# Can this textbook question on angular magnification be answered without a ruler?

This is a textbook question on angular magnification from a chapter on optics and imaging. Note that this comprehension check comes before telescopes are introduced.

12 Figure 15.12 [below] shows a partial eclipse of the Sun.

a What angle does this image of the Sun subtend at your eye when it is at the near point of your eye?

b The Sun has a diameter of $$1.4\times10^6\:\mathrm{km}$$ and it is $$1.5\times10^8\:\mathrm{km}$$ from Earth. What is the angular magnification of Figure 15.12?

Note that

Angular magnification, $$M$$, is defined as the angle subtended at the eye by the image, $$\theta_\mathrm{i}$$, divided by the angle subtended at the eye by the object, $$\theta_\mathrm{o}$$. $$\displaystyle M=\frac{\theta_\mathrm{i}}{\theta_\mathrm{o}}$$.

and that

If the lens is moved to obtain the largest clear image, the image will be formed at the near point and the angular magnification can be determined from $$\displaystyle M_{\text{near point}} = \frac Df+1$$

where $$D = 25\:\mathrm{cm}$$ is the distance to a normal eye’s near point. The question also comes right before this diagram of simple magnification:

Is it possible to answer these two questions without measuring the height of the image shown in Figure 15.12, or is it necessary to do so?

Edit: It is easy for me to see that the (planar) visual angle $$\psi$$ subtended by a linear object of of length $$l$$ whose centre is a distance $$d$$ from the eye is $$\displaystyle \psi = \arctan\frac d{2l}$$.