A small telescope has a concave mirror with a 2.00 mradius of curvature for its objective. Its eyepiece is a 4.00 cm focal length lens. (a) What is the telescope’s angular magnification? (b) What angle is subtended by a 25,000 kmdiameter sunspot? (c) What is the angle of its telescopic image?

Magnification tells us how much larger or smaller is the image as compared to the object.

a)

$M=\frac{f_{\circ }}{f_e}$

The angular magnification is supplied by, as we said before in the idea session. $$\begin{gathered} =-\frac{2m}{0.04m} \\ =-50 \end{gathered}$$

Therefore, the telescope's angular magnification is - 50

b)

The distance between the Sun and the Earth is $d=1.52\times {10}^{11}$, as we know. To calculate the angle that the sunspot occupies, we use the following formula:

$$\begin{gathered} \theta ={\tan }^{-1}\left(\frac{Sunspot\text{'}sdiameter}{d}\right) \\ ={\tan }^{-1}\left(\frac{2.5\times {10}^{7}m}{1.52\times {10}^{11}m}\right) \\ =9.424\times {10}^{-3\circ } \end{gathered}$$

Therefore, the angle subtended by the sunspot is $9.424\times {10}^{-3\circ }$.

c)

The angular magnification is supplied by, as we said before in the idea session.

$$\begin{gathered} M=\frac{{\theta }^{\text{'}}}{\theta }{\theta }^{\text{'}}=M\times \theta \\ ={\left(50\times 9.424\times {10}^{-3}\right)}^{\circ } \\ =0.{4712}^{\circ } \end{gathered}$$

Therefore, the angle subtended by the sunspot in the telescope's image is 0.4712^o.