In Young's double slit experiment, `5th` dark fringe is obtained at a point. If a thin transparent film is placed in the path of one of waves, then 7th bright is obtained at the same point. The thickness of the film in terms of wavelength `lambda` and refractive index `mu` will be

Options

A. `(1.5lambda)/((mu-1))`
B. `1.5(mu-1)lambda`
C. `2.5(mu-1)lambda`
D. `(2.5lambda)/((mu-1))`

Explanation

To solve the problem, we will follow these steps:

### Step 1: Understand the conditions of the problem
In Young's double slit experiment, we have two scenarios:
1. The 5th dark fringe is observed at a certain point.
2. After placing a thin transparent film in the path of one of the waves, the 7th bright fringe is observed at the same point.

### Step 2: Write the condition for the 5th dark fringe
The position of the nth dark fringe in Young's double slit experiment is given by the formula:
\[ y_n = \left( n - \frac{1}{2} \right) \frac{\lambda D}{d} \]
For the 5th dark fringe (n = 5):
\[ y_5 = \left( 5 - \frac{1}{2} \right) \frac{\lambda D}{d} = \frac{9}{2} \frac{\lambda D}{d} \]

### Step 3: Write the condition for the 7th bright fringe
The position of the mth bright fringe is given by:
\[ y_m = m \frac{\lambda D}{d} \]
For the 7th bright fringe (m = 7):
\[ y_7 = 7 \frac{\lambda D}{d} \]

### Step 4: Set the two fringe positions equal
Since both the 5th dark fringe and the 7th bright fringe occur at the same point, we can set their positions equal:
\[ \frac{9}{2} \frac{\lambda D}{d} = 7 \frac{\lambda D}{d} - \text{shift} \]
Let the shift due to the film be \( S \). Therefore, we can write:
\[ S = 7 \frac{\lambda D}{d} - \frac{9}{2} \frac{\lambda D}{d} \]
This simplifies to:
\[ S = \left( 7 - \frac{9}{2} \right) \frac{\lambda D}{d} = \frac{14 - 9}{2} \frac{\lambda D}{d} = \frac{5}{2} \frac{\lambda D}{d} \]

### Step 5: Relate the shift to the thickness of the film
The shift caused by introducing a thin film of thickness \( t \) and refractive index \( \mu \) is given by:
\[ S = \frac{D}{d} (\mu - 1) t \]
Equating the two expressions for shift:
\[ \frac{D}{d} (\mu - 1) t = \frac{5}{2} \frac{\lambda D}{d} \]
Cancelling \( \frac{D}{d} \) from both sides:
\[ (\mu - 1) t = \frac{5}{2} \lambda \]

### Step 6: Solve for the thickness \( t \)
Rearranging the equation gives:
\[ t = \frac{5}{2(\mu - 1)} \lambda \]

### Final Answer
Thus, the thickness of the film in terms of wavelength \( \lambda \) and refractive index \( \mu \) is:
\[ t = \frac{5}{2(\mu - 1)} \lambda \]