A concave mirror of radius of curvature R has a circular outline of radius r. A circular disc is to be placed normal to the axis at the focus so that it collects all the light that is reflected from the mirror from a beam parallel to the axis. For r `gt gt R`, the area of this disc has to be at least

Options

A. `(pi r^(6))/(4 R^(4))`
B. `(pi r^(4))/(4 R^(2))`
C. `(pi r^(5))/(4 R^(3))`
D. `(pi r^(4))/(R^(2))`

Answer

To solve the problem, we need to determine the minimum area of a circular disc that can be placed at the focus of a concave mirror, which collects all the light reflected from the mirror when a beam of light is parallel to the principal axis.

### Step-by-Step Solution:

1. **Understanding the Geometry of the Concave Mirror:**
- The radius of curvature of the concave mirror is \( R \).
- The focal length \( f \) of the concave mirror is given by the formula:
\[ f = \frac{R}{2} \]

2. **Positioning the Disc:**
- The disc is to be placed normal to the axis at the focus of the mirror, which is at a distance \( f = \frac{R}{2} \) from the mirror.

3. **Ray Diagram:**
- Consider rays of light coming parallel to the principal axis. When these rays strike the mirror, they will reflect and converge at the focal point.
- The rays that hit the edge of the mirror will create a cone of light that converges at the focus.

4. **Using Similar Triangles:**
- Let \( r \) be the radius of the circular outline of the mirror.
- The distance from the edge of the mirror to the focal point is \( \frac{R}{2} \).
- We can form similar triangles to relate the radius of the disc \( r_0 \) to the radius of the mirror \( r \).

5. **Setting Up the Similar Triangles:**
- The triangles formed by the radius of the mirror and the radius of the disc at the focus can be expressed as:
\[ \frac{r_0}{r} = \frac{\frac{R}{2}}{x} \]
- Here, \( x \) is the distance from the focal point to the point on the principal axis directly below the edge of the mirror.

6. **Expressing \( x \):**
- The distance \( x \) can be expressed in terms of \( r \) and \( R \):
\[ x = \frac{r}{2} - \frac{R}{2} \]

7. **Substituting and Simplifying:**
- Substitute \( x \) back into the triangle ratio:
\[ r_0 = r \cdot \frac{\frac{R}{2}}{\frac{r}{2} - \frac{R}{2}} \]
- Rearranging gives:
\[ r_0 = \frac{rR}{r - R} \]

8. **Finding the Area of the Disc:**
- The area \( A \) of the disc is given by:
\[ A = \pi r_0^2 \]
- Substituting for \( r_0 \):
\[ A = \pi \left(\frac{rR}{r - R}\right)^2 \]

9. **Considering the Condition \( r \gg R \):**
- Since \( r \gg R \), we can approximate \( r - R \approx r \):
\[ A \approx \pi \left(\frac{rR}{r}\right)^2 = \pi R^2 \]

10. **Final Result:**
- Therefore, the area of the disc must be at least:
\[ A \approx \pi R^2 \]