In `YDSE` when slab of thickness `t` and refractive index `mu` is placed in front of one slit then central maxima shifts by one fringe width. Find out `t` in terms of `lambda` and `mu`.

Options

A. `(lambda)/((2mu-1))`
B. `(2lambda)/((2mu-1))`
C. `(lambda)/((mu-1))`
D. `(2lambda)/((2mu+1))`

Answer

To solve the problem, we need to find the thickness \( t \) of the slab in terms of the wavelength \( \lambda \) and the refractive index \( \mu \) when a slab is placed in front of one slit in the Young's double slit experiment (YDSE), causing the central maxima to shift by one fringe width.

### Step-by-Step Solution:

1. **Understanding the Shift in Central Maxima:**
When a slab of thickness \( t \) and refractive index \( \mu \) is placed in front of one slit, it changes the optical path length of the light passing through that slit. This causes a shift in the interference pattern.

2. **Calculating the Optical Path Difference:**
The optical path length for the light passing through the slab is given by:
\[ \text{Optical Path Length} = \mu t \]
The light that does not pass through the slab travels a distance \( t \) without any change in refractive index, so the optical path length for that light remains \( t \).

The change in optical path due to the slab is:
\[ \Delta = \mu t - t = (\mu - 1)t \]

3. **Relating Shift to Fringe Width:**
The shift in the central maxima \( \Delta y \) due to this optical path difference is given by:
\[ \Delta y = \frac{(\mu - 1)t}{d} \]
where \( d \) is the distance between the slits and the screen.

4. **Fringe Width Calculation:**
The fringe width \( \beta \) in YDSE is given by:
\[ \beta = \frac{\lambda D}{d} \]
where \( D \) is the distance from the slits to the screen.

5. **Setting the Shift Equal to One Fringe Width:**
According to the problem, the central maxima shifts by one fringe width, so we set:
\[ \Delta y = \beta \]
Substituting the expressions we have:
\[ \frac{(\mu - 1)t}{d} = \frac{\lambda D}{d} \]

6. **Simplifying the Equation:**
We can cancel \( d \) from both sides:
\[ (\mu - 1)t = \lambda D \]

7. **Solving for Thickness \( t \):**
Rearranging the equation gives:
\[ t = \frac{\lambda D}{\mu - 1} \]

8. **Final Expression for Thickness \( t \):**
Since we need \( t \) in terms of \( \lambda \) and \( \mu \), we can express it as:
\[ t = \frac{\lambda}{\mu - 1} \]
(assuming \( D \) is taken as 1 for simplicity in terms of the fringe width).

### Final Answer:
\[ t = \frac{\lambda}{\mu - 1} \]