A microscope is focused on an object at the bottom of a bucket.If liquid with refractive index `(5)/(3)` is poured inside the bucket,then microscope have to be raised by `30cm` to focus the object again.The height of the liquid in the bucket is :

Options

A. `12cm`
B. `18cm`
C. `50cm`
D. `75cm`

Answer

To solve the problem, we need to determine the height of the liquid in the bucket when a microscope is focused on an object at the bottom and then raised by 30 cm after pouring a liquid with a refractive index of \( \frac{5}{3} \).

### Step-by-Step Solution:

1. **Understanding the Problem**:
- The microscope is initially focused on an object at the bottom of the bucket.
- When a liquid with a refractive index \( \mu = \frac{5}{3} \) is poured into the bucket, the microscope needs to be raised by 30 cm to refocus on the object.

2. **Using the Formula for Apparent Depth**:
- The formula relating the actual depth \( h \) and the apparent depth \( d \) when viewed through a medium with refractive index \( \mu \) is given by:
\[ d = h \left(1 - \frac{1}{\mu}\right) \]
- Here, \( d \) is the apparent depth (which changes when the microscope is raised), and \( h \) is the actual depth of the object.

3. **Setting Up the Equation**:
- Since the microscope is raised by 30 cm, we can say:
\[ d = h \left(1 - \frac{1}{\mu}\right) + 30 \]
- Rearranging gives us:
\[ 30 = h \left(1 - \frac{1}{\mu}\right) \]

4. **Substituting the Refractive Index**:
- Substitute \( \mu = \frac{5}{3} \) into the equation:
\[ 30 = h \left(1 - \frac{3}{5}\right) \]
- Simplifying \( 1 - \frac{3}{5} \) gives:
\[ 1 - \frac{3}{5} = \frac{2}{5} \]
- Therefore, the equation becomes:
\[ 30 = h \left(\frac{2}{5}\right) \]

5. **Solving for \( h \)**:
- To find \( h \), rearrange the equation:
\[ h = 30 \times \frac{5}{2} \]
- Calculating this gives:
\[ h = 75 \text{ cm} \]

### Final Answer:
The height of the liquid in the bucket is **75 cm**.