The apparent depth of a needle laying at the bottom of the tank, which is filled with water of refractive index `1.33` to a height of `12.5` cm is measured by a microscope to be `9.4 cm`. If water is replaced by a liquid of refractive index `1.63` upto the same height. What distance would the microscope have to be moved to focus on the needle again ?

Options

A. `1.73 cm`
B. `2.13 cm`
C. `1.5 cm`
D. `2.9 cm`

Explanation

To solve the problem, we will follow these steps:

### Step 1: Understand the given data
- The refractive index of water, \( n_1 = 1.33 \)
- The height of water in the tank, \( h = 12.5 \) cm
- The apparent depth of the needle in water, \( d_{A1} = 9.4 \) cm
- The refractive index of the new liquid, \( n_2 = 1.63 \)

### Step 2: Calculate the real depth of the needle
The real depth of the needle can be calculated using the formula for apparent depth:
\[ d_{A1} = \frac{h}{n_1} \]
Rearranging gives us:
\[ h = d_{A1} \cdot n_1 \]
Substituting the values:
\[ h = 9.4 \, \text{cm} \cdot 1.33 = 12.5 \, \text{cm} \]
This confirms that the real depth is indeed \( 12.5 \) cm.

### Step 3: Calculate the new apparent depth in the new liquid
Using the same formula for apparent depth for the new liquid:
\[ d_{A2} = \frac{h}{n_2} \]
Substituting the values:
\[ d_{A2} = \frac{12.5 \, \text{cm}}{1.63} \]
Calculating this gives:
\[ d_{A2} \approx 7.67 \, \text{cm} \]

### Step 4: Calculate the shift in the microscope
To find the distance the microscope has to be moved to focus on the needle again, we subtract the new apparent depth from the old apparent depth:
\[ \text{Shift} = d_{A1} - d_{A2} \]
Substituting the values:
\[ \text{Shift} = 9.4 \, \text{cm} - 7.67 \, \text{cm} = 1.73 \, \text{cm} \]

### Final Answer
The microscope would have to be moved by **1.73 cm** to focus on the needle again.

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