For a concave mirror, paraxial rays are focused at distance `R/2` from pole and marginal rays are focused at distance x from pole, then (R is radius of curvature)

Options

A. `x=R/2`
B. `x = -f`
C. `xgtR/2`
D. `xltR/2`

Answer

To solve the problem, we need to understand the behavior of paraxial rays and marginal rays in a concave mirror.

### Step-by-Step Solution:

1. **Understanding Paraxial Rays:**
- Paraxial rays are those rays that are close to the principal axis of the mirror. For a concave mirror, these rays converge at the focal point, which is located at a distance of \( \frac{R}{2} \) from the pole of the mirror, where \( R \) is the radius of curvature.

2. **Understanding Marginal Rays:**
- Marginal rays are those rays that are further away from the principal axis. They make a larger angle with the normal at the point of incidence on the mirror. These rays also converge to a point, but this point is different from the focal point for paraxial rays.

3. **Setting Up the Geometry:**
- Consider a concave mirror with its pole (P) and center of curvature (C). The focal point (F) for paraxial rays is at \( \frac{R}{2} \) from the pole.
- For marginal rays, we need to analyze the geometry involving the angles formed by these rays.

4. **Using Trigonometry:**
- For marginal rays, the angle of incidence is larger than \( 10^\circ \). We denote this angle as \( \theta \).
- The distance from the center of curvature to the point where the marginal rays focus can be expressed in terms of \( \theta \).

5. **Finding the Distance \( x \):**
- The distance \( x \) from the pole to the focus of the marginal rays can be derived using the cosine of the angle \( \theta \):
\[ CE = \frac{R}{2} \cos(\theta) \]
- The distance from the center of curvature to the pole is \( R \), so the distance from the pole to the focus of the marginal rays can be expressed as:
\[ x = R - CE = R - \frac{R}{2} \cos(\theta) = R \left(1 - \frac{1}{2} \cos(\theta)\right) \]

6. **Conclusion:**
- As \( \theta \) increases, \( \cos(\theta) \) decreases, which means \( x \) will be less than \( \frac{R}{2} \). Therefore, the focus for marginal rays will be less than \( \frac{R}{2} \).

### Final Answer:
The distance \( x \) from the pole to the focus of the marginal rays is given by:
\[ x = R \left(1 - \frac{1}{2} \cos(\theta)\right) \]
where \( x < \frac{R}{2} \).