A tank is filled with water to a height of `15.5 cm`. The apparent depth of a needle lying at the bottom of the tank is measureed by a microscope to be `8.5 cm`. If water is replaced by a liquid of refractive index `1.94` upto the same height by what distance would the microscope have to be moved to focus on the needle again ?

Options

A. `1.00 cm`
B. `2.37 cm`
C. `0.51 cm`
D. `3.93 cm`

Explanation

To solve the problem step by step, we will follow the principles of optics related to apparent depth and refractive index.

### Step 1: Understand the given data
- The height of the water in the tank (real depth, \( d_r \)) = 15.5 cm
- The refractive index of water (\( n_{water} \)) = 1.33
- The apparent depth of the needle in water (\( d_A \)) = 8.5 cm
- The refractive index of the new liquid (\( n_{new} \)) = 1.94

### Step 2: Calculate the apparent depth in the new liquid
The formula for apparent depth is given by:
\[ d_A = \frac{d_r}{n} \]
Where:
- \( d_A \) = apparent depth
- \( d_r \) = real depth
- \( n \) = refractive index of the liquid

For the new liquid, we can rearrange the formula to find the new apparent depth (\( d_{A,new} \)):
\[ d_{A,new} = \frac{d_r}{n_{new}} \]
Substituting the values:
\[ d_{A,new} = \frac{15.5 \, \text{cm}}{1.94} \]

### Step 3: Perform the calculation
Calculating the new apparent depth:
\[ d_{A,new} = \frac{15.5}{1.94} \approx 7.989 \, \text{cm} \]

### Step 4: Calculate the distance the microscope needs to be moved
To find out how much the microscope needs to be moved, we subtract the new apparent depth from the old apparent depth:
\[ \text{Distance to move} = d_A - d_{A,new} \]
Substituting the values:
\[ \text{Distance to move} = 8.5 \, \text{cm} - 7.989 \, \text{cm} \approx 0.511 \, \text{cm} \]

### Step 5: Final result
Thus, the microscope needs to be moved approximately \( 0.51 \, \text{cm} \) to focus on the needle again.

---