When a light wave passes from a rarer medium to a denser medium, there will be a phase change of `pi` radians. This difference brings change in the conditions for constructive and destructive interference. This phenomena also reasons the fromation of interference pattern in thin films like, oily layer, soap film, etc., but has no reason on the shifting of fringes from the central portion outward. The shift is dependent on the refractive index of the material as per the relation, `Delta y = (mu - 1) t`
On introducing a transparent slab `(mu)` the central fringe shifts to the point originally occupied by the fifth bright fringe. The thickness of the slab is

Options

A. `(5 lambda)/(mu - 1)`
B. `(4 lambda)/(mu - 1)`
C. `(mu -1)/(4 lambda)`
D. `(mu -1)/(5 lambda)`

Explanation

To solve the problem, we need to find the thickness of a transparent slab that causes the central fringe to shift to the position originally occupied by the fifth bright fringe. We will use the relationship given in the problem and apply it step by step.

### Step-by-Step Solution:

1. **Understand the Shift Due to the Slab**:
When a transparent slab of refractive index \( \mu \) and thickness \( t \) is introduced in the path of one of the rays, it causes a shift in the interference pattern. The shift \( \Delta y \) is given by the formula:
\[ \Delta y = \frac{(\mu - 1) t D}{d} \]
where \( D \) is the distance from the slits to the screen and \( d \) is the distance between the slits.

2. **Identify the Shift Corresponding to the Fifth Bright Fringe**:
The problem states that the central fringe shifts to the position of the fifth bright fringe. The position of the \( n \)-th bright fringe is given by:
\[ y_n = \frac{n \lambda D}{d} \]
For the fifth bright fringe (\( n = 5 \)):
\[ y_5 = \frac{5 \lambda D}{d} \]

3. **Set the Two Expressions for Shift Equal**:
Since the central fringe shifts to the position of the fifth bright fringe, we can equate the two expressions for the shift:
\[ \Delta y = y_5 \]
Therefore, we have:
\[ \frac{(\mu - 1) t D}{d} = \frac{5 \lambda D}{d} \]

4. **Cancel Common Terms**:
We can cancel \( D \) and \( d \) from both sides of the equation (assuming \( D \) and \( d \) are not zero):
\[ \mu - 1) t = 5 \lambda \]

5. **Solve for Thickness \( t \)**:
Rearranging the equation to solve for \( t \):
\[ t = \frac{5 \lambda}{\mu - 1} \]

Thus, the thickness of the slab is given by:
\[ t = \frac{5 \lambda}{\mu - 1} \]