`lambda=6000xx10^(-10) m` and width: `0.6xx10^(-4) m`. Find height of highest order of minima on both side central maxima.

Options

A. `10`
B. `20`
C. `100`
D. `200`

Answer

To solve the problem, we need to find the height of the highest order of minima on both sides of the central maxima in a single-slit diffraction pattern. We are given the wavelength (λ) and the width of the slit (d).

### Step-by-Step Solution:

1. **Identify the Given Values:**
- Wavelength, \( \lambda = 6000 \times 10^{-10} \, \text{m} \)
- Width of the slit, \( d = 0.6 \times 10^{-4} \, \text{m} \)

2. **Use the Condition for Minima:**
The condition for minima in single-slit diffraction is given by:
\[ d \sin \theta = n \lambda \]
where \( n \) is the order of the minima.

3. **Maximum Value of \( \sin \theta \):**
The maximum value of \( \sin \theta \) is 1. Therefore, we can write:
\[ n \lambda \leq d \]
Rearranging gives:
\[ n \leq \frac{d}{\lambda} \]

4. **Calculate the Maximum Order \( n \):**
Substitute the values of \( d \) and \( \lambda \):
\[ n \leq \frac{0.6 \times 10^{-4}}{6000 \times 10^{-10}} \]
Simplifying this:
\[ n \leq \frac{0.6 \times 10^{-4}}{6 \times 10^{-7}} = \frac{0.6}{6} \times 10^{3} = 0.1 \times 10^{3} = 100 \]
Thus, the maximum order of minima \( n \) is 100.

5. **Height of the Minima:**
The height of the minima on both sides of the central maxima can be calculated using the formula:
\[ y_n = \frac{n \lambda L}{d} \]
where \( L \) is the distance from the slit to the screen. Since \( L \) is not provided in the question, we can express the height in terms of \( L \):
\[ y_{100} = \frac{100 \cdot 6000 \times 10^{-10} \cdot L}{0.6 \times 10^{-4}} \]

6. **Simplifying the Expression for Height:**
\[ y_{100} = \frac{100 \cdot 6000 \cdot L \times 10^{-10}}{0.6 \times 10^{-4}} = \frac{100 \cdot 6000}{0.6} \cdot L \times 10^{-6} \]
\[ = 1000000 \cdot L \times 10^{-6} = 1000L \, \text{m} \]

Thus, the height of the highest order of minima on both sides of the central maxima is \( 1000L \, \text{m} \).