In Young's double-slit experiment, let A and B be the two slit. A thin film of thickness t and refractive index `mu` is placed in front of A. Let `beta = ` fringe width. Then the central maxima will shift

Options

A. toward A
B. toward B
C. by `t(mu - 1) (beta)/(lambda)`
D. by `mu t (beta)/(lambda)`

Explanation

To solve the problem of how much the central maxima will shift in Young's double-slit experiment when a thin film is placed in front of one of the slits, we can follow these steps:

### Step-by-Step Solution:

1. **Understanding the Setup**:
In Young's double-slit experiment, we have two slits A and B. A thin film of thickness \( t \) and refractive index \( \mu \) is placed in front of slit A.

2. **Path Difference Calculation**:
The introduction of the thin film changes the optical path length for light passing through slit A. The optical path length through the film is given by:
\[ \text{Optical Path Length} = \mu t \]
The geometrical path length is simply \( t \). Therefore, the path difference \( \Delta \) introduced by the thin film is:
\[ \Delta = \text{Optical Path Length} - \text{Geometrical Path Length} = \mu t - t = t(\mu - 1) \]

3. **Relating Path Difference to Fringe Shift**:
In Young's experiment, the path difference that leads to a shift in the position of the central maxima is given by:
\[ \Delta = \frac{X D}{d} \]
where \( X \) is the shift in the position of the central maxima, \( D \) is the distance from the slits to the screen, and \( d \) is the distance between the slits.

4. **Setting the Equations Equal**:
Equating the two expressions for path difference, we have:
\[ \frac{X D}{d} = t(\mu - 1) \]

5. **Solving for the Shift \( X \)**:
Rearranging the equation to solve for \( X \):
\[ X = \frac{t(\mu - 1) d}{D} \]

6. **Relating to Fringe Width**:
The fringe width \( \beta \) is given by:
\[ \beta = \frac{\lambda D}{d} \]
From this, we can express \( \frac{d}{D} \) in terms of fringe width:
\[ \frac{d}{D} = \frac{\beta}{\lambda} \]

7. **Substituting Back**:
Substituting \( \frac{d}{D} \) into the expression for \( X \):
\[ X = t(\mu - 1) \left(\frac{\beta}{\lambda}\right) \]

8. **Final Expression**:
Thus, the shift of the central maxima is given by:
\[ X = \frac{t(\mu - 1) \beta}{\lambda} \]

### Conclusion:
The central maxima will shift by \( \frac{t(\mu - 1) \beta}{\lambda} \) towards slit A.