Answer
a. 0.26 mm diameter.
b. 0.47 mm diameter.
c. 1.3 mm diameter.
d. $0.56\times$ for the wide-angle 28 mm lens, $2.7\times$ for the telephoto 135 mm lens.
Work Step by Step
a. The sun is very far away, so the image will fall at the focal point.
Find the size of the image from equation 23–9 for the magnification, $m=\frac{h_i}{h_o}=-\frac{d_i}{d_o}$
$$|h_i|=\frac{d_i}{d_o}h_o=\frac{28mm}{1.5\times10^8km}(1.4\times10^6km)=0.26mm$$
b. Repeat this, now for f = 50 mm.
$$|h_i|=\frac{d_i}{d_o}h_o=\frac{50mm}{1.5\times10^8km}(1.4\times10^6km)=0.47mm$$
c. Repeat this, now for f = 135 mm.
$$|h_i|=\frac{d_i}{d_o}h_o=\frac{135mm}{1.5\times10^8km}(1.4\times10^6km)=1.3mm$$
d. The above equations show that image height is directly proportional to the focal length of the lens. The relative magnification is the ratio of the focal lengths.
For the 28 mm lens, here is the relative magnification compared to the 50 mm lens:
$$\frac{28mm}{50mm}=0.56\times$$
For the 135 mm lens, here is the relative magnification compared to the 50 mm lens:
$$\frac{135mm}{50mm}=2.7\times$$
