In a biprism experiment, the fifth dark fringe is formed opposite to one of the slits. What is the wavelength of light?

Options

A. (a)`(d^(2))/(6D)`
B. (b)`(d^(2))/(5D)`
C. (c)`(d^(2))/(15D)`
D. (d)`(d^(2))/(9D)`

Answer

To solve the problem regarding the biprism experiment where the fifth dark fringe is formed opposite to one of the slits, we can follow these steps:

### Step-by-Step Solution:

1. **Understanding the Dark Fringe Position**:
The position of the nth dark fringe in an interference pattern can be given by the formula:
\[ y_n = \frac{(2n - 1) \lambda D}{2d} \]
where:
- \( y_n \) is the distance of the nth dark fringe from the center,
- \( n \) is the order of the dark fringe,
- \( \lambda \) is the wavelength of light,
- \( D \) is the distance from the slits to the screen,
- \( d \) is the distance between the slits.

2. **Substituting for the Fifth Dark Fringe**:
For the fifth dark fringe (\( n = 5 \)):
\[ y_5 = \frac{(2 \cdot 5 - 1) \lambda D}{2d} = \frac{9 \lambda D}{2d} \]

3. **Given Condition**:
It is given that this dark fringe is formed opposite to one of the slits. This means that the distance from the center to the fifth dark fringe can be expressed as:
\[ y_5 = \frac{9 \lambda D}{2d} \]

4. **Relating Fringe Width**:
The fringe width \( x \) is defined as:
\[ x = \frac{\lambda D}{d} \]
Therefore, substituting \( x \) into the equation for \( y_5 \):
\[ y_5 = \frac{9}{2} x \]

5. **Finding the Wavelength**:
Rearranging the equation gives:
\[ \frac{9}{2} x = \frac{9}{2} \left(\frac{\lambda D}{d}\right) \]
Since we know \( y_5 = \frac{9 \lambda D}{2d} \), we can equate this to \( \frac{d}{2} \) (the distance from the center to the slit) to find \( \lambda \):
\[ \frac{9 \lambda D}{2d} = \frac{d}{2} \]
Simplifying gives:
\[ 9 \lambda D = d^2 \]
Therefore, the wavelength \( \lambda \) can be expressed as:
\[ \lambda = \frac{d^2}{9D} \]

### Final Answer:
The wavelength of the light is given by:
\[ \lambda = \frac{d^2}{9D} \]