A spherical surface of radius of curvature R separates air (refractive index 1.0) from glass (refractive index 1.5). The center of curvature is in the glass. A point object P placed in air is found to have a real image Q in the glass. The ling PQ cuts the surface at a point O, and `PO=OQ`. The distance PO is equal to

Options

A. 5R
B. 3R
C. 2R
D. 1.5R

Explanation

To solve the problem, we need to analyze the situation involving the spherical surface separating air and glass, and the relationship between the object distance, image distance, and the radius of curvature.

### Step-by-Step Solution:

1. **Identify the Given Information:**
- Refractive index of air, \( \mu_1 = 1.0 \)
- Refractive index of glass, \( \mu_2 = 1.5 \)
- Radius of curvature of the spherical surface, \( R \)
- The center of curvature is in the glass.
- A point object \( P \) is placed in air, and it has a real image \( Q \) in the glass.
- The line \( PQ \) intersects the surface at point \( O \), and it is given that \( PO = OQ \).

2. **Set Up the Problem:**
- Let the distance \( PO = x \). Since \( PO = OQ \), we also have \( OQ = x \).
- Therefore, the total distance \( PQ = PO + OQ = x + x = 2x \).

3. **Use the Refraction Formula:**
- The refraction at the spherical surface can be described by the formula:
\[ \frac{\mu_2}{V} - \frac{\mu_1}{U} = \frac{\mu_2 - \mu_1}{R} \]
- Here, \( U \) is the object distance (negative in this case since the object is in air), and \( V \) is the image distance (positive since the image is in glass).

4. **Substituting Values:**
- Let \( U = -2x \) (since the object is on the left side of the surface) and \( V = 2x \) (since the image is on the right side of the surface).
- Substitute \( \mu_1 \), \( \mu_2 \), \( U \), and \( V \) into the formula:
\[ \frac{1.5}{2x} - \frac{1.0}{-2x} = \frac{1.5 - 1.0}{R} \]

5. **Simplifying the Equation:**
- This simplifies to:
\[ \frac{1.5}{2x} + \frac{1.0}{2x} = \frac{0.5}{R} \]
- Combine the fractions on the left:
\[ \frac{2.5}{2x} = \frac{0.5}{R} \]

6. **Cross-Multiplying:**
- Cross-multiply to solve for \( x \):
\[ 2.5R = 0.5 \cdot 2x \]
- This simplifies to:
\[ 2.5R = x \]

7. **Final Calculation:**
- Therefore, the distance \( PO \) is:
\[ PO = x = 5R \]

### Conclusion:
The distance \( PO \) is equal to \( 5R \).