There is a Young's double-slit experiment set-up completely dipped in water. Assume that the refractive index of water is `mu_1`. One of the slits is covered with a thin film of thickness t and refractive index `mu_2`. What will be the optical path difference at the centre of screen ?

Options

A. `|(mu_2 - mu_1)t|`
B. `|(mu_2 - 1)t|`
C. `|((mu_2)/(mu_1) - 1)t|`
D. `|((mu_1)/(mu_2) - 1)t|`

Explanation

To find the optical path difference at the center of the screen in a Young's double-slit experiment setup completely immersed in water, with one slit covered by a thin film, we can follow these steps:

### Step-by-Step Solution:

1. **Understanding the Setup**:
- The Young's double-slit experiment consists of two slits (let's call them slit 1 and slit 2) that emit light waves. The entire setup is submerged in water, which has a refractive index denoted as \( \mu_1 \).
- One of the slits (say slit 1) is covered with a thin film of thickness \( t \) and refractive index \( \mu_2 \).

2. **Optical Path Difference (OPD) Concept**:
- The optical path difference is defined as the difference in the effective distance traveled by light from the two slits to the screen.
- When light travels through different media, the effective distance is modified by the refractive index of the medium.

3. **Calculating the OPD for the Thin Film**:
- For the light passing through the thin film, the optical path length is given by the formula:
\[ \text{Optical Path Length} = \text{Refractive Index} \times \text{Thickness} \]
- In this case, the effective optical path length for the light passing through the thin film is:
\[ \text{OPL}_{\text{film}} = \mu_2 \cdot t \]

4. **Calculating the OPD in Water**:
- The light from slit 2, which does not pass through any thin film, travels through water, and its optical path length is:
\[ \text{OPL}_{\text{water}} = \mu_1 \cdot 0 = 0 \]
- Thus, the optical path difference (OPD) at the center of the screen, where the two waves meet, is given by:
\[ \text{OPD} = \text{OPL}_{\text{film}} - \text{OPL}_{\text{water}} = \mu_2 \cdot t - 0 = \mu_2 \cdot t \]

5. **Relative Refractive Index**:
- Since the entire setup is in water, we need to consider the relative refractive index of the thin film with respect to the surrounding water. The relative refractive index \( \mu_{\text{relative}} \) is given by:
\[ \mu_{\text{relative}} = \frac{\mu_2}{\mu_1} \]
- Therefore, the optical path difference can be expressed as:
\[ \text{OPD} = \left(\frac{\mu_2}{\mu_1} - 1\right) t \]

### Final Result:
The optical path difference at the center of the screen is:
\[ \text{OPD} = \left(\frac{\mu_2}{\mu_1} - 1\right) t \]