To solve the problem, we will analyze each statement one by one based on the principles of optics, particularly the lens maker's formula.
### Step-by-Step Solution:
**Step 1: Analyze Statement 1**
- **Statement**: When a lens is placed in a medium for which `μ` is less than that of the lens, its focal length is more than the focal length in air.
- **Analysis**: The lens maker's formula is given by:
\[
\frac{1}{f} = \frac{μ_l}{μ_m} - 1 \left( \frac{1}{r_1} - \frac{1}{r_2} \right)
\]
Where:
- \( f \) = focal length of the lens
- \( μ_l \) = refractive index of the lens
- \( μ_m \) = refractive index of the medium
- \( r_1 \) and \( r_2 \) are the radii of curvature of the lens surfaces.
If \( μ_m < μ_l \), then \( \frac{μ_l}{μ_m} > 1 \). Thus, \( \frac{μ_l}{μ_m} - 1 > 0 \) leads to a decrease in \( \frac{1}{f} \), which means \( f \) increases. Therefore, Statement 1 is **True**.
**Step 2: Analyze Statement 2**
- **Statement**: When the lens is placed in a medium for which `μ` is greater than that of the lens, the nature of the lens remains unchanged.
- **Analysis**: If \( μ_m > μ_l \), then \( \frac{μ_l}{μ_m} < 1 \). This means \( \frac{μ_l}{μ_m} - 1 < 0 \), which indicates that the effective focal length \( f \) becomes negative. This implies that a converging lens (positive focal length) becomes a diverging lens (negative focal length). Therefore, the nature of the lens changes. Thus, Statement 2 is **False**.
**Step 3: Analyze Statement 3**
- **Statement**: When a lens of focal length \( f \) is placed in a medium for which `μ` is the same as that of the lens, then the power of the lens becomes zero.
- **Analysis**: If \( μ_m = μ_l \), then \( \frac{μ_l}{μ_m} = 1 \). Therefore, \( \frac{μ_l}{μ_m} - 1 = 0 \), which leads to \( \frac{1}{f} = 0 \). Since power \( P \) of the lens is given by \( P = \frac{1}{f} \), it follows that \( P = 0 \). Thus, Statement 3 is **True**.
### Final Conclusion:
- Statement 1: True
- Statement 2: False
- Statement 3: True
The correct evaluation is: **True, False, True**.
