Animated Solution for Physics - Optics: In a YDSE bi-chromatic light of wavelengths 400 nm and 560 nm are used. The distance between the slits is 0.1 mm and the distance between the plane of the slits and the screen is 1 m. The minimum distance between two successive regions of complete darkness is

The phenomenon of interference is one of the most beautiful demonstrations of the wave nature of light. In a standard Young's Double Slit Experiment (YDSE), we usually deal with a single monochromatic light source. But what happens when we mix things up and use two different wavelengths simultaneously? The screen becomes a canvas of overlapping interference patterns!

In this problem, we are tasked with finding the minimum distance between two successive regions of complete darkness. Let's dive into the physics and mathematics behind this fascinating setup.

We are given a YDSE setup with the following parameters: - Wavelength 1, ${\lambda }_1=400\text{ nm}$ - Wavelength 2, ${\lambda }_2=560\text{ nm}$ - Slit separation, $d=0.1\text{ mm}$ - Distance to screen, $D=1\text{ m}$

When both wavelengths are incident on the slits, they each create their own independent interference pattern on the screen. The total intensity at any point is simply the sum of the intensities of the individual patterns.

For a region to be completely dark, there must be absolutely no light reaching that point. This means that both wavelengths must undergo destructive interference at that exact location. In other words, the minima of ${\lambda }_1$ must perfectly coincide with the minima of ${\lambda }_2$.

The position of the $n$-th minima for any wavelength $\lambda$ is given by the formula:

To find where the minima coincide, we equate the position formulas for the $n$-th minima of ${\lambda }_1$ and the $m$-th minima of ${\lambda }_2$:

Notice how the geometric constants $D$ and $2d$ beautifully cancel out from both sides, leaving us with a pure relationship between the order of the fringes and their wavelengths:

Substituting the given wavelengths:

Simplifying the equation gives us the ratio of the odd integers:

Since $(2n-1)$ and $(2m-1)$ must be odd integers, we need to find equivalent fractions where both the numerator and denominator are odd. Let's list the possible ratios by multiplying the numerator and denominator by integers:

We must discard the ratios with even numbers (like $14/10$ and $28/20$) because $(2n-1)$ and $(2m-1)$ cannot be even. Thus, the valid odd integer ratios are:

The first coincidence occurs when $(2n-1)=7$. Let's calculate its position $y_1$:

The second coincidence occurs when $(2n-1)=21$. Let's calculate its position $y_2$:

The minimum distance between two successive regions of complete darkness is simply the difference between these two positions: