A concave spherical refractive surface with radius R separates a medium of refractive index `5//2` from air. As an object is approaching the surface from far away from the surface along the central axis, its image

Options

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

Explanation

To solve the problem, we need to analyze the situation using the lens maker's formula for a spherical refractive surface. We will determine whether the image formed is real or virtual as the object approaches the surface.

### Step-by-Step Solution:

1. **Identify the Given Information:**
- Refractive index of the medium (μ2) = 5/2 = 2.5
- Refractive index of air (μ1) = 1
- Radius of curvature of the concave surface (R) = R (given)

2. **Sign Convention:**
- Since the object is approaching the surface from far away, we consider the object distance (U) to be negative (U < 0).
- The radius of curvature (R) for a concave surface is also considered negative (R < 0).

3. **Apply the Refraction Formula:**
The formula relating the refractive indices and distances is given by:
\[ \frac{\mu_2}{V} - \frac{\mu_1}{U} = \frac{\mu_2 - \mu_1}{R} \]
Substituting the values:
\[ \frac{2.5}{V} - \frac{1}{U} = \frac{2.5 - 1}{R} \]
Simplifying further:
\[ \frac{2.5}{V} - \frac{1}{U} = \frac{1.5}{R} \]

4. **Rearranging the Equation:**
Rearranging gives:
\[ \frac{2.5}{V} = \frac{1.5}{R} + \frac{1}{U} \]
Now, we can express V:
\[ \frac{1}{V} = \frac{1.5}{2.5R} + \frac{1}{2.5U} \]
Cross-multiplying gives:
\[ V = \frac{-2.5RU}{1.5U + R} \]

5. **Analyzing the Sign of V:**
Since both R and U are negative, the expression for V will also be negative:
\[ V < 0 \]
A negative value for V indicates that the image is virtual.

6. **Conclusion:**
As the object approaches the concave surface from far away, the image formed is always virtual.

### Final Answer:
The image is virtual.

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