Ray Diagrams for Astronomical Telescopes Revision Notes | A Level AQA Physics

1. Why Ray Diagrams Matter for Telescopes

Astronomers cannot bring distant stars closer — but they can make those objects subtend a larger angle at the eye. That is the entire purpose of a telescope, and a ray diagram is the clearest way to see exactly how it achieves this.

The AQA specification requires you to draw and interpret the ray diagram for an astronomical refracting telescope in normal adjustment. This single diagram encodes the positions of both lenses, the path of light, the location of the intermediate image, and the condition for the final image to form at infinity — all at once.

Before drawing anything, you need a clear mental picture of the two-lens system. Think of it as two jobs done in sequence: the objective lens collects light and creates a small real image, then the eyepiece lens magnifies that image for the observer. Understanding each job separately makes the full diagram straightforward to construct.

2. The Two Lenses and Their Roles

Because the object (a star or distant planet) is so far away, the rays arriving at the objective are treated as perfectly parallel. A converging lens brings parallel rays to a focus at its focal point, so the intermediate image forms exactly at the focal plane of the objective. This image is real — you could project it onto a screen placed there.

The intermediate image is also inverted and diminished relative to the original object. For astronomical use this is acceptable; the goal is angular magnification, not an upright image. The eyepiece then takes over from this point.

When an object is placed at the focal point of a converging lens, the refracted rays emerge parallel — meaning the image forms at infinity. This is precisely the condition for normal adjustment: the intermediate image sits at $F_e$, so the observer's eye receives parallel rays and views the final image at infinity without any strain.

3. Normal Adjustment: The Key Condition

Normal adjustment is defined by one geometric condition: the back focal point of the objective lens $F_o$ coincides with the front focal point of the eyepiece lens $F_e$. In other words, both focal points occupy the same position in space.

This means the separation between the two lenses equals the sum of their focal lengths. If the objective has focal length $f_o$ and the eyepiece has focal length $f_e$, then the tube length $L$ is given by $L = f_o + f_e$. This is a direct consequence of the coinciding focal points, not an arbitrary rule.

Notice that $f_o$ is always much larger than $f_e$ in a well-designed astronomical telescope. A large-diameter objective collects more light from faint astronomical objects; a large $f_o$ creates a larger intermediate image, increasing angular magnification. A small $f_e$ then magnifies that image strongly. The ratio $f_o / f_e$ gives the angular magnification — more on this shortly.

4. Drawing the Ray Diagram Step by Step

The ray diagram is the centrepiece of this topic. Examiners award marks for specific features, so every element must be placed deliberately. The walkthrough below builds the diagram from scratch, one ray at a time.

Worked example: Constructing the Ray Diagram for Normal Adjustment

Draw a fully labelled ray diagram for an astronomical refracting telescope in normal adjustment, showing how parallel rays from a distant object are brought to a final image at infinity.

1. Draw a horizontal principal axis. Place the objective lens (larger diameter) on the left and the eyepiece lens (smaller diameter) to its right. Label them 'Objective' and 'Eyepiece'. The separation between them must equal $f_o + f_e$. — The objective is always the larger lens because it needs to gather as much light as possible from faint astronomical objects. Relative sizes matter in a diagram — examiners notice if both lenses are drawn the same size.
2. Mark $F_o$ on the right side of the objective at distance $f_o$ from it. Mark $F_e$ on the left side of the eyepiece at distance $f_e$ from it. In normal adjustment, $F_o$ and $F_e$ are at the same point on the axis. — This shared focal point is the defining feature of normal adjustment. If you draw $F_o$ and $F_e$ at different positions, your diagram is not in normal adjustment and will lose marks.
3. Draw two rays arriving from the left, both parallel to the principal axis (representing light from a star on-axis). One ray travels along the principal axis; the other travels parallel to it but offset above (or below) the axis. — Parallel rays represent an object at infinity. For an on-axis object, all rays are parallel to the principal axis. You need at least two rays to define where the image forms.
4. The ray along the principal axis passes straight through the centre of the objective undeviated. The offset ray refracts and converges towards $F_o$. Both rays meet at $F_o$ (which coincides with $F_e$), forming the real, inverted intermediate image. — A ray through the optical centre of a thin lens is undeviated — this is a standard ray-tracing rule. The intermediate image forms at the shared focal point. Mark a small arrow at this point pointing downward (inverted) to represent the intermediate image.
5. The intermediate image is at $F_e$, so rays diverging from it hit the eyepiece and emerge parallel on the right-hand side. The ray that passed through the centre of the objective and arrived at $F_e$ on the axis exits the eyepiece along the axis. The offset ray exits parallel to the axis but displaced. — An object placed at the focal point of a converging lens produces parallel emergent rays — the image is at infinity. This is the optical definition of normal adjustment. Draw the emerging rays as parallel lines extending to the right.
6. Label: both lenses, $f_o$, $f_e$, $F_o = F_e$ (shared focal point), the intermediate image (real, inverted, at $F_o$), and the parallel emergent rays indicating the final image at infinity. Add arrows on all rays to show direction of travel. — Marks are awarded for labels, not just for drawing lines. A diagram without labels is incomplete in an AQA exam. The arrow on the intermediate image showing it is inverted is a detail that distinguishes a top-band answer.

Answer: A complete diagram shows: two converging lenses separated by $f_o + f_e$; two parallel incident rays converging at the shared focal point $F_o = F_e$ to form a real inverted intermediate image; rays then emerging parallel from the eyepiece, indicating the final virtual image at infinity.

One detail that students frequently omit: the intermediate image arrow. Drawing a small downward-pointing arrow at the shared focal point communicates that the image is real and inverted — both properties that the examiner is looking for.

5. Angular Magnification and What the Diagram Reveals

The ray diagram does more than show image formation — it encodes the angular magnification directly. Angular magnification $M$ is the ratio of the angle $\beta$ subtended at the eye by the final image to the angle $\alpha$ subtended by the object viewed with the naked eye.

You can read this result directly from the geometry of the diagram. The height of the intermediate image $h$ equals $f_o \tan\alpha \approx f_o \alpha$ for small angles. The same height $h$ is at distance $f_e$ from the eyepiece, so $\beta \approx h / f_e = f_o \alpha / f_e$. Dividing gives $M = f_o / f_e$.

This tells you something practically important: to achieve high magnification, you want a long-focal-length objective and a short-focal-length eyepiece. Swapping to a shorter eyepiece increases $M$; note that the tube length $L = f_o + f_e$ also decreases slightly, so the telescope must be adjusted accordingly — which is why astronomers carry several eyepieces.

Worked example: Calculating Angular Magnification and Tube Length

An astronomical 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 (a) the angular magnification and (b) the length of the telescope tube.

1. $f_o = 900 \text{ mm} = 0.900 \text{ m}$ $f_e = 25 \text{ mm} = 0.025 \text{ m}$ — Convert both focal lengths to the same unit before substituting. Using metres keeps everything in SI units.
2. $M = \frac{f_o}{f_e} = \frac{900}{25}$ — You can use millimetres here because the units cancel in the ratio — but be consistent.
3. $M = \frac{900}{25} = 36$ — Check: 25 × 36 = 900. Confirmed. The object appears 36 times larger in angle than with the naked eye. [A1]
4. $L = f_o + f_e = 900 \text{ mm} + 25 \text{ mm} = 925 \text{ mm}$ — In normal adjustment the lenses are separated by the sum of their focal lengths. [A1]
5. $M = 36$ (no units — it is a ratio) $L = 925 \text{ mm} = 0.925 \text{ m}$ — Angular magnification is dimensionless. Always state the unit for tube length. A sanity check: $L$ should be much larger than $f_e$ alone, which it is (925 mm >> 25 mm).

Answer: Angular magnification $M = 36$; tube length $L = 925$ mm (0.925 m).

6. Diagram Features That Win Marks

AQA mark schemes for ray diagram questions award marks for specific, identifiable features. Knowing these in advance means you can draw with purpose rather than hoping for the best.

• Two converging lenses drawn with the objective visibly larger in diameter than the eyepiece
• Incident rays drawn parallel to the principal axis (representing a distant on-axis object)
• Rays converging at the shared focal point $F_o = F_e$ to form the intermediate image
• Intermediate image marked as a real, inverted arrow at the shared focal point
• Rays emerging from the eyepiece as parallel lines (indicating the final image at infinity)
• Lens separation labelled as $f_o + f_e$
• All focal lengths and focal points clearly labelled

A diagram that has all rays correct but no labels will lose several marks. Treat labelling as part of the physics, not an afterthought.

7. Exam Technique and Common Mistakes

Ray diagram questions in AQA astrophysics papers follow predictable patterns. The mistakes below appear repeatedly in mark scheme reports — recognising them now saves marks in the exam.

For a 'describe' or 'explain' question about normal adjustment, the two-sentence structure below consistently earns full marks: state the geometric condition (shared focal points), then state the consequence (final image at infinity, eye relaxed).

Mastering the ray diagram for normal adjustment means understanding one central idea: the two focal points coincide, parallel rays enter, parallel rays leave, and the angular magnification is $f_o / f_e$. Every other detail in the diagram follows from this single condition.