Answer
a. 0.33 mm diameter.
b. 0.47 mm diameter.
c. 0.98 mm diameter.
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 $m=\frac{h_i}{h_o}=-\frac{d_i}{d_o}$
$$|h_i|=\frac{d_i}{d_o}h_o=\frac{35mm}{1.5\times10^8km}(1.4\times10^6km)=0.33mm$$
b. Repeat this, now for $f=d_i=50 mm$.
$$|h_i|=\frac{d_i}{d_o}h_o=\frac{50mm}{1.5\times10^8km}(1.4\times10^6km)=0.47mm$$
c. Repeat this, now for $f=d_i=105 mm$.
$$|h_i|=\frac{d_i}{d_o}h_o=\frac{105mm}{1.5\times10^8km}(1.4\times10^6km)=0.98mm$$
