Does a telescope magnify distant objects?

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A telescope's magnification refers to "angular magnification," not "linear magnification," meaning a 10x magnification does not make an object appear ten times taller. Instead, it creates the illusion of being closer to the object, increasing the subtended angle. While the observer may perceive the tree as larger, it does not change its actual height. The telescope effectively brings distant objects into clearer view, enhancing the observer's perspective. Thus, while the tree may seem more prominent, it does not literally increase in size.
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# An observer looks at a tree of height 15 meters with a telescope of magnifying power 10. How does the tree appear to the man?
I think the tree will appear to be 10 times taller. Is it right?
 
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Is that exactly how the question is stated?

In short, you are correct, but...

Telescope magnification is "angular magnification" not "linear magnification." A 10 X scope does not necessarily mean you will say "oh my god, that looks like a 150 meter tall tree!" You are more likely to have the impression that you are standing closer to the tree such that the subtended angle is increased ten times.
 
Chi Meson said:
Is that exactly how the question is stated?

In short, you are correct, but...

Telescope magnification is "angular magnification" not "linear magnification." A 10 X scope does not necessarily mean you will say "oh my god, that looks like a 150 meter tall tree!" You are more likely to have the impression that you are standing closer to the tree such that the subtended angle is increased ten times.
The tree is at a large distance from the man. The telescope brings the image of the tree nearer. Can’t we call that feature of the telescope as magnification?
 
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Starting with the mass considerations #m(t)# is mass of water #M_{c}# mass of container and #M(t)# mass of total system $$M(t) = M_{C} + m(t)$$ $$\Rightarrow \frac{dM(t)}{dt} = \frac{dm(t)}{dt}$$ $$P_i = Mv + u \, dm$$ $$P_f = (M + dm)(v + dv)$$ $$\Delta P = M \, dv + (v - u) \, dm$$ $$F = \frac{dP}{dt} = M \frac{dv}{dt} + (v - u) \frac{dm}{dt}$$ $$F = u \frac{dm}{dt} = \rho A u^2$$ from conservation of momentum , the cannon recoils with the same force which it applies. $$\quad \frac{dm}{dt}...

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