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 determine the value of \( n \) such that the \( 16^{th} \) order maximum for the wavelength \( \lambda_1 = 6000 \, \text{Å} \) coincides with the \( n^{th} \) order maximum for the wavelength \( \lambda_2 = 4800 \, \text{Å} \).

### Step-by-Step Solution:

1. **Understand the Condition for Bright Fringes**:
The condition for bright fringes in a double-slit experiment is given by:
\[ d \sin \theta = n \lambda \]
where \( d \) is the distance between the slits, \( \theta \) is the angle of the fringe, \( n \) is the order of the fringe, and \( \lambda \) is the wavelength of light used.

2. **Set Up the Equations for Both Wavelengths**:
For the \( 16^{th} \) order maximum of \( \lambda_1 = 6000 \, \text{Å} \):
\[ d \sin \theta = 16 \cdot 6000 \, \text{Å} \]
For the \( n^{th} \) order maximum of \( \lambda_2 = 4800 \, \text{Å} \):
\[ d \sin \theta = n \cdot 4800 \, \text{Å} \]

3. **Equate the Two Conditions**:
Since both equations equal \( d \sin \theta \), we can set them equal to each other:
\[ 16 \cdot 6000 = n \cdot 4800 \]

4. **Solve for \( n \)**:
Rearranging the equation gives:
\[ n = \frac{16 \cdot 6000}{4800} \]
Now, simplify the right-hand side:
\[ n = \frac{96000}{4800} = 20 \]

5. **Conclusion**:
The value of \( n \) is \( 20 \).

### Final Answer:
\[ n = 20 \]