A fringe width of 6 mm was produced for two slits separated by 1 mm apart. The screen is placed 10 m

Q: Why do we have to convert millimeters to meters? Can't we just calculate it directly?

If you don't convert to a consistent unit system (like SI meters), your final answer will have mixed, nonsensical units. For instance, multiplying mm by mm and dividing by meters gives $\text{mm}^2/\text{m}$, which is not a standard unit for wavelength. Converting everything to meters ensures your final answer is strictly in meters.

Q: What is the difference between slit separation ($d$) and screen distance ($D$)?

Slit separation ($d$) is the tiny microscopic distance between the two holes the light passes through. Screen distance ($D$) is the macroscopic distance from those slits to the wall where you observe the pattern. In YDSE, $D$ is always much, much larger than $d$.

Q: How did $6 \times 10^{-7} \text{ m}$ become $600 \text{ nm}$?

A nanometer is $10^{-9}$ meters. To convert $6 \times 10^{-7}$ to something times $10^{-9}$, we can multiply the number by $100$ and divide the power by $100$ (which is $10^{-2}$). So, $6 \times 100 = 600$, and $10^{-7} \times 10^{-2} = 10^{-9}$. Thus, $600 \times 10^{-9} \text{ m}$, which is $600 \text{ nm}$.

Q: Does the fringe width change if we use a different color of light?

Absolutely! The formula $\beta = \frac{\lambda D}{d}$ shows that fringe width $\beta$ is directly proportional to wavelength $\lambda$. Red light has a longer wavelength than blue light, so red fringes will be wider and more spread out than blue fringes.

Q: What happens if we move the screen further away?

If you increase the screen distance $D$, the formula $\beta = \frac{\lambda D}{d}$ tells us that the fringe width $\beta$ will also increase. The pattern will simply stretch out and become larger on the screen.

Question

A fringe width of $6 	ext{ mm}$ was produced for two slits separated by $1 	ext{ mm}$ apart. The screen is placed $10 	ext{ m}$ away. The wavelength of light used is $x 	ext{ nm}$. The value of $x$ to the nearest integer is .........

Sigma Percentile · Accepted Answer

### The Setup: Visualizing the Interference

Imagine you are standing in a dark room observing the classic Young's Double Slit Experiment (YDSE). Light passes through two incredibly narrow slits, $S_1$ and $S_2$, separated by a tiny distance $d = 1 	ext{ mm}$. This light travels a massive distance $D = 10 	ext{ m}$ to hit a screen, creating a beautiful pattern of alternating bright and dark bands.

We are given that the width of one of these fringes, denoted by $\beta$, is $6 	ext{ mm}$. Our mission is to work backward from this visible pattern to uncover the invisible: the exact wavelength $\lambda$ of the light being used.

### The Master Equation: Connecting the Dots

To solve this, we need the mathematical bridge that connects the macroscopic world of the screen to the microscopic world of the light waves. That bridge is the fringe width formula:

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

This elegant equation tells us that the fringe width $\beta$ is directly proportional to the wavelength $\lambda$ and the screen distance $D$, but inversely proportional to the slit separation $d$. Since we are hunting for the wavelength, let's rearrange this formula to make $\lambda$ the subject:

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

### The Crucial Step: Unit Conversion

Before we rush into plugging in the numbers, we must heed a critical warning: **Unit Consistency**. Physics equations demand that all variables speak the same language, which in the SI system is meters.

Let's translate our given values:
- Fringe width: $\beta = 6 	ext{ mm} = 6 	imes 10^{-3} 	ext{ m}$
- Slit separation: $d = 1 	ext{ mm} = 10^{-3} 	ext{ m}$
- Screen distance: $D = 10 	ext{ m}$ (already in meters!)

### The Final Calculation

Now, we substitute these pristine, meter-converted values into our rearranged equation:

$$ \lambda = \frac{(6 	imes 10^{-3}) 	imes (10^{-3})}{10} $$

Multiplying the terms in the numerator, we add the exponents: $-3 + (-3) = -6$. This gives us:

$$ \lambda = \frac{6 	imes 10^{-6}}{10} $$

Dividing by $10$ simply subtracts another $1$ from the exponent, yielding:

$$ \lambda = 6 	imes 10^{-7} 	ext{ m} $$

We have the wavelength, but the question specifically asks for it in nanometers ($	ext{nm}$). Recall that $1 	ext{ nm} = 10^{-9} 	ext{ m}$. To convert our answer, we can mathematically manipulate the expression by multiplying and dividing by $100$:

$$ \lambda = 600 	imes 10^{-9} 	ext{ m} = 600 	ext{ nm} $$

Comparing this to the given format $x 	ext{ nm}$, we can confidently state that **$x = 600$**.

Animated Solution for Physics - Optics: A fringe width of $6 mm$ was produced for two slits separated by $1 mm$ apart. The screen is placed $10 m$ away. The wavelength of light used is $x nm$ . The value of $x$ to the nearest integer is .........

Enter Numerical Value:

Visualized Solution

\text{YDSE Setup}

\text{Fringe Width Formula}

\text{Substitution}

\text{Calculation}

\text{Conversion to nm}

\text{Food for Thought}

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

The Setup

The Master Equation

The Crucial Step

The Final Calculation

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