In a double slit experiment, the two slits are `1mm` apart and the screen is placed `1m` away. A monochromatic light of wavelength `500nm` is used. What will be the width of each slit for obtaining ten maxima of double slit within the central maxima of single-slit pattern?

Options

A. `0.2 mm`
B. `0.1 mm`
C. `0.5 mm`
D. `0.02 mm`

Explanation

To solve the problem, we need to determine the width of each slit in a double slit experiment such that there are ten maxima of the double slit pattern within the central maximum of the single slit pattern. Here’s a step-by-step solution:

### Step 1: Understand the given parameters
- Distance between the slits, \( d = 1 \, \text{mm} = 1 \times 10^{-3} \, \text{m} \)
- Distance from the slits to the screen, \( D = 1 \, \text{m} \)
- Wavelength of light, \( \lambda = 500 \, \text{nm} = 500 \times 10^{-9} \, \text{m} \)

### Step 2: Calculate the width of the central maximum of the single slit pattern
The width of the central maximum \( W \) of a single slit is given by the formula:
\[ W = \frac{2\lambda D}{a} \]
where \( a \) is the width of each slit.

### Step 3: Calculate the position of the maxima for the double slit pattern
The position of the \( n \)-th maximum in the double slit pattern is given by:
\[ y_n = \frac{n \lambda D}{d} \]
We need to find the width of the central maximum that can accommodate 10 maxima, meaning we need to consider \( n = 0, 1, 2, \ldots, 9 \) (10 maxima).

### Step 4: Find the total width for 10 maxima
The position of the 10th maximum (i.e., \( n = 9 \)) is:
\[ y_{9} = \frac{9 \lambda D}{d} \]

### Step 5: Set the width of the central maximum equal to the total width of the 10 maxima
The width of the central maximum of the single slit should be equal to the distance between the first and the tenth maxima:
\[ W = y_{9} - y_{-1} = y_{9} - (-y_{9}) = 2y_{9} = 2 \left(\frac{9 \lambda D}{d}\right) \]

### Step 6: Substitute the expression for \( W \)
Equating the two expressions for the width:
\[ \frac{2\lambda D}{a} = 2 \left(\frac{9 \lambda D}{d}\right) \]

### Step 7: Simplify the equation
Cancel \( 2 \) and \( \lambda D \) from both sides:
\[ \frac{1}{a} = \frac{9}{d} \]
Thus,
\[ a = \frac{d}{9} \]

### Step 8: Substitute the value of \( d \)
Substituting \( d = 1 \times 10^{-3} \, \text{m} \):
\[ a = \frac{1 \times 10^{-3}}{9} = \frac{1}{9} \times 10^{-3} \approx 0.1111 \times 10^{-3} \, \text{m} \]

### Step 9: Convert to millimeters
Convert \( a \) to millimeters:
\[ a \approx 0.1111 \, \text{mm} \]

### Final Answer
The width of each slit should be approximately \( 0.1111 \, \text{mm} \).

---