe of ${\lambda }_1$ to the position of the $4^{\text{th}}$ bright fringe of ${\lambda }_2$:

Substituting our formula into this equality, we get:

Notice how elegantly the geometry of the setup cancels out. The terms $D$ and $d$ appear on both sides, so we can divide them away. We are left with a pure relationship between the integers and the wavelengths:

Our goal is to unmask the unknown wavelength, ${\lambda }_2$. Rearranging the equation to isolate ${\lambda }_2$, we find:

Now, we simply plug in the known value of ${\lambda }_1=590\text{ nm}$:

And there we have it! The unknown wavelength is exactly $442.5\text{ nm}$.

This is a high-yield concept for competitive exams. The logic we used here is universal. What if the examiner throws a curveball and says the $n^{\text{th}}$ bright fringe of one light coincides with the $m^{\text{th}}$ dark fringe of another?

You don't need to panic. You simply write down the position formula for a bright fringe on one side, and the position formula for a dark fringe ($y_m=(2m-1)\frac{\lambda D}{2d}$) on the other side, and equate them. The physical reality dictates the math. Always visualize the overlap, and the equations will naturally follow.