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 where a concave mirror is half-dipped in water. We are given that the focal length of the portion of the mirror in air is \( f_1 \) and the focal length of the portion of the mirror in water is \( f_2 \). We also know the refractive index of water is \( \frac{4}{3} \).

### Step-by-Step Solution:

1. **Understanding Focal Length**:
The focal length \( f \) 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 is directly related to the radius of curvature.

2. **Identifying the Situation**:
The mirror is half-dipped in water. This means that the upper half of the mirror is in air, and the lower half is in water.

3. **Focal Length in Air**:
For the portion of the mirror in air, the focal length is:
\[ f_1 = \frac{R}{2} \]

4. **Focal Length in Water**:
When the lower half of the mirror is submerged in water, the effective focal length \( f_2 \) can be affected by the refractive index of the medium. However, the radius of curvature \( R \) remains the same for both portions of the mirror.

5. **Using the Refractive Index**:
The focal length of a mirror does not change when it is submerged in a medium like water, because the focal length depends on the geometry of the mirror rather than the medium it is in. The effective focal length remains the same:
\[ f_2 = \frac{R}{2} \]

6. **Equating the Focal Lengths**:
Since both portions of the mirror have the same radius of curvature, we can equate the focal lengths:
\[ f_1 = f_2 \]

7. **Conclusion**:
Therefore, we conclude that:
\[ f_1 = f_2 \]

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