Light of wavelength `6000A^(0)` is incident on a single slit. The first minimum of the diffraction pattern is obtained at 4 mm from the centre. The screen is at a distance of 2m from the slit. The slit width will be

Options

A. `0.3 mm`
B. `0.2 mm`
C. `0.15 mm`
D. `0.1 mm`

Answer

To find the slit width \( d \) for the given problem, we can use the formula for the position of the first minimum in a single slit diffraction pattern:

\[ x = \frac{n \lambda D}{d} \]

Where:
- \( x \) = position of the first minimum from the center of the pattern
- \( n \) = order of the minimum (for the first minimum, \( n = 1 \))
- \( \lambda \) = wavelength of the light
- \( D \) = distance from the slit to the screen
- \( d \) = width of the slit

### Step 1: Identify the given values
- Wavelength \( \lambda = 6000 \, \text{Å} = 6000 \times 10^{-10} \, \text{m} \)
- Distance to the screen \( D = 2 \, \text{m} \)
- Position of the first minimum \( x = 4 \, \text{mm} = 4 \times 10^{-3} \, \text{m} \)
- Order of the minimum \( n = 1 \)

### Step 2: Rearrange the formula to solve for \( d \)
We can rearrange the formula to find \( d \):

\[ d = \frac{n \lambda D}{x} \]

### Step 3: Substitute the known values into the equation
Now we substitute the known values into the equation:

\[ d = \frac{1 \times (6000 \times 10^{-10} \, \text{m}) \times (2 \, \text{m})}{4 \times 10^{-3} \, \text{m}} \]

### Step 4: Calculate the value of \( d \)
Calculating the numerator:

\[ 1 \times (6000 \times 10^{-10}) \times 2 = 12000 \times 10^{-10} \, \text{m} = 1.2 \times 10^{-6} \, \text{m} \]

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

\[ d = \frac{1.2 \times 10^{-6}}{4 \times 10^{-3}} = \frac{1.2}{4} \times 10^{-3} = 0.3 \times 10^{-3} \, \text{m} = 0.3 \, \text{mm} \]

### Final Answer
The width of the slit \( d \) is \( 0.3 \, \text{mm} \).

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