`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 using the given wavelength (\( \lambda \)) and width (\( d \)).

### Step-by-Step Solution:

1. **Identify the Given Values:**
- Wavelength, \( \lambda = 6000 \times 10^{-10} \, \text{m} = 6 \times 10^{-7} \, \text{m} \)
- Width, \( d = 0.6 \times 10^{-4} \, \text{m} = 6 \times 10^{-5} \, \text{m} \)

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

3. **Approximate \( \sin \theta \):**
For small angles, \( \sin \theta \approx \theta \) (in radians). Thus, we can rewrite the equation as:
\[ d \theta = n \lambda \]

4. **Solve for \( n \):**
Rearranging the equation gives:
\[ n = \frac{d}{\lambda} \theta \]
Since we are looking for the highest order of minima, we can find \( n \) when \( \theta \) approaches its maximum value, which is \( \frac{\pi}{2} \) (or 90 degrees). However, we need to consider the practical limit of \( n \).

5. **Calculate \( n \):**
Substituting the values of \( d \) and \( \lambda \):
\[ n = \frac{6 \times 10^{-5}}{6 \times 10^{-7}} = 100 \]

6. **Determine the Number of Minima:**
Since \( n = 100 \) corresponds to the first minimum on the edge of the slit, the number of minima on one side of the central maximum is \( n - 1 = 99 \). Therefore, the total number of minima on both sides is:
\[ \text{Total minima} = 99 + 99 = 198 \]

7. **Final Answer:**
The height of the highest order of minima on both sides of the central maxima is approximately 200.