A real object is kept in air and refraction occurs at the convex boundry of a spherical glass surface. For the image to be real, the real object distance `(n_(g)=3//2)`

Options

A. Should be greater than three times the radius of curvature of the refracting surface.
B. Should be lesser than two times the radius of curvature of the refracting surface
C. Should be greater than the radius of curvature of the refracting surface
D. Is independent of the radius of curvature of the refracting surface

Answer

To solve the problem, we need to find the condition for the object distance \( u \) such that the image formed by refraction at the convex boundary of a spherical glass surface is real. We will use the refraction formula for a spherical surface and the sign conventions for the distances involved.

### Step-by-Step Solution:

1. **Identify the Given Values**:
- Refractive index of glass, \( n_g = \frac{3}{2} \)
- Refractive index of air, \( n_a = 1 \)

2. **Understand the Geometry**:
- We have a convex spherical surface. The center of curvature is to the right of the surface, which is considered the positive direction.
- Let \( R \) be the radius of curvature of the spherical surface.

3. **Apply the Refraction Formula**:
The refraction formula for a spherical surface is given by:
\[ \frac{n_2}{V} - \frac{n_1}{U} = \frac{n_2 - n_1}{R} \]
Where:
- \( n_1 \) is the refractive index of the medium from which light is coming (air, \( n_a = 1 \)).
- \( n_2 \) is the refractive index of the medium into which light is entering (glass, \( n_g = \frac{3}{2} \)).
- \( U \) is the object distance (taken as negative in this case).
- \( V \) is the image distance (positive for real images).

4. **Substituting the Values**:
Substituting the values into the formula:
\[ \frac{3/2}{V} - \frac{1}{-U} = \frac{3/2 - 1}{R} \]
This simplifies to:
\[ \frac{3/2}{V} + \frac{1}{U} = \frac{1/2}{R} \]

5. **Rearranging the Equation**:
Rearranging gives:
\[ \frac{1}{U} = \frac{1/2}{R} - \frac{3/2}{V} \]
To find the condition for a real image, we need \( V \) to be positive.

6. **Finding the Condition for Real Image**:
For the image to be real, we need:
\[ U - 2R > 0 \quad \text{(from the rearranged equation)} \]
This implies:
\[ U > 2R \]

7. **Conclusion**:
The condition for the object distance \( U \) for the image to be real is:
\[ U > 2R \]