Animated Solution for Physics - Optics: In a Young's double slit experiment, two wavelengths of $500\text{ nm}$ and $700\text{ nm}$ were used. What is the minimum distance from the central maximum where their maximas coincide again? Take $D/d=10^3$. Symbols have their usual meanings.

Imagine you are standing in front of a Young's Double Slit setup. But instead of a single monochromatic light, you fire two different colors simultaneously—one with a wavelength of $500\text{ nm}$ and another with $700\text{ nm}$.

Each color will independently create its own beautiful interference pattern on the screen. You will see a series of bright and dark fringes for the $500\text{ nm}$ light, and a separate, slightly more spread-out series of fringes for the $700\text{ nm}$ light.

At the exact center of the screen, the central maxima of both colors perfectly overlap. But as you move away from the center, the fringes start to drift apart because they have different fringe widths. The question asks: at what minimum distance from the center will a bright fringe of the first color perfectly align with a bright fringe of the second color again?

To solve this, we need to recall the formula for the position of the $n^{\text{th}}$ bright fringe from the central maximum. For any wavelength $\lambda$, the position $y$ is given by:

For the two bright fringes to coincide, their physical distance $y$ from the central maximum must be exactly the same. Let's say the $n_1^{\text{th}}$ maxima of the first light coincides with the $n_2^{\text{th}}$ maxima of the second light. We can set their position equations equal to each other:

Notice how the geometry of the setup—the distance to the screen $D$ and the slit separation $d$—is the same for both colors. These terms beautifully cancel out from both sides, leaving us with a pure, elegant relationship between the fringe numbers and their wavelengths:

We can rearrange this equation to find the ratio of the fringe numbers:

Now, let's plug in the given wavelengths. We have ${\lambda }_1=500\text{ nm}$ and ${\lambda }_2=700\text{ nm}$:

This ratio tells us that the $7^{\text{th}}$ maxima of the first light will overlap with the $5^{\text{th}}$ maxima of the second light. Because we are looking for the minimum distance from the central maximum, we need the smallest possible integer values for $n_1$ and $n_2$. The fraction $7/5$ is already in its simplest form, so our smallest integers are simply $n_1=7$ and $n_2=5$.

Now that we know exactly which fringes are overlapping, we can calculate their physical distance from the center. We can use the position formula for either the first or the second light; the result will be identical. Let's use the first light ($n_1=7$):

We are given the ratio $D/d=10^3$. Let's substitute all our known values. Remember to convert the wavelength from nanometers to meters to keep our units consistent:

Let's multiply the numbers:

To make this number more intuitive, let's convert it into millimeters. Since $1\text{ mm}=10^{-3}\text{ m}$, we can rewrite our answer as:

And there we have it! At exactly $3.5\text{ mm}$ from the central maximum, the two interference patterns will perfectly synchronize again. This beautiful periodicity will continue further up the screen, with the next coincidence happening at $7.0\text{ mm}$, and so on.