Questions on telescopes beyond the visible

Question 1

Explain why ultraviolet and X-ray telescopes must be placed in orbit, while radio telescopes work at ground level.

Accepted Answer

The atmosphere absorbs ultraviolet and X-rays high above the ground, so they never reach a ground-based instrument (1). It is transparent to radio wavelengths, so a dish at sea level receives them day and night, even through cloud (1).

Question 2

State the Rayleigh criterion for two point sources to be just resolved by a telescope of aperture D observing at wavelength λ.

Accepted Answer

The sources are just resolved when the central maximum of one diffraction pattern falls on the first minimum of the other (1): their angular separation is then θ ≈ λ/D, with θ in radians (1).

Question 3

Define the quantum efficiency of a detector, and give typical values for a CCD and for the human eye.

Accepted Answer

Quantum efficiency is the fraction (or percentage) of the photons arriving at the detector that are actually registered (1). A CCD achieves about 80%; the eye manages about 1% (1).

Question 4

The reflecting surface of a large radio dish can be open wire mesh, yet an optical telescope's mirror must be polished smooth. Explain the difference.

Accepted Answer

A surface reflects cleanly provided its irregularities are small compared with the wavelength (1). Radio wavelengths are centimetres to metres, so mesh serves, while visible light at about 5 × 10⁻⁷ m demands a far smoother surface (1).

Question 5

State the two regions of the electromagnetic spectrum for which the atmosphere is transparent down to sea level.

Accepted Answer

Visible light (1) and radio waves (1).

Question 6

Give two reasons a radio telescope can observe for more hours in a year than an optical telescope at the same site.

Accepted Answer

Radio waves pass through cloud (1), and observing can continue in daylight, because the daytime sky is not bright at radio wavelengths (1).

Question 7

Calculate the smallest angle a telescope with a 5.0 m mirror can resolve when observing at 550 nm.

Accepted Answer

θ ≈ λ/D = 5.5 × 10⁻⁷/5.0 (1) θ = 1.1 × 10⁻⁷ rad (1)

Question 8

A dark-adapted pupil is 5.0 mm across and works at about 500 nm. A radio dish 64 m across observes at 0.21 m. Calculate the resolving power of each and state which can resolve the finer detail.

Accepted Answer

Eye: θ ≈ 5.0 × 10⁻⁷/(5.0 × 10⁻³) (1) θ = 1.0 × 10⁻⁴ rad (1) Dish: θ ≈ 0.21/64 = 3.3 × 10⁻³ rad (1) The eye's angle is the smaller, so the eye resolves the finer detail (1)

Question 9

Calculate the ratio of the collecting powers of a 10 m telescope and a 2.0 m telescope.

Accepted Answer

Collecting power ∝ D², so the ratio = (10/2.0)² (1) Ratio = 25 (1)

Question 10

A space telescope has a 2.4 m mirror and observes at 550 nm. The Moon is 3.8 × 10⁸ m away. Estimate the smallest feature it can resolve on the lunar surface.

Accepted Answer

θ ≈ λ/D = 5.5 × 10⁻⁷/2.4 (1) θ = 2.3 × 10⁻⁷ rad (1) Smallest feature ≈ θd = 2.3 × 10⁻⁷ × 3.8 × 10⁸ (1) ≈ 87 m (1)

Question 11

An amateur telescope has an objective 130 mm in diameter and observes at 550 nm. The two stars of a binary are separated by an angle of 2.2 × 10⁻⁶ rad. Determine whether the telescope can resolve the pair.

Accepted Answer

θ ≈ λ/D = 5.5 × 10⁻⁷/0.130 (1) = 4.2 × 10⁻⁶ rad (1). This is larger than the stars' separation of 2.2 × 10⁻⁶ rad, so the telescope cannot resolve them (1).

Question 12

A telescope of aperture 0.50 m collects enough light from a faint galaxy in a 1.0 hour exposure. Calculate the exposure a 0.25 m telescope of the same design would need to collect the same amount of light.

Accepted Answer

Collecting power ∝ D² (1), so the smaller telescope gathers light (0.50/0.25)² = 4 times more slowly (1). Exposure needed = 4 × 1.0 = 4.0 hours (1).

Question 13

A telescope of aperture 0.30 m can observe through a blue filter at 400 nm or a red filter at 700 nm. Calculate the smallest resolvable angle with each filter, and state which filter shows the finer detail.

Accepted Answer

Blue: θ ≈ 4.0 × 10⁻⁷/0.30 = 1.3 × 10⁻⁶ rad (1). Red: θ ≈ 7.0 × 10⁻⁷/0.30 = 2.3 × 10⁻⁶ rad (1). The blue angle is the smaller, so the blue filter resolves the finer detail (1).

Question 14

During one exposure, 6.0 × 10⁴ photons from a faint galaxy arrive at a detector. Calculate the number registered by a CCD of quantum efficiency 80%, and by an eye of quantum efficiency 1%.

Accepted Answer

CCD: 0.80 × 6.0 × 10⁴ = 4.8 × 10⁴ photons (1). Eye: 0.01 × 6.0 × 10⁴ = 600 photons (1).

Question 15

Calculate the dish diameter a radio telescope observing at 0.21 m would need to match the resolving power of a 15 cm optical telescope working at 500 nm. Comment on your answer.

Accepted Answer

Optical: θ ≈ 5.0 × 10⁻⁷/0.15 = 3.33 × 10⁻⁶ rad (1)
Radio: D = λ/θ = 0.21/(3.33 × 10⁻⁶) (1)
D = 63 km (1)
A single dish 63 km across cannot be built, which is why radio astronomers link widely separated dishes to act as one large aperture (1)

Question 16

An astronomy society is debating whether to image a faint galaxy by eye or with a CCD camera. Give four distinct reasons the CCD will out-perform the eye.

Accepted Answer

Its quantum efficiency is far higher, about 80% against about 1% (1); it can accumulate light over an exposure of minutes or hours while the eye refreshes many times a second (1); it produces a permanent, shareable record whose brightness can be measured (1); and it responds beyond the visible, into the infrared and ultraviolet (1).

Question 17

Describe one way the structure or siting of an infrared telescope and of an X-ray telescope must each differ from an ordinary optical telescope, giving the reason in each case.

Accepted Answer

Infrared: sited on high, dry mountains or in space because water vapour absorbs infrared (1), and the detectors are cooled so the instrument's own warmth does not swamp the signal (1). X-ray: placed in orbit, and focused by grazing-incidence nested mirrors, because X-rays penetrate an ordinary mirror surface instead of reflecting (1).

Question 18

An astronomy society may borrow one instrument to try to split a double star of angular separation 3.0 × 10⁻⁶ rad: an optical telescope of aperture 0.20 m used at 550 nm, or a radio dish of diameter 25 m used at 0.21 m. Deduce which instrument, if either, can resolve the pair, and determine the dish diameter a radio observatory would need to match it.

Accepted Answer

Optical: θ ≈ 5.5 × 10⁻⁷/0.20 = 2.75 × 10⁻⁶ rad, smaller than 3.0 × 10⁻⁶ rad, so the optical telescope just resolves the pair (1). Radio: θ ≈ 0.21/25 = 8.4 × 10⁻³ rad, far too coarse (1). Required dish: D = λ/θ = 0.21/(3.0 × 10⁻⁶) (1) = 70 km, which is why single dishes never match optical resolution (1).

Question 19

Two radio dishes 320 km apart are linked so that they act as a single aperture equal to their separation, observing at 0.060 m. Calculate the resolving power of the linked pair and of a single 64 m dish at the same wavelength, and state the improvement factor.

Accepted Answer

Linked pair: θ ≈ 0.060/(3.2 × 10⁵) = 1.9 × 10⁻⁷ rad (1). Single dish: θ ≈ 0.060/64 = 9.4 × 10⁻⁴ rad (1). Improvement factor = 3.2 × 10⁵/64 = 5000 (1): the linked pair resolves detail 5000 times finer (1).

Question 20

Estimate the greatest distance at which a dark-adapted eye, pupil diameter 4.0 mm working at 500 nm, could resolve a car's two headlamps, which are 1.5 m apart.

Accepted Answer

θ ≈ λ/D = 5.0 × 10⁻⁷/(4.0 × 10⁻³) = 1.25 × 10⁻⁴ rad (1). The lamps are just resolved when they subtend this angle, so d ≈ s/θ (1) = 1.5/(1.25 × 10⁻⁴) ≈ 1.2 × 10⁴ m, about 12 km (1).