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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:

Fig 15.11 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 $\displaystyle \psi = \arctan\frac d{2l}$.

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1 Answer 1

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No, for part a you need to measure the diameter of the image in millimetres and then divide it by the near point distance (250 mm).

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