The angular width of the central maximum in a single slit diffraction pattern is `60^(@)`. The width of the slit is `1 mu 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. `100 mum`
B. `25mum`
C. `50mum`
D. `75mum`

Answer

To solve the problem, we will follow these steps:

### Step 1: Understanding the Angular Width of the Central Maximum
The angular width of the central maximum is given as \(60^\circ\). The half-angle is:
\[ \theta = \frac{60^\circ}{2} = 30^\circ \]

### Step 2: Relating Angular Width to Wavelength
The angular width of the central maximum in a single slit diffraction pattern is related to the width of the slit (B) and the wavelength (\(\lambda\)) by the formula:
\[ B \sin(\theta) = n \lambda \]
where \(n = 1\) for the first minimum.

### Step 3: Converting Units
The width of the slit is given as \(1 \mu m = 1 \times 10^{-6} m\).

### Step 4: Calculate Wavelength
Substituting the known values into the equation:
\[ 1 \times 10^{-6} \sin(30^\circ) = 1 \lambda \]
Since \(\sin(30^\circ) = \frac{1}{2}\), we have:
\[ 1 \times 10^{-6} \cdot \frac{1}{2} = \lambda \]
\[ \lambda = 0.5 \times 10^{-6} m = 5 \times 10^{-7} m \]

### Step 5: Using Young's Fringe Width Formula
The fringe width (\(w\)) in Young's double slit experiment is given by:
\[ w = \frac{\lambda D}{d} \]
where:
- \(w\) = fringe width
- \(D\) = distance from the slits to the screen
- \(d\) = distance between the slits

Given:
- \(w = 1 cm = 0.01 m\)
- \(D = 50 cm = 0.5 m\)

### Step 6: Rearranging to Find Slit Separation (d)
Rearranging the fringe width formula to solve for \(d\):
\[ d = \frac{\lambda D}{w} \]

### Step 7: Substituting Known Values
Substituting the values we have:
\[ d = \frac{5 \times 10^{-7} m \cdot 0.5 m}{0.01 m} \]
\[ d = \frac{2.5 \times 10^{-7}}{0.01} = 2.5 \times 10^{-5} m = 25 \mu m \]

### Final Answer
The slit separation distance is \(25 \mu m\).

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