Fraunhoffer diffraction pattern of a single slit is obtained in the focal plane of lens of focal length `1m`. If third minimum is formed at a distance of 5mm from the central maximum and wavelength of light used is `5000Å`, then width of the slit will be –

Answer

To find the width of the slit in a single slit Fraunhofer diffraction pattern, we can use the formula for the position of the minima in the diffraction pattern. The position of the nth minimum in a single slit diffraction pattern is given by:

\[ y_n = \frac{n \lambda D}{a} \]

where:
- \(y_n\) is the distance of the nth minimum from the central maximum,
- \(n\) is the order of the minimum (for the third minimum, \(n = 3\)),
- \(\lambda\) is the wavelength of light,
- \(D\) is the distance from the slit to the screen (which is the focal length of the lens in this case),
- \(a\) is the width of the slit.

### Step-by-Step Solution:

1. **Identify the Given Values**:
- Wavelength, \(\lambda = 5000 \, \text{Å} = 5000 \times 10^{-10} \, \text{m} = 5 \times 10^{-7} \, \text{m}\)
- Distance of the third minimum from the central maximum, \(y_3 = 5 \, \text{mm} = 5 \times 10^{-3} \, \text{m}\)
- Focal length of the lens, \(D = 1 \, \text{m}\)
- Order of the minimum, \(n = 3\)

2. **Use the Formula for the nth Minimum**:
\[ y_n = \frac{n \lambda D}{a} \]
Rearranging the formula to find the slit width \(a\):
\[ a = \frac{n \lambda D}{y_n} \]

3. **Substitute the Values**:
\[ a = \frac{3 \times (5 \times 10^{-7} \, \text{m}) \times (1 \, \text{m})}{5 \times 10^{-3} \, \text{m}} \]

4. **Calculate the Numerator**:
\[ 3 \times 5 \times 10^{-7} \times 1 = 15 \times 10^{-7} \, \text{m} \]

5. **Calculate the Slit Width**:
\[ a = \frac{15 \times 10^{-7} \, \text{m}}{5 \times 10^{-3} \, \text{m}} = \frac{15}{5} \times 10^{-4} \, \text{m} = 3 \times 10^{-4} \, \text{m} \]

6. **Convert to mm**:
\[ a = 3 \times 10^{-4} \, \text{m} = 0.3 \, \text{mm} \]

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