Questions on the astronomical refracting telescope

Question 1

State what is meant by a refracting telescope in normal adjustment, and where the final image is formed.

Accepted Answer

The focal planes of the objective and the eye lens coincide (1). The final image is formed at infinity, viewed by a relaxed eye (1).

Question 2

Define the angular magnification of an astronomical telescope.

Accepted Answer

The angle subtended at the eye by the final image divided by the angle subtended at the unaided eye by the object (1). It is a ratio of angles, not of sizes (1).

Question 3

Name the two mirrors of a Cassegrain reflecting telescope, giving the shape of each, and state where the light comes to its final focus.

Accepted Answer

A concave parabolic primary mirror and a convex secondary mirror (1). The secondary reflects the converging light back through a central hole in the primary, to a focus just behind it (1).

Question 4

Explain why the angular magnification of a telescope has no unit.

Accepted Answer

It is the ratio of two angles: the angle subtended at the eye by the final image divided by the angle subtended by the object at the unaided eye (1). Both angles carry the same unit, which cancels in the ratio (1).

Question 5

A refracting telescope is pointed at a distant star. State where the objective lens forms an image of the star, and the nature of that image.

Accepted Answer

In the objective's focal plane, one focal length behind the lens, because the light arrives as (near-)parallel rays (1). The image is real (and inverted) (1).

Question 6

A telescope in normal adjustment has its eye lens replaced by one of half the focal length. State the effect on the angular magnification.

Accepted Answer

M = f_o/f_e, so halving f_e doubles the angular magnification (1).

Question 7

A telescope's objective is a mirror. State which of chromatic aberration and spherical aberration can still spoil the image it forms, giving a reason in each case.

Accepted Answer

Spherical aberration can still occur (1): a mirror ground to a spherical surface brings rays striking near its edge to a focus closer than rays near its axis, and only a parabolic surface removes that. Chromatic aberration cannot occur (1): reflection is independent of wavelength, so the mirror sends every colour to the same focus, and no light is refracted in forming the image.

Question 8

A refractor has an objective of focal length 1.5 m and an eye lens of focal length 30 mm. For normal adjustment, calculate the angular magnification and the length of the telescope.

Accepted Answer

M = f_o/f_e = 1.5/0.030 (1) M = 50 (1) Length = f_o + f_e = 1.53 m (1)

Question 9

The Moon subtends an angle of 9.0 × 10⁻³ rad to the naked eye. Calculate the angle it subtends through a telescope with a 1.0 m objective and a 25 mm eye lens in normal adjustment.

Accepted Answer

M = f_o/f_e = 1.0/0.025 = 40 (1) Angle through the telescope = 40 × 9.0 × 10⁻³ (1) = 0.36 rad (1)

Question 10

A refractor in normal adjustment has an angular magnification of 60 and a tube 1.22 m long. Calculate both focal lengths.

Accepted Answer

f_o = 60f_e (1) f_o + f_e = 1.22 m (1) 61f_e = 1.22 m, so f_e = 20 mm (1) f_o = 1.20 m (1)

Question 11

Explain what causes chromatic aberration in a refracting telescope, and why a reflecting telescope does not suffer from it.

Accepted Answer

Glass refracts blue light more than red, so each colour has a different focal length (1) and no single sharp focus exists; images carry coloured fringes (1). Reflection is the same for every wavelength, so mirrors focus all colours together (1).

Question 12

Explain what is meant by spherical aberration, and how the design of the Cassegrain primary mirror avoids it.

Accepted Answer

For a spherical surface, rays striking near the edge focus closer than rays near the axis (1), smearing the focus (1). The primary is ground to a parabola, which brings all rays parallel to the axis to a single focus (1).

Question 13

Two stars are separated by an angle of 1.5 × 10⁻⁴ rad when viewed with the unaided eye. Through a refractor in normal adjustment fitted with a 20 mm eye lens, they appear separated by 6.0 × 10⁻³ rad. Calculate the focal length of the objective.

Accepted Answer

M = 6.0 × 10⁻³/(1.5 × 10⁻⁴) = 40 (1). f_o = Mf_e (1) = 40 × 0.020 = 0.80 m (1).

Question 14

An astronomer removes the eyepiece of a refractor and places a CCD in the focal plane of the objective, which has a focal length of 1.4 m. Mars subtends an angle of 1.2 × 10⁻⁴ rad to the naked eye. The pixels of the CCD are 5.0 μm wide. Determine the number of pixels spanned by the diameter of the image of Mars.

Accepted Answer

Diameter of the real image = f_oθ (1) = 1.4 × 1.2 × 10⁻⁴ = 1.7 × 10⁻⁴ m (1). Number of pixels = 1.7 × 10⁻⁴/(5.0 × 10⁻⁶) (1) ≈ 34 pixels (accept 33 to 34) (1).

Question 15

Explain how the Cassegrain design gives a long effective focal length in a short tube, and give one practical benefit of the short tube.

Accepted Answer

The convex secondary reflects the converging beam back down the tube and through the hole in the primary, folding the light path so the effective focal length is much longer than the tube (1). A short tube is stiffer, cheaper to mount and easier to steer (any one) (1).

Question 16

A refractor in normal adjustment has an objective of focal length 1.25 m in a tube 1.30 m long. Show that its angular magnification is about 25. Go on to calculate the angle subtended through the telescope by a lunar crater that subtends 6.0 × 10⁻⁵ rad to the naked eye.

Accepted Answer

f_e = 1.30 − 1.25 = 0.050 m (1). M = f_o/f_e = 1.25/0.050 = 25 (1). Angle through the telescope = Mθ (1) = 25 × 6.0 × 10⁻⁵ = 1.5 × 10⁻³ rad (1).

Question 17

Sketch the path taken by light from a distant star on the axis of a Cassegrain reflecting telescope, from its arrival at the tube to the point where it comes to a focus. Label the primary mirror, the secondary mirror and the position of the eyepiece.

Accepted Answer

Parallel rays arrive along the axis and are drawn converging after reflection at the large concave primary mirror, which is labelled (1). A small convex secondary mirror, labelled, is drawn on the axis in front of the primary and before the primary's focus, reflecting the converging beam back down the tube (1). The beam is drawn passing through a central hole in the primary (1). The focus, marked as the position of the eyepiece, lies just behind the primary mirror (1).

Question 18

An uncorrected single-lens objective has a focal length of 1.200 m for red light and 1.176 m for blue light. It is used with a 24 mm eye lens. Calculate the angular magnification the telescope gives in each colour, and explain what the observer sees when the eyepiece is set to give a sharp red image.

Accepted Answer

Red: M = f_o/f_e = 1.200/0.024 = 50 (1). Blue: M = 1.176/0.024 = 49 (1). The two focal planes lie 24 mm apart, so the eyepiece cannot be in normal adjustment for both colours at once (1). With the red image sharp the blue image is out of focus and slightly smaller, so each star carries a coloured fringe: this is chromatic aberration (1).

Question 19

A Cassegrain telescope has an effective focal length of 3.2 m and is used with an 8.0 mm eyepiece. A lunar crater subtends 4.0 × 10⁻⁵ rad to the naked eye. Calculate the magnification and the angle the crater subtends through the telescope.

Accepted Answer

M = f_o/f_e = 3.2/0.0080 (1) M = 400 (1) The crater subtends 400 × 4.0 × 10⁻⁵ (1) = 0.016 rad (1)

Question 20

A school is choosing between a large refractor and a Cassegrain reflector of the same aperture. Give two advantages of the reflector and one practical advantage of the refractor.

Accepted Answer

Reflector, any two of the following, one mark each (2): no chromatic aberration; the parabolic primary removes spherical aberration; a mirror can be supported across its back, so large apertures are practical; less light is lost than in thick glass. Refractor: a sealed tube needs little maintenance (1), while mirror coatings need occasional renewal (1).

Question 21

A magazine advert praises a small telescope for its “500× magnification”. Explain why this figure alone says little about the telescope's quality.

Accepted Answer

Magnification only stretches the light already collected; it can be made large with a short-focus eyepiece (1). Image quality depends on the objective's diameter, which sets how much light is gathered and how fine the detail that can be resolved (1).

Question 22

A catalogue states that a refractor gives an angular magnification of 60 in normal adjustment with the 25 mm eyepiece supplied, in a tube 1.05 m long. Deduce whether the claim is consistent with the stated dimensions.

Accepted Answer

In normal adjustment f_o = tube length − f_e = 1.05 − 0.025 = 1.025 m (1). M = f_o/f_e = 1.025/0.025 = 41 (1). A magnification of 60 with this eyepiece would need f_o = 1.5 m, so a tube 1.525 m long (OR 41 falls well short of 60) (1). The claim is not consistent with the dimensions (1).

Question 23

A refractor in normal adjustment is pointed at a distant star cluster. Explain fully how the two lenses produce the final image, and show how the angular magnification is related to the two focal lengths.

Accepted Answer

Rays from the distant cluster arrive effectively parallel, and the objective converges them to a real image in its focal plane (1). In normal adjustment this plane is also the focal plane of the eye lens (1), so the light leaves the eye lens as parallel beams and the final image is at infinity, viewed by a relaxed eye (1). An image of height h in the shared focal plane subtends the object's unaided angle θ_o ≈ h/f_o at the objective (1) and the larger angle θ_i ≈ h/f_e at the eye (1). Dividing: M = θ_i/θ_o = f_o/f_e (1).

Question 24

The largest refracting telescope used for research has an objective 1.0 m in diameter, while reflecting telescopes with mirrors 8 m across are in everyday use. Explain, in terms of how each objective has to be supported and the thickness of glass the light meets, why a mirror can be built to this size and a lens cannot.

Accepted Answer

Light has to pass through the whole of a lens, so a lens can be held only around its rim (1). At a large diameter it then sags under its own weight, distorting its surfaces and spoiling the image (1). A mirror forms its image at the front surface alone, so it can be supported across the whole of its back and keeps its shape at any diameter (1). A large lens must also be thick, and the light crosses all of that glass, so an appreciable fraction is absorbed and the glass must be flawless throughout its volume, whereas only the surface of the mirror has to be right (1).

Question 25

In a Cassegrain reflector the light is reflected first at the primary mirror and then at the secondary, and each coating returns 92% of the light that falls on it. In a refractor of the same aperture the light instead crosses a single thick objective lens: 4% is reflected away at each of its two glass surfaces, and 18% of the light entering the glass is absorbed on the way through. Calculate the fraction of the collected light that reaches the eyepiece in each instrument, and compare the two designs on this point.

Accepted Answer

Reflector: the light meets two coated surfaces, so the fraction surviving is 0.92 × 0.92 (1) = 0.846, about 85% (1).
Refractor: 96% gets past the first surface, 82% of that survives the glass and 96% of what is left gets past the second surface, so the fraction is 0.96 × 0.82 × 0.96 = 0.756, about 76% (1).
The reflector delivers about 1.12 times as much of the light it collects. The refractor's largest single loss is the absorption in the glass, and that grows as the lens is made thicker, whereas a mirror loses light only at its coating however large it is built (1).