A concave refractive surface of a medium having refractive index `mu` produces a real image of an object (located outside the medium) irrespective of its location. Choose the correct option from the following.

Options

A. Always
B. May be if refractive index of surrounding medium is greater than `mu`
C. May be if refractive index of surrounding medium is less than `mu`
D. None of the above

Explanation

To solve the problem, we need to analyze the situation involving a concave refractive surface and its ability to produce a real image of an object located outside the medium.

### Step-by-Step Solution:

1. **Understanding the Setup**:
- We have a concave refractive surface with a refractive index \( \mu \).
- An object is placed outside the medium, and we need to determine the conditions under which a real image is produced.

2. **Using the Refraction Formula**:
- The formula for refraction at a curved surface is given by:
\[ \frac{\mu_2}{v} - \frac{\mu_1}{u} = \frac{\mu_2 - \mu_1}{r} \]
- Here, \( \mu_1 \) is the refractive index of the surrounding medium (which we can take as 1 for air), \( \mu_2 \) is the refractive index of the medium (which is \( \mu \)), \( u \) is the object distance (taken as negative in the sign convention), and \( v \) is the image distance.

3. **Applying the Sign Convention**:
- For a concave surface, the radius of curvature \( r \) is negative.
- Therefore, we rewrite the equation as:
\[ \frac{\mu}{v} - \frac{1}{-u} = \frac{\mu - 1}{-r} \]
- This simplifies to:
\[ \frac{\mu}{v} + \frac{1}{u} = \frac{1 - \mu}{r} \]

4. **Finding the Image Distance**:
- Rearranging the equation gives:
\[ \frac{\mu}{v} = \frac{1 - \mu}{r} - \frac{1}{u} \]
- From this, we can express \( v \):
\[ v = \frac{\mu u r}{u(1 - \mu) - r} \]

5. **Condition for a Real Image**:
- For the image to be real, \( v \) must be positive:
\[ \frac{\mu u r}{u(1 - \mu) - r} > 0 \]
- This implies that the numerator and denominator must have the same sign.

6. **Analyzing the Signs**:
- The numerator \( \mu u r \) is positive if \( \mu > 0 \) and \( u, r > 0 \).
- The denominator \( u(1 - \mu) - r \) must also be positive for \( v \) to be positive.

7. **Conclusion**:
- For \( v \) to be positive, we need \( 1 - \mu > 0 \) or \( \mu < 1 \). This means the refractive index of the medium must be less than that of the surrounding medium (air).
- Thus, the correct option is that the refractive index of the surrounding medium is greater than \( \mu \).

### Final Answer:
The correct option is: **The refractive index of the surrounding medium is greater than \( \mu \)**.

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