In Young's double slit experiment the y-coordinates of central maxima and 10th maxima are `2cm` and `5cm` respectively. When the YDSE apparatus is immersed in a liquid of refractive index `1.5` the corresponding y-coordintates will be

Options

A. `2cm, 7.5 cm`
B. `3 cm, 6cm`
C. `2 cm, 4 cm`
D. `4//3 cm, 10//3 cm`

Explanation

To solve the problem, we need to determine the new y-coordinates of the central maxima and the 10th maxima when the Young's double-slit experiment (YDSE) apparatus is immersed in a liquid with a refractive index of 1.5.

### Step-by-Step Solution:

1. **Understanding the Initial Conditions:**
- The y-coordinate of the central maxima (y₀) is given as 2 cm.
- The y-coordinate of the 10th maxima (y₁₀) is given as 5 cm.

2. **Refractive Index and Wavelength:**
- The refractive index (n) of the liquid is given as 1.5.
- The wavelength of light in the medium (λ') can be calculated using the formula:
\[ \lambda' = \frac{\lambda}{n} \]
- Since n = 1.5, we have:
\[ \lambda' = \frac{\lambda}{1.5} \]

3. **Effect on the Maxima:**
- The position of the maxima in YDSE is given by:
\[ y = \frac{m \lambda D}{d} \]
where m is the order of the maxima, D is the distance from the slits to the screen, and d is the distance between the slits.
- When the apparatus is immersed in the liquid, the wavelength changes, which affects the position of the maxima.

4. **Finding the New Position of the 10th Maxima:**
- The y-coordinate of the 10th maxima in the liquid (y'₁₀) can be expressed as:
\[ y'_{10} = \frac{10 \lambda' D}{d} \]
- Substituting λ' into this equation gives:
\[ y'_{10} = \frac{10 \left(\frac{\lambda}{1.5}\right) D}{d} = \frac{10 \lambda D}{1.5 d} = \frac{2}{3} \cdot 10 \frac{\lambda D}{d} = \frac{20}{3} \]

5. **Calculating the New y-coordinate:**
- The original y-coordinate of the 10th maxima was 5 cm, which corresponds to:
\[ y_{10} = 5 \text{ cm} \]
- The ratio of the new wavelength to the original wavelength is:
\[ \frac{\lambda}{\lambda'} = \frac{3}{2} \]
- Therefore, the new y-coordinate of the 10th maxima can be calculated as:
\[ y'_{10} = y_{10} \cdot \frac{3}{2} = 5 \cdot \frac{3}{2} = 7.5 \text{ cm} \]

6. **Finding the New Position of the Central Maxima:**
- The central maxima does not change its position when the refractive index changes:
\[ y'_{0} = y_{0} = 2 \text{ cm} \]

### Final Results:
- The new y-coordinate of the central maxima is **2 cm**.
- The new y-coordinate of the 10th maxima is **7.5 cm**.