Average Intensity of Sunlight Passed Through Convex Lens

Let the intensity of the incident sunlight be $I_0 = 1.3\text{ kW m}^{-2}$.

Since sunlight arrives from a very large distance, the incident rays are parallel to the principal axis of the thin convex lens.

Let the radius of the circular aperture of the lens be $R$. The total power of the sunlight incident on the lens (and transmitted through it without loss) is: $P = I_0 \cdot A_{\text{lens}} = I_0 \left(\pi R^2\right)$

The convex lens has a focal length of $f = 20\text{ cm}$. Therefore, the parallel rays converge at the focal point, located at a distance of $20\text{ cm}$ on the other side of the lens.

Beyond the focal point, the light rays diverge and form a circular cross-section at a distance of $d = 22\text{ cm}$ from the lens. The distance from the focal point to this observation plane is: $\Delta x = d - f = 22\text{ cm} - 20\text{ cm} = 2\text{ cm}$

From the geometry of similar triangles formed by the converging and diverging light cones: $\frac{r}{\Delta x} = \frac{R}{f}$

where $r$ is the radius of the light spot at $d = 22\text{ cm}$. Substituting the given values: $\frac{r}{2\text{ cm}} = \frac{R}{20\text{ cm}} \implies r = \frac{R}{10}$

The cross-sectional area of the light beam at this distance is: $A = \pi r^2 = \pi \left(\frac{R}{10}\right)^2 = \frac{\pi R^2}{100}$

Since energy is conserved, the total power $P$ passes through this area $A$. The average intensity $I$ of the light at a distance of $22\text{ cm}$ is given by: $I = \frac{P}{A} = \frac{I_0 \pi R^2}{\frac{\pi R^2}{100}} = 100 I_0$

Substituting $I_0 = 1.3\text{ kW m}^{-2}$: $I = 100 \times 1.3\text{ kW m}^{-2} = 130\text{ kW m}^{-2}$