A double-slit interference pattern is formed on a screen in a Young's double slit experiment performed with light consisting of two wavelengths `lambda_(1)` = 6000A and  `lambda_(2)` = 4800A. It is observed that the maximum of the `16^(th)` order corresponding to `lambda`= 6000 A coincides with a maximum of the n order corresponding to `lambda`= 4800 A The value of n is

Options

A. 10
B. 20
C. 5
D. none of these

Answer

To solve the problem, we need to find the value of \( n \) such that the 16th order maximum for wavelength \( \lambda_1 = 6000 \) Å coincides with the \( n \)th order maximum for wavelength \( \lambda_2 = 4800 \) Å in a Young's double slit experiment.

### Step-by-step Solution:

1. **Understanding the Condition for Maxima**:
The condition for maxima in a double-slit experiment is given by:
\[ \Delta x = n \lambda \]
where \( \Delta x \) is the path difference, \( n \) is the order of the maximum, and \( \lambda \) is the wavelength of light used.

2. **Setting Up the Equations**:
For the 16th order maximum for \( \lambda_1 = 6000 \) Å:
\[ \Delta x_1 = 16 \lambda_1 = 16 \times 6000 \, \text{Å} \]
For the \( n \)th order maximum for \( \lambda_2 = 4800 \) Å:
\[ \Delta x_2 = n \lambda_2 = n \times 4800 \, \text{Å} \]

3. **Equating the Two Conditions**:
Since the maxima coincide, we can set the two path differences equal to each other:
\[ 16 \times 6000 = n \times 4800 \]

4. **Solving for \( n \)**:
Rearranging the equation gives:
\[ n = \frac{16 \times 6000}{4800} \]
Simplifying this:
\[ n = \frac{16 \times 6000}{4800} = \frac{16 \times 6}{4.8} = \frac{96}{4.8} = 20 \]

5. **Conclusion**:
Thus, the value of \( n \) is:
\[ n = 20 \]

### Final Answer:
The value of \( n \) is **20**.