In a Young's double slit experiment, slits are separated by 0.5 mm and the screen is placed 150 cm a

Q: Why do we equate $y_1$ and $y_2$?

Coincidence means the fringes are located at the exact same physical spot on the screen. Since $y$ represents the distance from the central maximum, equating $y_1$ and $y_2$ ensures we are looking at the same location.

Q: Why did we choose $n_1=4$ and $n_2=5$ instead of $8$ and $10$?

The question specifically asks for the *least* distance from the central maximum. The smallest distance corresponds to the smallest possible integer values for $n_1$ and $n_2$ that satisfy the ratio.

Q: Can we use the formula for dark fringes instead?

No, the question explicitly states we are looking for the point where the *bright* fringes due to both wavelengths coincide.

Q: What happens to the central maximum? Don't they coincide there too?

Yes, the central maximum for all wavelengths is always at $y=0$, so they perfectly coincide there. However, the question asks for the least distance *from* the central maximum, implying a non-zero distance.

Q: Why is the ratio $n_1/n_2$ inversely proportional to the wavelengths?

Because the position $y$ is proportional to $n \times \lambda$. For $y$ to be the same for both colors, a larger wavelength (which creates wider fringes) requires a smaller fringe number to reach the same spot.

Question

In a Young's double slit experiment, slits are separated by $0.5 	ext{ mm}$ and the screen is placed $150 	ext{ cm}$ away. A beam of light consisting of two wavelengths, $650 	ext{ nm}$ and $520 	ext{ nm}$, is used to obtain interference fringes on the screen. The least distance from the common central maximum to the point where the bright fringes due to both the wavelengths coincide, is

Options:

(A) $7.8 	ext{ mm}$

(B) $9.75 	ext{ mm}$

(C) $15.6 	ext{ mm}$

(D) $1.56 	ext{ mm}$

Sigma Percentile · Accepted Answer

Imagine standing in front of a classic Young's Double Slit Experiment setup. Usually, we shine a single, pure monochromatic light—say, a laser pointer—and watch as a beautiful, alternating pattern of bright and dark fringes paints the screen. But what happens if we get a little adventurous? What if we mix two different colors of light and fire them through the slits simultaneously?

This is exactly the scenario we are dealing with in this fascinating problem. We have a beam of light consisting of two distinct wavelengths: a reddish light at $650 	ext{ nm}$ and a greenish light at $520 	ext{ nm}$.

### The Race on the Screen

Think of the interference pattern as a race track. The central maximum, located exactly at the center of the screen ($y = 0$), is the starting line. Because the path difference for both wavelengths is zero at the center, both the red light and the green light will form a bright fringe right there. They start together.

But as we move away from the center, their paths diverge. The position of any bright fringe on the screen is dictated by the formula:
$$y = \frac{n \lambda D}{d}$$

Here, $D$ is the distance to the screen, $d$ is the slit separation, $\lambda$ is the wavelength, and $n$ is the integer number of the fringe (1st, 2nd, 3rd, etc.).

Because the red light has a longer wavelength ($650 	ext{ nm}$) than the green light ($520 	ext{ nm}$), its fringes are spaced further apart. The red fringes take 'longer strides' across the screen, while the green fringes take 'shorter strides'.

### The Point of Coincidence

The question asks for the **least distance** from the central maximum where the bright fringes of both wavelengths coincide again. In our race track analogy, we are looking for the first time the runner with long strides (red) and the runner with short strides (green) land their feet on the exact same spot.

For their bright fringes to overlap perfectly at some distance $y$, their position coordinates must be identical. Let's say the $n_1^{	ext{th}}$ bright fringe of the red light ($\lambda_1$) coincides with the $n_2^{	ext{th}}$ bright fringe of the green light ($\lambda_2$). We can set their position equations equal to each other:
$$y_{n_1} = y_{n_2}$$
$$\frac{n_1 \lambda_1 D}{d} = \frac{n_2 \lambda_2 D}{d}$$

### The Elegance of Ratios

This is where the physics simplifies beautifully. The macroscopic parameters of our setup—the screen distance $D$ and the slit separation $d$—are the same for both colors. They cancel out completely!
$$n_1 \lambda_1 = n_2 \lambda_2$$

This tells us that the product of the fringe number and the wavelength must be constant for a coincidence to occur. Rearranging this to find the ratio of the fringe numbers, we get:
$$\frac{n_1}{n_2} = \frac{\lambda_2}{\lambda_1}$$

Now, let's plug in our specific wavelengths. We have $\lambda_1 = 650 	ext{ nm}$ and $\lambda_2 = 520 	ext{ nm}$.
$$\frac{n_1}{n_2} = \frac{520}{650}$$

To find the *first* point of coincidence (the least distance), we need to reduce this fraction to its simplest integer form. Both 520 and 650 are divisible by 130.
$$\frac{n_1}{n_2} = \frac{4}{5}$$

What a profound result! This simple ratio tells us a complete physical story. It means that exactly at the location of the 4th bright fringe of the red light, the 5th bright fringe of the green light will be sitting right on top of it.

### The Final Calculation

Now that we know *which* fringes coincide, finding *where* they coincide is just a matter of plugging the numbers back into our original position formula. We can use either the red light's parameters or the green light's parameters; they will both give the exact same distance $y$. Let's use the red light ($n_1 = 4$, $\lambda_1 = 650 	ext{ nm}$).

Before we calculate, we must be absolutely rigorous with our units. The slit separation $d$ is given in millimeters, the screen distance $D$ in centimeters, and the wavelength $\lambda$ in nanometers. Let's convert everything to standard SI units (meters) to avoid any silly mistakes.
- $n_1 = 4$
- $\lambda_1 = 650 	imes 10^{-9} 	ext{ m}$
- $D = 150 	ext{ cm} = 1.5 	ext{ m}$
- $d = 0.5 	ext{ mm} = 0.5 	imes 10^{-3} 	ext{ m}$

Substituting these into the formula:
$$y = \frac{n_1 \lambda_1 D}{d}$$
$$y = \frac{4 	imes (650 	imes 10^{-9}) 	imes 1.5}{0.5 	imes 10^{-3}}$$

Let's group the numbers and the powers of 10 to make the mental math easier:
$$y = \left( \frac{4 	imes 650 	imes 1.5}{0.5} ight) 	imes \frac{10^{-9}}{10^{-3}}$$

Notice that $1.5 / 0.5$ is exactly 3.
$$y = (4 	imes 650 	imes 3) 	imes 10^{-6}$$
$$y = (12 	imes 650) 	imes 10^{-6}$$
$$y = 7800 	imes 10^{-6} 	ext{ m}$$

To make this number more intuitive, let's convert it back to millimeters by multiplying by $10^3$:
$$y = 7800 	imes 10^{-6} 	imes 10^3 	ext{ mm}$$
$$y = 7800 	imes 10^{-3} 	ext{ mm}$$
**$$y = 7.8 	ext{ mm}$$**

### The Bigger Picture

The least distance from the central maximum where the two bright fringes coincide is $7.8 	ext{ mm}$. But the physics doesn't stop there. Because the ratio $4/5$ is constant, the next coincidence will happen at the next equivalent fraction, $8/10$. This means the 8th red fringe will coincide with the 10th green fringe at exactly double the distance, $15.6 	ext{ mm}$.

The interference pattern is a beautiful, periodic tapestry. By understanding the simple mathematical ratios that govern the waves, we can predict exactly where these moments of perfect alignment will occur across the entire screen.

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Visualized Solution

$YDSE Setup$

$Condition for Coincidence$

$Equating Path Differences$

$Finding the Fringe Ratio$

$Calculating the Distance y$

$Final Result$

$Further Exploration$

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

The Race on the Screen

The Point of Coincidence

The Elegance of Ratios

The Final Calculation

The Bigger Picture

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Visualized Solution

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

.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; } Equating Path Differences

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

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

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

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