The distance of the sun from the earth is 1.5 × 10^ 11 m , and its...

The angle subtended by the sun at the optical centre of the lens is given by $\alpha = \dfrac{D}{u}$, where $D$ is the diameter of the sun and $u$ is the distance of the sun from the earth.

Since the sun is at a very large distance, its image is formed at the focal plane of the lens. The angle subtended by the image at the optical centre is equal to the angle subtended by the sun.

The diameter of the image $d$ is given by $d = f \alpha = \dfrac{f D}{u}$, where $f$ is the focal length of the lens.

The radius of the image is $r = \dfrac{d}{2} = \dfrac{f D}{2 u}$.

Substituting the given values, $f = 0.2 \text{ m}$, $D = 1.4 \times 10^9 \text{ m}$, and $u = 1.5 \times 10^{11} \text{ m}$:

$r = \dfrac{0.2 \times 1.4 \times 10^9}{2 \times 1.5 \times 10^{11}}$

$r = \dfrac{0.28 \times 10^9}{3.0 \times 10^{11}} \text{ m}$

$r = \dfrac{0.28}{300} \text{ m} = \dfrac{280}{300} \text{ mm}$

$r = \dfrac{14}{15} \text{ mm} \approx 0.933 \text{ mm}$