The angular width of the central maximum in a single slit diffraction pattern is 60 degrees. The wid

Q: Why is the angular width of the central maximum taken as $2\theta$?

The central maximum in a single slit diffraction pattern extends from the first minimum on one side to the first minimum on the other side. If the angular position of the first minimum is $\theta$, the total angular spread is $\theta - (-\theta) = 2\theta$.

Q: Why do we use the formula $a \sin\theta = \lambda$ instead of $a \sin\theta = 1.22\lambda$?

The formula $a \sin\theta = \lambda$ is the exact condition for the first minimum in single slit diffraction. The factor $1.22$ appears in the Rayleigh criterion for the resolution of a circular aperture, which is a different scenario.

Q: Can we assume the wavelength remains the same when the second slit is added?

Yes! The problem states 'another slit of same width is made near it', which implies we are modifying the physical setup while keeping the same monochromatic light source. Therefore, the wavelength $\lambda$ is constant.

Q: What is the difference between $a$ and $d$ in this problem?

$a$ represents the width of a single slit, which is responsible for the diffraction envelope. $d$ represents the distance between the centers of the two slits, which determines the spacing of the interference fringes in the YDSE pattern.

Q: Why did we convert $1 \text{ cm}$ to $10^{-2} \text{ m}$ and $1 \mu\text{m}$ to $10^{-6} \text{ m}$?

In physics, it is crucial to maintain dimensional consistency. Mixing units like centimeters, micrometers, and meters in the same equation will lead to incorrect results. Converting everything to the base SI unit (meters) ensures accuracy.

Question

The angular width of the central maximum in a single slit diffraction pattern is $60^\circ$. The width of the slit is $1\mu	ext{m}$. The slit is illuminated by monochromatic plane waves. If another slit of same width is made near it, Young's fringes can be observed on a screen placed at a distance 50 cm from the slits. If the observed fringe width is 1 cm, what is slit separation distance? (i.e. distance between the centres of each slit.)

Options:

(A) $25\mu	ext{m}$

(B) $50\mu	ext{m}$

(C) $75\mu	ext{m}$

(D) $100\mu	ext{m}$

Sigma Percentile · Accepted Answer

This problem is a beautiful symphony of two fundamental wave optics phenomena: Single Slit Diffraction and Young's Double Slit Experiment (YDSE). It tests your ability to extract a hidden parameter from one setup and seamlessly apply it to another. Let's break down the physics step-by-step.

### Phase 1: Decoding the Single Slit

The problem begins by describing a single slit diffraction pattern. We are given that the angular width of the central maximum is $60^\circ$.

What exactly is the central maximum? It is the bright, wide band of light in the center of the diffraction pattern, bounded on both sides by the first dark fringes (minima). If the angular position of the first minimum is $	heta$, the total angular width is $2	heta$.

Therefore, we can write:

$$2	heta = 60^\circ \implies 	heta = 30^\circ$$

Now, we use the fundamental condition for the first minimum in a single slit diffraction pattern:

$$a \sin	heta = \lambda$$

Here, $a$ is the width of the slit, which is given as $1\mu	ext{m}$ or $10^{-6} 	ext{ m}$. Substituting our known values, we can find the wavelength $\lambda$ of the monochromatic light:

$$\lambda = (10^{-6} 	ext{ m}) 	imes \sin(30^\circ)$$

Since $\sin(30^\circ) = 0.5$, we get:

$$\lambda = 0.5 	imes 10^{-6} 	ext{ m}$$

This wavelength is the critical bridge that connects the first part of the problem to the second.

### Phase 2: Transition to YDSE

Next, the problem states that another identical slit is placed near the first one. This transforms our single slit setup into a classic Young's Double Slit Experiment (YDSE).

Because the light source hasn't changed, the wavelength $\lambda$ remains exactly the same. We are given two new pieces of information for this setup:
1. The distance to the screen, $D = 50 	ext{ cm} = 0.5 	ext{ m}$.
2. The observed fringe width, $\beta = 1 	ext{ cm} = 10^{-2} 	ext{ m}$.

Our goal is to find the separation distance between the centers of the two slits, denoted by $d$.

### Phase 3: The Final Calculation

In YDSE, the fringe width $\beta$ is governed by the formula:

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

We can rearrange this equation to solve for the unknown slit separation $d$:

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

Now, we carefully substitute all our values, ensuring everything is in standard SI units (meters) to avoid any silly mistakes:

$$d = \frac{(0.5 	imes 10^{-6} 	ext{ m}) 	imes (0.5 	ext{ m})}{10^{-2} 	ext{ m}}$$

Let's compute the numerator first:

$$d = \frac{0.25 	imes 10^{-6}}{10^{-2}}$$

Dividing by $10^{-2}$ is equivalent to multiplying by $10^2$:

$$d = 0.25 	imes 10^{-4} 	ext{ m}$$

To match the options provided, we convert this back into micrometers ($\mu	ext{m}$):

$$d = 25 	imes 10^{-6} 	ext{ m} = 25 \mu	ext{m}$$

**The distance between the centers of the two slits is $25 \mu	ext{m}$.** This perfectly matches option (a). By methodically extracting the wavelength from the diffraction data and applying it to the interference formula, we've elegantly solved the problem!

Select Answer:

Visualized Solution

Single Slit Setup

First Minima Condition

Calculating Wavelength

YDSE Setup

Fringe Width Formula

Substituting Values

Final Calculation

Conclusion

The Sigma Insight: Diffraction due to Single Slit

Phase 1

Phase 2

Phase 3

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