A beam of light consisting of two wavelengths, 6500 A and 5200 A is used to obtain interference frin

Q: Why do we equate $n_1 \lambda_1 = n_2 \lambda_2$ for coincidence?

When two fringes coincide, they are at the exact same physical position $y$ on the screen. Since $y = \frac{n \lambda D}{d}$, equating $y_1$ and $y_2$ allows us to cancel out the constants $D$ and $d$, leaving us with the fundamental condition that their path differences must be equal: $n_1 \lambda_1 = n_2 \lambda_2$.

Q: Can I use the second wavelength ($\lambda_2$) to calculate the least distance instead of $\lambda_1$?

Absolutely! Since the fringes coincide at that exact spot, calculating the position using the $5^{\text{th}}$ fringe of $\lambda_2$ will give you the exact same result as using the $4^{\text{th}}$ fringe of $\lambda_1$. You can try it yourself: $y = \frac{5 \times 5200 \times 10^{-10} \times 1.2}{2 \times 10^{-3}} = 1.56 \text{ mm}$.

Q: What if the ratio $\frac{n_1}{n_2}$ doesn't simplify to a neat integer fraction?

In physical reality, $n_1$ and $n_2$ must be integers because they represent the order of the bright fringes (1st, 2nd, 3rd, etc.). If the ratio of the wavelengths is irrational, the fringes will technically never perfectly coincide anywhere other than the central maximum! However, in textbook problems, they are always designed to form a clean rational fraction.

Q: Why does the central maximum always coincide for all wavelengths?

At the central maximum, the path difference between the waves from the two slits is exactly zero ($\Delta x = 0$). Since the condition for a bright fringe is $\Delta x = n\lambda$, setting $n=0$ works for *any* wavelength $\lambda$. Thus, all colors perfectly overlap at the center, which is why white light produces a white central fringe.

Q: Does changing the distance between the slits ($d$) change which fringes coincide?

No, it doesn't! The condition for coincidence is $\frac{n_1}{n_2} = \frac{\lambda_2}{\lambda_1}$. Notice that $d$ and $D$ are completely absent from this equation. Changing the slit separation will change *where* the coincidence happens on the screen (the physical distance $y$), but it will still be the $4^{\text{th}}$ fringe of $\lambda_1$ overlapping with the $5^{\text{th}}$ fringe of $\lambda_2$.

Question

A beam of light consisting of two wavelengths, $6500	ext{ \AA}$ and $5200	ext{ \AA}$ is used to obtain interference fringe in a Young's double slit experiment.
(a) Find the distance of the third bright fringe on the screen from the central maximum for wavelength $6500	ext{ \AA}$.
(b) What is the least distance from the central maximum where the bright fringes due to both the wavelengths coincide?
The distance between the slits is $2	ext{ mm}$ and the distance between the plane of the slits and the screen is $120	ext{ cm}$.

Sigma Percentile · Accepted Answer

The Dance of Wavelengths: Unraveling the Young's Double-Slit Experiment

Imagine standing at the edge of a calm pond and dropping two pebbles into the water simultaneously. The ripples spread out, intersecting and creating a beautiful, complex pattern of peaks and troughs. This is the essence of interference, a phenomenon that Thomas Young brilliantly demonstrated with light in his famous Double-Slit Experiment. But what happens when we don't just drop two identical pebbles, but instead introduce two different types of waves into the mix?

In this classic JEE problem, we are tasked with analyzing an interference pattern created not by a single monochromatic light source, but by a beam containing two distinct wavelengths: $6500	ext{ \AA}$ and $5200	ext{ \AA}$. This adds a fascinating layer of complexity to the standard setup. Let's dive into the physics and mathematics behind this beautiful phenomenon.

### Analyzing the Setup

Before we jump into the calculations, let's visualize the physical reality of the setup. We have two narrow slits separated by a tiny distance, $d = 2	ext{ mm}$. A screen is placed far away at a distance $D = 120	ext{ cm}$. When light passes through these slits, it diffracts and overlaps on the screen, creating a series of bright and dark bands known as interference fringes.

The position of any bright fringe on the screen, measured from the central maximum, is governed by a beautifully simple equation:
$$y_n = \frac{n \lambda D}{d}$$
Here, $n$ represents the order of the fringe (1st, 2nd, 3rd, etc.), $\lambda$ is the wavelength of the light, $D$ is the distance to the screen, and $d$ is the slit separation.

To ensure our math is flawless, we must first convert all our given parameters into standard SI units (meters). 
- $\lambda_1 = 6500	ext{ \AA} = 6500 	imes 10^{-10}	ext{ m}$
- $\lambda_2 = 5200	ext{ \AA} = 5200 	imes 10^{-10}	ext{ m}$
- $d = 2	ext{ mm} = 2 	imes 10^{-3}	ext{ m}$
- $D = 120	ext{ cm} = 1.2	ext{ m}$

### The Master Equation: Finding the Third Bright Fringe

The first part of our problem asks for the distance of the third bright fringe for the first wavelength, $\lambda_1$. This is a straightforward application of our master equation. We simply plug in $n = 3$ and the parameters for $\lambda_1$.

$$y_3 = \frac{3 \lambda_1 D}{d}$$

Now, let's substitute the raw values into the structure:
$$y_3 = \frac{3 	imes (6500 	imes 10^{-10}) 	imes 1.2}{2 	imes 10^{-3}}$$

Executing the arithmetic carefully:
$$y_3 = \frac{3 	imes 6500 	imes 1.2}{2} 	imes 10^{-7}	ext{ m}$$
$$y_3 = 11700 	imes 10^{-7}	ext{ m}$$
$$y_3 = 1.17 	imes 10^{-3}	ext{ m}$$

Converting this back into a more intuitive unit, we get **$1.17	ext{ mm}$**. This tells us exactly where to look on the screen to find the third bright red band!

### The Race of the Fringes: Condition for Coincidence

Now comes the truly thrilling part of the problem. We have two different wavelengths creating their own independent fringe patterns on the same screen. Because $\lambda_1$ is longer than $\lambda_2$, its fringes are spaced further apart.

Think of it like two runners on a track. Runner 1 takes long strides ($\lambda_1$), while Runner 2 takes shorter, quicker strides ($\lambda_2$). They both start at the exact same point (the central maximum, where $y=0$). As they run, their footprints will occasionally land on the exact same spot. We want to find the *first* time this happens.

For the bright fringes to perfectly coincide, their physical distance from the central maximum must be identical. Therefore, we set their position equations equal to each other:
$$y_{n_1} = y_{n_2}$$
$$\frac{n_1 \lambda_1 D}{d} = \frac{n_2 \lambda_2 D}{d}$$

Notice how the geometry of the setup ($D$ and $d$) completely cancels out! The condition for coincidence depends purely on the intrinsic properties of the light itself:
$$n_1 \lambda_1 = n_2 \lambda_2$$

### Finding the Overlap

To find out which specific fringes overlap, we rearrange the equation to find the ratio of their orders:
$$\frac{n_1}{n_2} = \frac{\lambda_2}{\lambda_1}$$

Substituting our wavelengths:
$$\frac{n_1}{n_2} = \frac{5200}{6500} = \frac{4}{5}$$

This elegant fraction reveals the secret of the pattern. It tells us that the **$4^{	ext{th}}$ bright fringe of $\lambda_1$** perfectly overlaps with the **$5^{	ext{th}}$ bright fringe of $\lambda_2$**.

### Final Calculation: The Least Distance

To find the physical distance where this magical overlap occurs, we simply take the order we just found ($n_1 = 4$) and plug it back into our original position formula for $\lambda_1$. (We could also use $n_2 = 5$ with $\lambda_2$ and get the exact same result!).

$$Y_{	ext{min}} = \frac{4 \lambda_1 D}{d}$$
$$Y_{	ext{min}} = \frac{4 	imes (6500 	imes 10^{-10}) 	imes 1.2}{2 	imes 10^{-3}}$$

Crunching the final numbers:
$$Y_{	ext{min}} = 2 	imes 6500 	imes 1.2 	imes 10^{-7}	ext{ m}$$
$$Y_{	ext{min}} = 15600 	imes 10^{-7}	ext{ m}$$
$$Y_{	ext{min}} = 1.56 	imes 10^{-3}	ext{ m}$$

And there we have it! The least distance from the central maximum where the two bright fringes coincide is **$1.56	ext{ mm}$**.

This problem beautifully illustrates how abstract mathematical ratios directly govern the physical phenomena we observe in the universe. The next time you see a rainbow of colors reflecting off a thin film of oil, remember the precise, mathematical dance of wavelengths happening right before your eyes!

Visualized Solution

$Given Parameters$

$Position of n^{th} Bright Fringe$

$Distance of 3^{rd} Bright Fringe$

$Calculating y_{3}$

$Condition for Coincidence$

$Finding the Ratio \frac{n _{1}}{n _{2}}$

$Least Distance for Coincidence$

$Final Calculation$

The Sigma Insight: Interference and Young's Double-Slit Experiment

Analyzing the Setup

The Master Equation

The Race of the Fringes

Finding the Overlap

Final Calculation

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Visualized Solution

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Given Parameters

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Position of nth Bright Fringe

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Distance of 3rd Bright Fringe

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Calculating y3​

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Condition for Coincidence

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Finding the Ratio n2​n1​​

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Least Distance for Coincidence

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Final Calculation