A convex spherical refracting surfaces separates two media glass and air `(mu_g=1.5).` If the image is to be real, at what minimum distance u should the object be placed in air if R is the radius of curvature

Options

A. `ugt3R`
B. `ugt2R`
C. `ugt4R`
D. `ugtR`

Explanation

To solve the problem step-by-step, we will use the refraction formula for a curved surface and analyze the conditions for a real image.

### Step-by-Step Solution:

1. **Understand the Setup**:
- We have a convex spherical refracting surface separating two media: glass (with refractive index \( \mu_g = 1.5 \)) and air (with refractive index \( \mu_a = 1.0 \)).
- The object is placed in air, and we need to find the minimum distance \( u \) from the surface where the object should be placed to ensure that the image formed is real.

2. **Sign Convention**:
- According to the sign convention, the radius of curvature \( R \) for a convex surface is taken as positive.
- The object distance \( u \) is negative since the object is placed on the side from which light is coming (left side of the surface).

3. **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, \( \mu_2 = 1.5 \) (glass), \( \mu_1 = 1.0 \) (air), and \( R \) is positive.

4. **Substituting Values**:
- Substitute the values into the formula:
\[ \frac{1.5}{v} - \frac{1.0}{-u} = \frac{1.5 - 1.0}{R} \]
- This simplifies to:
\[ \frac{1.5}{v} + \frac{1.0}{u} = \frac{0.5}{R} \]

5. **Rearranging the Equation**:
- Rearranging gives:
\[ \frac{1.5}{v} = \frac{0.5}{R} - \frac{1.0}{u} \]
- This can be rewritten as:
\[ \frac{1.5}{v} = \frac{1}{2R} - \frac{1}{u} \]

6. **Condition for Real Image**:
- For the image to be real, \( v \) must be positive. Therefore, we require:
\[ \frac{1}{2R} > \frac{1}{u} \]
- This implies:
\[ u > 2R \]

7. **Conclusion**:
- The minimum distance \( u \) at which the object should be placed in air for the image to be real is:
\[ u > 2R \]

### Final Answer:
The object should be placed at a distance \( u \) greater than \( 2R \) from the refracting surface.

---