Determine the diameter of the image of the moon produced by a...

Given, focal length of the mirror, $f = 7.6 \text{ m}$

Diameter of the moon, $D_o = 3450 \text{ km} = 3.45 \times 10^6 \text{ m}$

Distance of the moon from the mirror, $u = 3.8 \times 10^5 \text{ km} = 3.8 \times 10^8 \text{ m}$

Since the moon is at a very large distance, its image is formed at the focus of the concave mirror.

Image distance, $v \approx f = 7.6 \text{ m}$

The angle subtended by the object at the pole of the mirror is equal to the angle subtended by the image at the pole.

$\theta = \dfrac{D_o}{u} = \dfrac{D_i}{v}$

Substituting the values:

$\dfrac{3.45 \times 10^6}{3.8 \times 10^8} = \dfrac{D_i}{7.6}$

$D_i = 7.6 \times \dfrac{3.45 \times 10^6}{3.8 \times 10^8}$

$D_i = 2 \times 3.45 \times 10^{-2} \text{ m}$

$D_i = 6.9 \times 10^{-2} \text{ m}$

$D_i = 6.9 \text{ cm}$