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 \(60^\circ\). The angular width is defined as \(2\theta\), where \(\theta\) is the angle to the first minimum. Thus, we have:

\[ \theta = \frac{60^\circ}{2} = 30^\circ \]

Using the formula for the first minimum in single slit diffraction:

\[ A \sin \theta = n \lambda \]

For the first minimum (\(n = 1\)), the width of the slit \(A\) is given as \(1 \, \mu m = 1 \times 10^{-6} \, m\). Therefore:

\[ 1 \times 10^{-6} \sin(30^\circ) = \lambda \]

Since \(\sin(30^\circ) = \frac{1}{2}\):

\[ \lambda = 1 \times 10^{-6} \times \frac{1}{2} = 5 \times 10^{-7} \, m = 500 \, nm \]

### Step 2: Use the fringe width formula to find the slit separation distance.
In a double slit experiment, the fringe width \(w\) is given by:

\[ w = \frac{\lambda D}{d} \]

Where:
- \(w\) 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 distance between the slits (which we need to find).

Rearranging the formula to solve for \(d\):

\[ d = \frac{\lambda D}{w} \]

### Step 3: Substitute the known values into the equation.
Substituting the values we have:

\[ d = \frac{(5 \times 10^{-7} \, m)(0.5 \, m)}{0.01 \, m} \]

Calculating the numerator:

\[ 5 \times 10^{-7} \times 0.5 = 2.5 \times 10^{-7} \, m \]

Now substituting this back into the equation for \(d\):

\[ d = \frac{2.5 \times 10^{-7}}{0.01} = 2.5 \times 10^{-5} \, m = 25 \, \mu m \]

### Final Answer:
The distance between the centers of each slit (slit separation distance) is \(25 \, \mu m\).

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