A convex spherical refracting surface with radius R separates a medium having refractive index `5//2` from air. As an object is moved towards the surface from far away from the surface along the principle axis, its image

Options

A. changes from real to virtual when it is at a distance R from the surface
B. changes from virtual to real when it is at a distance R from the surface
C. changes from real to virtual when it is at a distance `2R//3` from the surface
D. changes from virtual to real when it is at a distance `2R//3` from the surface

Explanation

To solve the problem, we need to analyze the behavior of the image formed by a convex spherical refracting surface as an object approaches it. The surface separates a medium with a refractive index of \( \frac{5}{2} \) (or 2.5) from air (with a refractive index of 1).

### Step-by-Step Solution:

1. **Identify the Refractive Indices:**
- Let \( \mu_1 = 1 \) (refractive index of air).
- Let \( \mu_2 = \frac{5}{2} = 2.5 \) (refractive index of the medium).

2. **Sign Convention:**
- For a convex surface, the radius of curvature \( R \) is positive.
- The object distance \( u \) is considered negative as per the sign convention (since the object is on the same side as the incoming light).

3. **Use the Refraction Formula:**
The formula for refraction at a spherical surface is given by:
\[ \frac{\mu_2}{v} - \frac{\mu_1}{u} = \frac{\mu_2 - \mu_1}{R} \]
Substituting the values of \( \mu_1 \) and \( \mu_2 \):
\[ \frac{2.5}{v} - \frac{1}{u} = \frac{2.5 - 1}{R} \]
Simplifying:
\[ \frac{2.5}{v} - \frac{1}{u} = \frac{1.5}{R} \]

4. **Rearranging the Equation:**
Rearranging gives:
\[ \frac{2.5}{v} = \frac{1.5}{R} + \frac{1}{u} \]
Taking the LCM:
\[ \frac{2.5}{v} = \frac{1.5u + R}{Ru} \]
Thus,
\[ v = \frac{2.5Ru}{1.5u + R} \]

5. **Determine the Image Nature:**
- Initially, when the object is far away (i.e., \( u \to -\infty \)), the image is real and formed on the opposite side of the surface.
- As the object moves closer, we need to check the nature of the image at specific distances.

6. **Check for \( u = R \):**
Substituting \( u = -R \):
\[ v = \frac{2.5R(-R)}{1.5(-R) + R} = \frac{-2.5R^2}{-0.5R} = 5R \]
Since \( v \) is positive, the image is real.

7. **Check for \( u = \frac{2R}{3} \):**
Substituting \( u = -\frac{2R}{3} \):
\[ v = \frac{2.5R\left(-\frac{2R}{3}\right)}{1.5\left(-\frac{2R}{3}\right) + R} = \frac{-\frac{5R^2}{3}}{-R} = \frac{5R}{3} \]
Here, \( v \) is positive, indicating a real image.

8. **Conclusion:**
The image changes from real to virtual when the object is at a distance of \( \frac{2R}{3} \) from the surface. Thus, the correct option is that the image changes from real to virtual at \( u = \frac{2R}{3} \).

### Final Answer:
The image changes from real to virtual when the object is at a distance of \( \frac{2R}{3} \) from the surface.