# What is the physical meaning of magnifying power of a telescope?

So the following question was given in the JEE Mains 2016 conducted throughout India on 3rd April.

An observer looks at a distant tree of height 10 m with a telescope of magnifying power of 20. To the observer the tree appears

(1) 10 times taller (2) 10 times nearer (3) 20 times taller (4) 20 times nearer

The question has created a lot of controversy as many of the famous physics teachers of India have different opinions. Some go with the third and some with the fourth one. I am looking for a second opinion regarding the answer.

• This multiple choice question (apparently featuring in an Indian national level entrance test for admission to top institute STEM studies) demonstrates what is seriously wrong in the Indian educational system: it tests for memorization of definitions, and fails to test for real insight. – Johannes Apr 5 '16 at 16:12
• They actually do on the second level perhaps. This is first level. Out of 1.2 million students 200,000 people qualify (earlier 100,000) for the second level known as "JEE Advanced" Out of those 200,000 students only 5000 get into the top institutes (IITs) (You can check the paper here - cms.fiitjee.co/Resources/DownloadCentre/Document_Pdf_183.pdf ) – Dhiraj Barnwal Apr 5 '16 at 16:59

I think probably_someone's answer is already very good, but I wanted to do some math instead of studying, so here we go:

Consider the tree with height $h=10 \mathrm{m}$ at a distance $d$. The apparent size of the object can be expressed in terms of the angle $\theta$ it occupies, that is given by \begin{align} \tan \theta &= h/d \quad (\text{opposite over adjacent)} \\ \Rightarrow \theta &\simeq h/d \quad \text{(small angle approximation)} \end{align} Let's call this initial angle $\theta_1$.

Then consider the two cases $h'= h \cdot 20$ or $d'=d/20$. In case the height is 20 times larger, $\theta_2$ becomes (in small angle approximation) $$\theta_2 = h'/d = 20 \cdot \frac{h}{d} = 20 \cdot \theta_1$$ and $\theta_3$ becomes $$\theta_3 = h/d' = \frac{h}{\frac{d}{20}} = 20 \cdot \frac{h}{d} = 20 \cdot \theta_1$$ and answers 3 and 4 are equivalent, as long as the small angle approximation holds. Given that the question states "...a distant tree..." I think this is a valid approximation to make.

• Yeah, thanks for the additional clarification. I was implicitly assuming the small-angle approximation when I conflated the object's height with its arclength. Usually in astrophysics (which I study) nobody makes this assumption clear, because we deal exclusively with faraway objects, but you're absolutely correct. – probably_someone Dec 21 '16 at 20:07

I believe the reason for the confusion is that choices 3 and 4 actually refer to the same physical situation and in fact are both correct. They both are linked to the following geometric law: an object with height $s$ and distance $r$ from the observer will have an apparent angular size $\theta$ given by:

$\theta=s/r$

A telescope magnifies the apparent angular size of objects; in this case, an object of size $\theta$ is magnified to $20\times\theta$. Since $\theta$ corresponds to the above ratio, one could say that the numerator $s$ had been made 20 times larger (i.e. the object is 20 times taller). However, one could also say that the denominator $r$ was 20 times smaller (i.e. the object is 20 times nearer). Either one leads to the correct angular size.

May be interesting to note that $$\theta=s/r$$ is dependent on the small angle approximation $$\sin(\theta)=\theta$$ which would be a good approximation in most applications of telescopes.

A object that is 20 times nearer would require re-focusing of near field optics. For terrestrial telescopes, cameras and binoculars the intuition of focusing according to r could seem to be at odds with magnification "bringing objects nearer".

Additionally an object that is 20 times nearer or 20 times taller will be 20 times brighter (- if only we had telescopes like that!). In many cases the naked eye won't notice that a magnified image is too dim, but it is relevant in photography exposure times and probably to other "measurements". So I find I would go for answer e) (none of the above) and instead say that the image of the object is 20 times larger.

Maybe the reason people can't agree on one of the options provided is because none of them are really very good.

I once looked at the moon through a big optical telescope (at Abastumani, Georgia). The feeling was of being real close to the surface of the moon. That was a visceral sensensation, something I will always remember.

When the exam question is asking about how things appear, there is also psychology involved. Our brain knows the size of things (moon, trees, birds). In ordinary circumstances, only by getting closer we can achieve an object to take up a larger part of the eye's field of view. Blackbirds 20 times as large do not exist. The brain interprets angular magnification of familiar things not as growth but as a reduced distance.

i think it is the fourth one cause magnifying power means how much times near you want the object to be. so, the tree has been seen through 20 magnifying power so it will be twenty times nearer