A concave mirror is half dipped in water. Focal length of portion of mirror is the air is `f_(1)` and length of mirror in water is `f_(2)`, then (Take refractive index of water = `4/3`)

Options

A. `f_(1)gtf_(2)`
B. `f_(1)ltf_(2)`
C. `f_(1)=f_(2)`
D. `f_(1)ge4/3f_(2)`

Answer

To solve the problem, we need to analyze the situation of a concave mirror that is half dipped in water. We will derive the relationship between the focal lengths of the mirror when it is in air and when it is in water.

### Step-by-Step Solution:

1. **Understanding Focal Length**:
The focal length of a concave mirror is given by the formula:
\[ f = \frac{R}{2} \]
where \( R \) is the radius of curvature of the mirror. This relationship indicates that the focal length depends only on the geometry of the mirror and not on the medium in which it is placed.

2. **Focal Length in Air**:
When the concave mirror is in air, we denote its focal length as \( f_1 \). According to the formula, we can express this as:
\[ f_1 = \frac{R}{2} \]

3. **Focal Length in Water**:
When the concave mirror is half dipped in water, the portion of the mirror that is in water will have a different effective focal length due to the change in medium. The focal length of the portion of the mirror in water is denoted as \( f_2 \).

4. **Refractive Index Consideration**:
The refractive index of water is given as \( n = \frac{4}{3} \). The effective focal length of a mirror in a medium can be modified by the refractive index. However, for mirrors, the focal length remains dependent on the radius of curvature and does not change due to the medium.

5. **Conclusion**:
Since both portions of the mirror (in air and in water) are parts of the same concave mirror, and the focal length is a function of the radius of curvature (which remains constant), we conclude that:
\[ f_1 = f_2 \]

### Final Answer:
The relationship between the focal lengths is:
\[ f_1 = f_2 \]