A parallel beam of light of wavelength 600 nm is incident normally on a slit of width d. If the distance between the slits and the screen is 0.8 m and the distance of `2^(nd)` order maximum from the centre of the screen is 15 mm. The width of the slit is

Options

A. `40 mu m`
B. `80 mu m`
C. `160 mu m`
D. `200 mu m`

Explanation

To solve the problem, we need to find the width of the slit (d) given the parameters of the light wave and the diffraction pattern observed on the screen.

### Step-by-Step Solution:

1. **Understand the Given Information:**
- Wavelength of light, \( \lambda = 600 \, \text{nm} = 600 \times 10^{-9} \, \text{m} \)
- Distance from the slit to the screen, \( D = 0.8 \, \text{m} \)
- Distance of the second order maximum from the center of the screen, \( x = 15 \, \text{mm} = 15 \times 10^{-3} \, \text{m} \)
- Order of maximum, \( n = 2 \)

2. **Use the Formula for the Position of Maxima:**
The position of the nth order maximum in a single-slit diffraction pattern is given by:
\[ x = \frac{n \lambda D}{d} \]
where:
- \( x \) is the distance from the center to the nth order maximum,
- \( n \) is the order of the maximum,
- \( \lambda \) is the wavelength of light,
- \( D \) is the distance from the slit to the screen,
- \( d \) is the width of the slit.

3. **Rearranging the Formula to Find d:**
We can rearrange the formula to solve for \( d \):
\[ d = \frac{n \lambda D}{x} \]

4. **Substituting the Known Values:**
Now we substitute the known values into the equation:
\[ d = \frac{2 \times (600 \times 10^{-9} \, \text{m}) \times (0.8 \, \text{m})}{15 \times 10^{-3} \, \text{m}} \]

5. **Calculating the Value:**
- Calculate the numerator:
\[ 2 \times 600 \times 10^{-9} \times 0.8 = 960 \times 10^{-9} \, \text{m} \]
- Now divide by \( 15 \times 10^{-3} \):
\[ d = \frac{960 \times 10^{-9}}{15 \times 10^{-3}} = \frac{960}{15} \times 10^{-6} = 64 \times 10^{-6} \, \text{m} = 64 \, \mu m \]

6. **Final Result:**
The width of the slit is:
\[ d = 64 \, \mu m \]