Light of wavelength `5000 A^(@)` is diffracted by a slit. In diffraction pattern fifth minimum is at a distance of 5 mm from central maximum. If the distance between the screen and the slit is 1m. The slit width is

Options

A. 0.5 mm
B. 0.55mm
C. 0.25mm
D. 0.6 mm

Answer

To solve the problem of finding the slit width given the diffraction pattern, we will follow these steps:

### Step 1: Understand the Given Data
- Wavelength of light, \( \lambda = 5000 \, \text{Å} = 5000 \times 10^{-10} \, \text{m} = 5 \times 10^{-7} \, \text{m} \)
- Distance of the 5th minimum from the central maximum, \( x_5 = 5 \, \text{mm} = 5 \times 10^{-3} \, \text{m} \)
- Distance from the slit to the screen, \( d = 1 \, \text{m} \)
- Order of the minimum, \( n = 5 \)

### Step 2: Use the Formula for Minima in Single Slit Diffraction
The condition for the minima in a single slit diffraction pattern is given by:
\[ a \sin \theta = n \lambda \]
where \( a \) is the slit width, \( n \) is the order of the minimum, and \( \theta \) is the angle of diffraction.

For small angles, \( \sin \theta \) can be approximated as:
\[ \sin \theta \approx \tan \theta = \frac{x_n}{d} \]
Thus, we can rewrite the equation as:
\[ a \frac{x_n}{d} = n \lambda \]

### Step 3: Rearrange the Formula to Solve for Slit Width \( a \)
Rearranging the equation gives:
\[ a = \frac{d \cdot n \cdot \lambda}{x_n} \]

### Step 4: Substitute the Known Values
Substituting the known values into the equation:
\[ a = \frac{1 \, \text{m} \cdot 5 \cdot (5 \times 10^{-7} \, \text{m})}{5 \times 10^{-3} \, \text{m}} \]

### Step 5: Calculate the Slit Width
Calculating the above expression:
\[ a = \frac{1 \cdot 5 \cdot 5 \times 10^{-7}}{5 \times 10^{-3}} = \frac{25 \times 10^{-7}}{5 \times 10^{-3}} = 5 \times 10^{-4} \, \text{m} \]
Converting to mm:
\[ a = 5 \times 10^{-4} \, \text{m} = 0.5 \, \text{mm} \]

### Final Answer
The slit width \( a \) is \( 0.5 \, \text{mm} \).