Visible light of wavelength 500 nm falls normally on a single slit and produces a diffraction pattern. It is found that the diffraction pattern is on a screen 1 m away from slit. If the first minimum is produced at a distance of 2.5 mm from the centre of screen, then the width of the slit is

Options

A. 0.1mm
B. 0.4mm
C. 0.3mm
D. 0.2mm

Explanation

To solve the problem, we will follow these steps:

### Step 1: Understand the given data
- Wavelength of light, \( \lambda = 500 \, \text{nm} = 500 \times 10^{-9} \, \text{m} \)
- Distance from the slit to the screen, \( D = 1 \, \text{m} \)
- Distance of the first minimum from the center of the screen, \( y = 2.5 \, \text{mm} = 2.5 \times 10^{-3} \, \text{m} \)

### Step 2: Use the formula for the position of the first minimum
For a single slit diffraction pattern, the position of the first minimum is given by:
\[ a \sin \theta = n \lambda \]
where \( n = 1 \) for the first minimum.

### Step 3: Relate \( \sin \theta \) to \( y \) and \( D \)
Using the small angle approximation, we can say:
\[ \sin \theta \approx \tan \theta = \frac{y}{D} \]
Thus, we can rewrite the equation as:
\[ a \frac{y}{D} = \lambda \]

### Step 4: Rearranging the equation to find \( a \)
From the equation above, we can solve for \( a \):
\[ a = \frac{\lambda D}{y} \]

### Step 5: Substitute the known values
Now, substituting the values we have:
\[ a = \frac{(500 \times 10^{-9} \, \text{m})(1 \, \text{m})}{2.5 \times 10^{-3} \, \text{m}} \]

### Step 6: Calculate \( a \)
Calculating the above expression:
\[ a = \frac{500 \times 10^{-9}}{2.5 \times 10^{-3}} = 200 \times 10^{-6} \, \text{m} \]

### Step 7: Convert to millimeters
To convert to millimeters:
\[ a = 200 \times 10^{-6} \, \text{m} = 0.2 \, \text{mm} \]

### Final Answer
The width of the slit is \( 0.2 \, \text{mm} \).

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