Interference fringes are observed on a screen by illuminating two thin slits 1 mm apart with a light

Q: Why didn't we use the wavelength $\lambda$ in the main calculation?

The wavelength is not needed to find the path difference if the physical position $y$, slit separation $d$, and screen distance $D$ are already given. The wavelength can be used to verify which order of bright fringe it is.

Q: What if the point was a dark fringe instead of a bright fringe?

The geometric formula $\Delta x = \frac{y \cdot d}{D}$ still perfectly holds for the path difference regardless of whether it's a bright or dark fringe. The nature of the fringe only determines if $\Delta x$ is a whole or half-integer multiple of $\lambda$.

Q: Why do we use the approximation $\sin\theta \approx \tan\theta$?

In a typical YDSE setup, the distance to the screen $D$ is much larger than the distance $y$ on the screen ($D \gg y$). This makes the angle $\theta$ very small, allowing us to safely use the small-angle approximation.

Q: How do we convert the final answer to micrometers?

The prefix 'micro' ($\mu$) stands for $10^{-6}$. So, if your answer is $1.27 \times 10^{-6} \text{ m}$, you can directly replace the $10^{-6} \text{ m}$ part with $\mu\text{m}$, giving $1.27 \mu\text{m}$.

Q: What does the path difference physically mean?

It represents the extra distance that the light wave from one slit has to travel compared to the light wave from the other slit to reach the exact same point on the screen.

Question

Interference fringes are observed on a screen by illuminating two thin slits $1	ext{ mm}$ apart with a light source ($\lambda =632.8	ext{ nm}$). The distance between the screen and the slits is $100	ext{ cm}$. If a bright fringe is observed on a screen at a distance of $1.27	ext{ mm}$ from the central bright fringe, then the path difference between the waves, which are reaching this point from the slits is close to

Options:

(A) $2	ext{ nm}$

(B) $1.27\mu	ext{m}$

(C) $2.87	ext{ nm}$

(D) $2.05\mu	ext{m}$

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Imagine you are standing in a dark room, observing the classic Young's Double Slit Experiment. We have two tiny slits, $S_1$ and $S_2$, separated by a microscopic distance $d = 1 	ext{ mm}$.

A screen is placed far away at a distance $D = 100 	ext{ cm}$ (which is $1 	ext{ m}$). When coherent light passes through these slits, it creates a beautiful pattern of bright and dark fringes on the screen.

We are told to focus on a specific bright fringe located at point $P$, which is at a distance $y = 1.27 	ext{ mm}$ from the central maximum. Our goal is to find the path difference between the two light waves reaching this exact point.

### The Master Equation

To find the path difference, denoted as $\Delta x$, we need to understand the geometry of the setup. The path difference is simply the extra distance the wave from the lower slit ($S_2$) has to travel compared to the wave from the upper slit ($S_1$).

Mathematically, this is written as $\Delta x = S_2P - S_1P$.

For a setup where the screen is very far away compared to the slit separation ($D \gg d$), the rays are almost parallel. This allows us to use a brilliant geometric approximation:

$$ \Delta x = d \sin	heta \approx d 	an	heta $$

Since $	an	heta = \frac{y}{D}$, our master equation becomes:

$$ \Delta x = \frac{y \cdot d}{D} $$

### Final Calculation

Now, it's time to plug in the numbers. But beware, this is where silly mistakes happen! We must ensure all our units are consistent. Let's convert everything into standard SI units (meters).

We have $y = 1.27 	imes 10^{-3} 	ext{ m}$, $d = 1 	imes 10^{-3} 	ext{ m}$, and $D = 1 	ext{ m}$. Substituting these into our master equation:

$$ \Delta x = \frac{(1.27 	imes 10^{-3}) \cdot (10^{-3})}{1} $$

Multiplying the terms in the numerator, we get:

$$ \Delta x = 1.27 	imes 10^{-6} 	ext{ m} $$

To match our options, we need to convert this back into a more convenient unit. We know that $10^{-6} 	ext{ m}$ is exactly $1 \mu	ext{m}$ (one micrometer).

**Therefore, the path difference is $1.27 \mu	ext{m}$.**

### The Redundant Wavelength

You might be wondering, why did the problem provide the wavelength $\lambda = 632.8 	ext{ nm}$? Was it a trick?

In physics problems, sometimes extra information is given to test your confidence in the core concepts. We didn't need the wavelength to find the path difference because we already knew the exact physical location ($y$) of the fringe.

However, we can use it to verify our answer! Since it's a bright fringe, the path difference must be an integer multiple of the wavelength ($\Delta x = n\lambda$).

If we divide our path difference by the wavelength:

$$ n = \frac{1.27 	imes 10^{-6} 	ext{ m}}{632.8 	imes 10^{-9} 	ext{ m}} \approx 2 $$

This tells us that the fringe at $y = 1.27 	ext{ mm}$ is exactly the **second bright fringe**. The physics is perfectly consistent!

Select Answer:

Visualized Solution

$YDSE Setup$

$Path Difference Formula$

$Substitution$

$Calculation$

$Final Result$

$Checking the Bright Fringe Condition$

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

Analyzing the Setup

The Master Equation

Final Calculation

The Redundant Wavelength

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