Solution

Problem

$13.4.16.$ Is it possible to use a lens or a mirror to create an image of the Sun that is brighter than the Sun itself? What is the maximum illuminance of a Sun image that can be obtained using a concave mirror of diameter $D$ and radius of curvature $R$? The surface brightness of the Sun is $B$.

Solution

Brightness is the specific luminance — the energy emitted in a given direction per unit area of a surface. When forming an image, whether by a lens or a mirror, the brightness does **not** increase — it remains the same as that of the source (neglecting losses). That is, the brightness of the Sun's image cannot exceed the brightness of the Sun’s surface. In general, if it were possible to get a brighter image, it would violate the law of conservation of energy.

Now let's find the **maximum illuminance** of the Sun's image. Let $\theta$ be the angular size of the Sun as seen from Earth:

$$\theta = \frac{D_C}{R_{orb}}$$

where:

• $D_C$ is the diameter of the Sun ($1.4 \cdot 10^9 \, \text{m}$),
• $R_{orb}$ is the radius of Earth's orbit ($150 \cdot 10^9 \, \text{m}$).

Numerically, the Sun’s angular size is:

$$\theta \approx 0.0093 \, \text{rad}$$

This is a small angle, so we can approximate $\sin\theta \approx \theta$ and $\tan\theta \approx \theta$.

The incident solar power through the mirror's surface:

$$\Phi = q \cdot S = q \cdot \frac{\pi D^2}{4} \tag{1}$$

where $q$ is the energy flux density (solar irradiance).
The radius of the Sun's image at the mirror's focus:

$$r = F \tan\frac{\theta}{2} \approx F \cdot \frac{\theta}{2} = \frac{R}{2} \cdot \frac{\theta}{2} = \frac{R\theta}{4}$$

since the focal length of a spherical mirror is $F = \frac{R}{2}$.
The area of the Sun's image:

$$S_c = \pi r^2 = \pi \left( \frac{R\theta}{4} \right)^2 = \frac{\pi R^2 \theta^2}{16} \tag{2}$$

Illuminance at the focus is defined as:

$$E = \frac{\Phi}{S_c} \tag{3}$$

Substituting (1) and (2) into (3):

$$E = \frac{q \cdot \frac{\pi D^2}{4}}{\frac{\pi R^2 \theta^2}{16}} = \frac{4q D^2}{R^2 \theta^2} \tag{4}$$

Solar energy flux density $q$ on Earth can be expressed in terms of brightness $B$ (by the definition of brightness):

$$q = B \cdot \Omega$$

where $\Omega$ is the solid angle subtended by the Sun as seen from Earth.
Let’s express the solid angle in terms of the angular size:

$$\Omega = \frac{\pi (R_{orb} \cdot \frac{\theta}{2})^2}{R_{orb}^2} = \frac{\pi \theta^2}{4} \tag{5}$$

So,

$$q = B \cdot \frac{\pi \theta^2}{4} \tag{6}$$

Substitute (6) into (4):

$$E = \frac{4 D^2}{R^2 \theta^2} \cdot B \cdot \frac{\pi \theta^2}{4} = \frac{\pi D^2}{R^2} B$$