White light is passed through a double slit and interference is observed on a screen 1.5 m away. The

Q: Why is the central fringe white?

At the central position, the path difference for all wavelengths is zero. Since all colors interfere constructively at this point, they recombine to form white light.

Q: Why does the red fringe form further away than the violet fringe?

The position of a bright fringe is directly proportional to its wavelength ($y \propto \lambda$). Since red light has a longer wavelength than violet light, its fringes are spread further apart.

Q: Do I have to convert all units to meters?

Yes! Mixing millimeters and meters is a classic trap. Always convert everything to standard SI units (meters) before calculating, and then convert to the requested unit (nanometers) at the very end.

Q: What does $n=1$ mean in this context?

The integer $n$ represents the order of the fringe. $n=0$ is the central maximum, $n=1$ is the first bright fringe on either side, $n=2$ is the second, and so on.

Q: Could I have calculated the wavelengths separately and then subtracted them?

Absolutely! However, factoring out the common terms $\frac{d}{D}$ first makes the calculation much faster and reduces the chance of arithmetic errors.

Question

White light is passed through a double slit and interference is observed on a screen $1.5	ext{ m}$ away. The separation between the slits is $0.3	ext{ mm}$. The first violet and red fringes are formed $2.0	ext{ mm}$ and $3.5	ext{ mm}$ away from the central white fringes. The difference in wavelengths of red and violet light is ......... $	ext{nm}$.

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Imagine you are standing in a dark room, observing a classic Young's Double Slit Experiment. But instead of a single monochromatic laser, we are shining pure **white light** through the two slits.

Because white light is a beautiful mixture of all the colors in the visible spectrum, something fascinating happens on the screen. At the exact center, the path difference for every single color is zero. They all interfere constructively, recombining to form a brilliant **central white fringe**.

However, as we move away from the center, the colors begin to separate. Why? Because each color has a different wavelength, and therefore, a different fringe width. The problem tells us that the first violet fringe appears at $2.0 	ext{ mm}$, and the first red fringe appears further out at $3.5 	ext{ mm}$.

### The Master Equation

To unlock the wavelengths, we need our trusty logic bridge. The position of the $n^{	ext{th}}$ bright fringe from the central maximum is given by the formula:

$$y_n = \frac{n\lambda D}{d}$$

Since we are dealing with the **first** fringes for both colors, we can simply set $n = 1$. Rearranging this equation to solve for the wavelength $\lambda$, we get:

$$\lambda = \frac{y d}{D}$$

This elegant little equation tells us exactly how to find the wavelength if we know where the fringe is located!

### Setting Up the Difference

The question doesn't just ask for one wavelength; it asks for the **difference** between the red and violet wavelengths, which we can call $\Delta\lambda$.

$$\Delta\lambda = \lambda_{	ext{red}} - \lambda_{	ext{violet}}$$

Instead of calculating them separately and doing twice the work, let's be smart and substitute our rearranged formula:

$$\Delta\lambda = \frac{y_{	ext{red}} d}{D} - \frac{y_{	ext{violet}} d}{D}$$

Notice that both terms share $\frac{d}{D}$. Let's factor that out to make our lives easier:

$$\Delta\lambda = \frac{(y_{	ext{red}} - y_{	ext{violet}})d}{D}$$

### Final Calculation

Now, it's time to plug in the raw numbers. But beware of the classic physics trap: **units!** We must convert everything to standard SI units (meters) before we calculate.

The difference in positions is $3.5 	ext{ mm} - 2.0 	ext{ mm} = 1.5 	ext{ mm}$, which is $1.5 	imes 10^{-3} 	ext{ m}$. The slit separation $d$ is $0.3 	ext{ mm}$, or $0.3 	imes 10^{-3} 	ext{ m}$. The screen distance $D$ is already in meters: $1.5 	ext{ m}$.

Let's substitute these into our equation:

$$\Delta\lambda = \frac{(1.5 	imes 10^{-3}) 	imes (0.3 	imes 10^{-3})}{1.5}$$

Look at how beautifully the math works out! The $1.5$ in the numerator perfectly cancels the $1.5$ in the denominator. We are left with:

$$\Delta\lambda = 0.3 	imes 10^{-6} 	ext{ m}$$

The question asks for the answer in nanometers ($	ext{nm}$), which is $10^{-9} 	ext{ m}$. To convert, we can multiply by $1000$ and adjust the exponent:

$$\Delta\lambda = 300 	imes 10^{-9} 	ext{ m} = 300 	ext{ nm}$$

And there we have it! A perfectly clean integer answer. The difference in wavelengths between the red and violet light is exactly **$300 	ext{ nm}$**.

Enter Numerical Value:

Visualized Solution

$YDSE Setup with White Light$

$Position of Fringes$

$Bright Fringe Formula$

$Wavelength Expression$

$Difference in Wavelengths$

$Substitution of Values$

$Calculation$

$Final Answer$

$The Way Forward$

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

Analyzing the Setup

The Master Equation

Setting Up the Difference

Final Calculation

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Enter Numerical Value:

Visualized Solution

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

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

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

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

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

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

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

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

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