A concave mirror is placed on a horizontal table with its axis directed vertically upwards.Let O be the pole of the mirror and C its centre of curvature.A point object is placed at C. It has a real image, also located at C.If the mirror is now filled with water, the image will be

Options

A. Real and will remain at C
B. Rela and located at a point between C and `infty`
C. Virtual and located at a point between C and 0.
D. Real and located at a point between C and 0.

Answer

To solve the problem step by step, we will analyze the situation of a concave mirror filled with water and how it affects the image formation.

### Step-by-Step Solution:

1. **Understanding the Setup**:
- A concave mirror is placed with its axis directed vertically upwards. The pole of the mirror is denoted as O, and the center of curvature is C.
- A point object is placed at C. In air, the image formed by the concave mirror at this position is also at C (real image).

**Hint**: Visualize the setup with the mirror and the object to understand the initial conditions.

2. **Effect of Water on the Mirror**:
- When the concave mirror is filled with water, the refractive index of the medium changes. The effective focal length of the mirror will change due to the water.
- The focal length of a concave mirror in air is given by \( F = -\frac{R}{2} \). When submerged in water (with refractive index \( n \)), the new focal length \( F' \) can be calculated using the formula:
\[ F' = \frac{F}{n} = -\frac{R}{2n} \]

**Hint**: Remember that the focal length is affected by the medium surrounding the mirror.

3. **Finding the New Image Position**:
- Since the object is still at C, we need to find the new image position using the mirror formula:
\[ \frac{1}{f} = \frac{1}{v} + \frac{1}{u} \]
- Here, \( u = -R \) (the object is at the center of curvature), and substituting the new focal length:
\[ \frac{1}{F'} = \frac{1}{v} + \frac{1}{(-R)} \]
- Rearranging gives:
\[ \frac{1}{v} = \frac{1}{F'} + \frac{1}{R} \]

**Hint**: Use the mirror formula correctly by substituting the values for \( u \) and \( F' \).

4. **Calculating the New Image Distance**:
- Substitute \( F' = -\frac{R}{2n} \):
\[ \frac{1}{v} = -\frac{2n}{R} + \frac{1}{R} \]
- Simplifying gives:
\[ \frac{1}{v} = \frac{1 - 2n}{R} \]
- Thus, the new image distance \( v \) becomes:
\[ v = \frac{R}{1 - 2n} \]

**Hint**: Ensure to simplify the equation correctly to find \( v \).

5. **Analyzing the Image Position**:
- Depending on the value of \( n \) (the refractive index of water is approximately 1.33), we can determine whether the image is real or virtual.
- If \( n < 0.5 \), the image will be real and located beyond C. If \( n \geq 0.5 \), the image will be virtual and located between C and O.

**Hint**: Consider the implications of the refractive index on the image formation.

6. **Conclusion**:
- Since water has a refractive index greater than 1, the image will be virtual and located between C and O.

**Final Answer**: The image will be virtual and located between C and O.