Angular Magnification of Telescopes Revision Notes | A Level AQA Physics

1. Why Angular Magnification?

When you look at the Moon with the naked eye, it subtends a tiny angle at your eye — roughly half a degree. A telescope cannot physically bring the Moon closer, so talking about how many times 'bigger' it appears in terms of length is meaningless. What the telescope actually does is increase the angle that the image subtends at your eye.

That is the core idea behind angular magnification: we compare angles, not lengths. A larger angle at the eye means more detail is resolved, which is exactly what an astronomer needs.

In normal adjustment, angular magnification can be written in two equivalent AQA forms: M = β/α and M = f_o/f_e. The first definition compares the angle subtended by the final image at the eye with the angle subtended by the object at the unaided eye, while the second follows from the telescope geometry when the intermediate image lies at the eyepiece focal plane.

Notice that $M$ has no units. This catches students out in exams — do not attach degrees or radians to your final value of $M$. The ratio of two angles in the same unit always cancels to a pure number.

2. How a Refracting Telescope Works

Picture parallel rays arriving from a very distant star — so distant that the rays are effectively parallel when they reach Earth. The objective lens, with its long focal length $f_o$, converges these rays to form a real, inverted intermediate image at its focal point $F_o$.

The eyepiece lens then acts as a magnifying glass. Its front focal point $F_e$ is placed at the same position as $F_o$, so the intermediate image sits exactly at the front focal point of the eyepiece. Rays leaving the eyepiece emerge as a parallel beam — the final image is at infinity and the observer's eye is fully relaxed. This is the condition called normal adjustment.

The total length of the telescope tube in normal adjustment equals $f_o + f_e$, because the two focal points coincide at the same physical location inside the tube. This is a useful cross-check in exam questions asking you to find the tube length.

3. Deriving $M = f_o / f_e$

Here is where the geometry pays off. The intermediate image has a height $h$ and forms at the common focal point. From the objective side, the angle $\alpha$ that the original object subtends at the unaided eye is approximately equal to the angle the object subtends at the objective (valid for distant objects). Using the small-angle approximation:

From the eyepiece side, the same intermediate image of height $h$ sits at the front focal point of the eyepiece. The angle $\beta$ that the final image subtends at the eye is therefore:

Dividing $\beta$ by $\alpha$ gives the angular magnification. The image height $h$ appears in both numerator and denominator, so it cancels cleanly:

The result is elegant: to increase magnification, use a longer objective focal length or a shorter eyepiece focal length — or both. This directly shapes telescope design choices.

4. Worked Calculations

The formula $M = f_o / f_e$ looks straightforward, but exam questions test it in several directions — finding $M$, finding an unknown focal length, or checking whether a given design is physically consistent. Work through each type below.

Worked example: Calculate angular magnification from focal lengths

A refracting telescope has an objective lens of focal length 900 mm and an eyepiece lens of focal length 25 mm. The telescope is used in normal adjustment. Calculate the angular magnification. [2 marks]

1. $M = \frac{f_o}{f_e}$ — This is the standard formula for angular magnification in normal adjustment. Writing it first earns the method mark.
2. $f_o = 900 \text{ mm}, \quad f_e = 25 \text{ mm}$ — Both focal lengths are already in the same unit (mm), so no conversion is needed. If one were in metres and the other in mm, you would need to convert first.
3. $M = \frac{900}{25} = 36$ — 900 divided by 25: 25 times 36 = 900. Confirmed. $M$ is dimensionless — do not write units.

Answer: $M = 36$ (no units)

Worked example: Find an unknown focal length given magnification

A telescope in normal adjustment produces an angular magnification of 45. The objective lens has a focal length of 1.35 m. Calculate the focal length of the eyepiece lens. [2 marks]

1. $M = \frac{f_o}{f_e} \Rightarrow f_e = \frac{f_o}{M}$ — Rearranging for the unknown before substituting keeps the working clear and earns the method mark.
2. $f_e = \frac{1.35}{45}$ — $f_o = 1.35$ m and $M = 45$. Both values substituted with units shown.
3. $f_e = 0.030 \text{ m} = 30 \text{ mm}$ — 1.35 divided by 45: 45 times 0.03 = 1.35. Confirmed. The eyepiece focal length is 0.030 m (30 mm). A short eyepiece focal length is physically reasonable for a high-magnification instrument.

Answer: $f_e = 0.030 \text{ m}$ (30 mm)

Worked example: Find the tube length in normal adjustment

Using the telescope from the first worked example ($f_o = 900$ mm, $f_e = 25$ mm), calculate the minimum tube length required when the telescope is in normal adjustment. [1 mark]

1. Tube length $= f_o + f_e$ — In normal adjustment the back focal point of the objective and the front focal point of the eyepiece coincide. The lenses are therefore separated by exactly $f_o + f_e$.
2. Tube length $= 900 + 25 = 925 \text{ mm}$ — 900 + 25 = 925. Confirmed. The tube length is 925 mm (0.925 m). This is a useful sanity check: the tube must be longer than $f_o$ alone.

Answer: Tube length $= 925$ mm

5. Comparing Telescope Designs

Now that the formula is established, it is worth analysing what it tells us about design trade-offs — the kind of reasoning that earns marks on 'explain' and 'evaluate' questions.

A key point for evaluation questions: increasing magnification by lengthening the objective also increases the tube length, making the instrument less portable. There is always a design compromise between magnification, physical size, and field of view.

6. Exam Technique and Common Mistakes

The angular magnification topic is reliable marks territory — the formula is short and the calculations are straightforward. Marks are lost almost entirely through avoidable errors in method and presentation, not through lack of knowledge.

For extended-response questions asking you to 'evaluate' a telescope design, structure your answer as: state the magnification formula, identify which parameter changes, state the effect on $M$, then address a second consequence such as tube length or field of view, and finish with a clear conclusion about the trade-off.

Bringing it all together: angular magnification in normal adjustment is governed by the single relationship $M = f_o / f_e$, which follows directly from the geometry of two lenses sharing a common focal point — and every exam question on this topic traces back to that one equation.