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 mu m`
B. `25mu m`
C. `50mu m`
D. `75 mu m`

Answer

To solve the problem, we will follow these steps:

### Step 1: Determine the wavelength of the light used
The angular width of the central maximum in a single slit diffraction pattern is given as \(60^\circ\). This means that the angle \(\theta\) for the first minimum on either side is:
\[ \theta = \frac{60^\circ}{2} = 30^\circ \]

Using the formula for the first minimum in single slit diffraction:
\[ A \sin(\theta) = n \lambda \]
where \(A\) is the width of the slit, \(n\) is the order of the minimum (1 for the first minimum), and \(\lambda\) is the wavelength.

Given:
- Width of the slit \(A = 1 \, \mu m = 1 \times 10^{-6} \, m\)
- For the first minimum, \(n = 1\)

Substituting the values:
\[ 1 \times 10^{-6} \sin(30^\circ) = 1 \lambda \]
\[ 1 \times 10^{-6} \cdot \frac{1}{2} = \lambda \]
\[ \lambda = \frac{1 \times 10^{-6}}{2} = 0.5 \times 10^{-6} \, m = 0.5 \, \mu m \]

### Step 2: Use the wavelength to find the slit separation distance
In Young's double slit experiment, the fringe width \(\beta\) is given by:
\[ \beta = \frac{\lambda D}{d} \]
where:
- \(\beta\) is the fringe width (given as \(1 \, cm = 0.01 \, m\))
- \(D\) is the distance from the slits to the screen (given as \(50 \, cm = 0.5 \, m\))
- \(d\) is the separation between the slits (which we need to find)

Rearranging the formula to find \(d\):
\[ d = \frac{\lambda D}{\beta} \]

Substituting the known values:
\[ d = \frac{(0.5 \times 10^{-6} \, m)(0.5 \, m)}{0.01 \, m} \]
\[ d = \frac{0.25 \times 10^{-6}}{0.01} \]
\[ d = 0.25 \times 10^{-4} \, m = 2.5 \times 10^{-5} \, m = 25 \, \mu m \]

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

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