In Young's experiment the wavelenght of red light is `7.8xx10^(-5) cm ` and that of blue light is `5.2xx10^(-5)` cm. The value of n for which `(n+1)^(th)` blue bright band coincides with `n^(th)` red band is

Options

A. 1
B. 2
C. 3
D. 4

Answer

To find the value of \( n \) for which the \( (n+1)^{th} \) blue bright band coincides with the \( n^{th} \) red band in Young's double-slit experiment, we can use the formula for the position of bright fringes.

### Step-by-Step Solution:

1. **Identify the Wavelengths**:
- Wavelength of red light, \( \lambda_R = 7.8 \times 10^{-5} \) cm.
- Wavelength of blue light, \( \lambda_B = 5.2 \times 10^{-5} \) cm.

2. **Use the Condition for Coincidence**:
The position of the \( n^{th} \) bright fringe for red light is given by:
\[ y_n = n \cdot \lambda_R \]
The position of the \( (n+1)^{th} \) bright fringe for blue light is given by:
\[ y_{n+1} = (n+1) \cdot \lambda_B \]

3. **Set the Two Positions Equal**:
Since the \( (n+1)^{th} \) blue bright band coincides with the \( n^{th} \) red band, we can set the two equations equal to each other:
\[ n \cdot \lambda_R = (n+1) \cdot \lambda_B \]

4. **Substitute the Wavelengths**:
Plugging in the values of the wavelengths:
\[ n \cdot (7.8 \times 10^{-5}) = (n+1) \cdot (5.2 \times 10^{-5}) \]

5. **Eliminate the Common Factor**:
We can divide both sides by \( 10^{-5} \):
\[ n \cdot 7.8 = (n+1) \cdot 5.2 \]

6. **Expand and Rearrange the Equation**:
Expanding the right-hand side:
\[ 7.8n = 5.2n + 5.2 \]
Rearranging gives:
\[ 7.8n - 5.2n = 5.2 \]
\[ 2.6n = 5.2 \]

7. **Solve for \( n \)**:
Dividing both sides by 2.6:
\[ n = \frac{5.2}{2.6} = 2 \]

### Final Answer:
Thus, the value of \( n \) for which the \( (n+1)^{th} \) blue bright band coincides with the \( n^{th} \) red band is:
\[ \boxed{2} \]