Focal length of a lens in air is f. Refractive index of the lens is `mu`. Focal length changes to `f_(1)` if lens is immersed in a liquid of refractive index `(mu)/(2)` and it becomes `f_(2)` if the lens is immersed in a liquid of refractive index `2 mu`. Then

Options

A. `f_(1) = (f)/(2)`
B. `f_(2) = -2f`
C. `f_(2) = -(3f)/(2)`
D. Data is insufficient

Answer

To solve the problem, we will use the lens maker's formula, which relates the focal length of a lens to the refractive indices of the lens and the surrounding medium. The formula is given by:

\[ \frac{1}{f} = \left( \frac{\mu - 1}{\mu_{\text{medium}}} \right) \left( \frac{1}{R_1} - \frac{1}{R_2} \right) \]

where:
- \( f \) is the focal length of the lens,
- \( \mu \) is the refractive index of the lens,
- \( \mu_{\text{medium}} \) is the refractive index of the medium in which the lens is placed,
- \( R_1 \) and \( R_2 \) are the radii of curvature of the lens surfaces.

### Step 1: Focal length in air
For the lens in air, the refractive index of air is approximately 1. Thus, we have:

\[ \frac{1}{f} = \left( \frac{\mu - 1}{1} \right) \left( \frac{1}{R_1} - \frac{1}{R_2} \right) \]

This simplifies to:

\[ \frac{1}{f} = (\mu - 1) \left( \frac{1}{R_1} - \frac{1}{R_2} \right) \quad \text{(Equation 1)} \]

### Step 2: Focal length in liquid with refractive index \( \frac{\mu}{2} \)
When the lens is immersed in a liquid with a refractive index of \( \frac{\mu}{2} \), we apply the lens maker's formula again:

\[ \frac{1}{f_1} = \left( \frac{\mu - 1}{\frac{\mu}{2}} \right) \left( \frac{1}{R_1} - \frac{1}{R_2} \right) \]

This simplifies to:

\[ \frac{1}{f_1} = \left( \frac{2(\mu - 1)}{\mu} \right) \left( \frac{1}{R_1} - \frac{1}{R_2} \right) \quad \text{(Equation 2)} \]

### Step 3: Focal length in liquid with refractive index \( 2\mu \)
Now, for the liquid with a refractive index of \( 2\mu \):

\[ \frac{1}{f_2} = \left( \frac{\mu - 1}{2\mu} \right) \left( \frac{1}{R_1} - \frac{1}{R_2} \right) \]

This simplifies to:

\[ \frac{1}{f_2} = \left( \frac{\mu - 1}{2\mu} \right) \left( \frac{1}{R_1} - \frac{1}{R_2} \right) \quad \text{(Equation 3)} \]

### Step 4: Relating \( f_1 \) to \( f \)
To find the relationship between \( f \) and \( f_1 \), we can divide Equation 2 by Equation 1:

\[ \frac{\frac{1}{f_1}}{\frac{1}{f}} = \frac{\frac{2(\mu - 1)}{\mu} \left( \frac{1}{R_1} - \frac{1}{R_2} \right)}{(\mu - 1) \left( \frac{1}{R_1} - \frac{1}{R_2} \right)} \]

This simplifies to:

\[ \frac{f}{f_1} = \frac{2}{\mu} \implies f_1 = \frac{f \cdot \mu}{2} \quad \text{(Equation 4)} \]

### Step 5: Relating \( f_2 \) to \( f \)
Now, we will find the relationship between \( f \) and \( f_2 \) by dividing Equation 3 by Equation 1:

\[ \frac{\frac{1}{f_2}}{\frac{1}{f}} = \frac{\frac{\mu - 1}{2\mu} \left( \frac{1}{R_1} - \frac{1}{R_2} \right)}{(\mu - 1) \left( \frac{1}{R_1} - \frac{1}{R_2} \right)} \]

This simplifies to:

\[ \frac{f}{f_2} = \frac{1}{2\mu} \implies f_2 = 2\mu f \quad \text{(Equation 5)} \]

### Conclusion
From the above relationships, we have:

1. \( f_1 = \frac{f \cdot \mu}{2} \)
2. \( f_2 = 2\mu f \)

Since the values of \( \mu \) are not provided, we cannot determine the exact numerical relationships between \( f, f_1, \) and \( f_2 \). Thus, the data is insufficient to find a direct relation.

### Final Answer
The correct option is **D: Data is insufficient**.