A parallel beam of light of wavelenght `600 nm` is incident normally on a slit of width `'d'`. If the distance between the slit and screen is `0.8 m` and distance of 2nd order maximum from the centre of the screen is `15mm`, calculate the width of the slit.

Explanation

To solve the problem, we need to use the formula for the position of the maxima in a single-slit diffraction pattern. The position of the m-th order maximum on the screen can be described by the formula:

\[ D \sin \theta = \frac{(2n + 1) \lambda}{2} \]

where:
- \( D \) is the distance from the slit to the screen,
- \( \theta \) is the angle of the maximum,
- \( n \) is the order of the maximum,
- \( \lambda \) is the wavelength of the light.

### Step 1: Identify the given values
- Wavelength, \( \lambda = 600 \, \text{nm} = 600 \times 10^{-9} \, \text{m} \)
- Distance from the slit to the screen, \( D = 0.8 \, \text{m} \)
- Distance of the 2nd order maximum from the center of the screen, \( x = 15 \, \text{mm} = 15 \times 10^{-3} \, \text{m} \)
- Order of maximum, \( n = 2 \)

### Step 2: Calculate the angle \( \theta \)
Using the small angle approximation, where \( \sin \theta \approx \tan \theta \approx \frac{x}{D} \), we can find \( \sin \theta \):

\[ \sin \theta \approx \frac{x}{D} = \frac{15 \times 10^{-3}}{0.8} \]

Calculating this gives:

\[ \sin \theta \approx \frac{15 \times 10^{-3}}{0.8} = 0.01875 \]

### Step 3: Substitute into the maxima formula
Now we substitute \( n = 2 \) into the maxima formula:

\[ D \sin \theta = \frac{(2 \cdot 2 + 1) \lambda}{2} \]

This simplifies to:

\[ D \sin \theta = \frac{5 \lambda}{2} \]

### Step 4: Solve for the slit width \( d \)
Now, substituting \( \sin \theta \) back into the equation:

\[ 0.8 \cdot 0.01875 = \frac{5 \cdot (600 \times 10^{-9})}{2} \]

Calculating the left side:

\[ 0.8 \cdot 0.01875 = 0.015 \]

And the right side:

\[ \frac{5 \cdot (600 \times 10^{-9})}{2} = \frac{3000 \times 10^{-9}}{2} = 1500 \times 10^{-9} = 1.5 \times 10^{-6} \]

### Step 5: Equate and solve for \( d \)
Now we equate both sides:

\[ 0.015 = 1.5 \times 10^{-6} \cdot d \]

Solving for \( d \):

\[ d = \frac{0.015}{1.5 \times 10^{-6}} = 10000 \, \text{m} = 10 \, \mu m \]

### Final Answer
Thus, the width of the slit \( d \) is:

\[ d = 10 \, \mu m \]