Non-Optical Telescopes and Telescope Performance

SPEC 3.9.1.3 & 3.9.1.4

Radio telescopes, resolving power, collecting power, and CCDs

Beyond Visible Light

The visible spectrum is only a tiny fraction of the electromagnetic spectrum. Many astronomical objects emit radiation across all wavelengths.

Key Concept

Different types of telescope are needed to detect different parts of the electromagnetic spectrum. Each region reveals different information about the universe.

Resolving Power and Diffraction

When light passes through a telescope aperture, it diffracts. This limits the telescope's ability to distinguish between two closely-spaced objects.

Definition

Resolving power is the ability of a telescope to produce separate images of close-together objects. It equals the minimum angular separation at which two objects can be distinguished.

Key Insight

Even a perfect point source appears as a diffraction pattern called an Airy disk. Two sources are just resolved when the central maximum of one pattern falls on the first minimum of the other.

The Rayleigh Criterion

The Rayleigh criterion defines when two objects can just be resolved. This occurs when the central maximum of one Airy pattern falls on the first minimum of the other.

Rayleigh Criterion

θ = \frac{1.22 λ}{D} \approx \frac{λ}{D}

θ = minimum angular resolution (radians), λ = wavelength, D = aperture diameter

Exam Tip

Smaller θ = better resolution. To improve: use shorter wavelength (smaller λ) or larger aperture (bigger D). θ must be in radians. 1 rad = 206,265 arcseconds.

Worked Example01

A radio telescope has a dish diameter of 76 m and operates at a wavelength of 21 cm. Calculate its angular resolution in: (a) radians, (b) arcseconds.

1

Identify values: D = 76 m, λ = 21 cm = 0.21 m

2

Apply Rayleigh criterion: θ ≈ λ/D = 0.21/76 = 2.76 × 10⁻³ rad

3

Convert to arcseconds: θ = 2.76 × 10⁻³ × 206,265 = 570 arcseconds

Answer: (a) 2.8 × 10⁻³ rad (b) 570 arcseconds (≈ 9.5 arcminutes)

Single-Dish Radio Telescopes

Radio telescopes detect radio waves (λ ~ 1 mm to 30 m) from astronomical sources. A key example is detecting the 21 cm hydrogen emission line from interstellar gas.

Key Features

Wire mesh can replace solid dish (if mesh size < λ/20)

Can operate in daytime

Surface accuracy: ±0.005 m acceptable

Interference

Radio telescopes suffer man-made interference: mobile phones, microwave ovens, radio transmissions. Must be in remote areas.

Why Radio Telescopes Must Be Large

Radio wavelengths are much longer than visible light. Since θ = λ/D, to achieve comparable resolution, radio telescopes need much larger diameters.

The Scale Problem

TypeTypical λRatio to VisibleVisible light500 nm = 5 × 10⁻⁷ m1Radio (21 cm)0.21 m420,000×

Key Comparison

To achieve the same resolving power as an optical telescope, a radio telescope would need to be about 50,000 times larger. This is why radio telescopes have much lower resolving power.

Surface Accuracy

Radio: ±0.005 m acceptable. Optical: ±0.0000001 m (100 nm) needed. Surface imperfections must be small compared to λ.

Radio vs Optical Telescopes

Below is a table with the similarities and differences of radio telescopes and optical telescopes:

SimilaritiesDifferences

Both types of telescopes function in the same way : they intercept and focus incoming radiation to detect its intensity .Since radio wavelengths are much larger than visible wavelengths , radio telescopes have to be much larger in diameter to achieve the same resolving power. They also have a larger collecting power .

Both radio and optical telescopes can be moved to focus on different sources , or to track a moving source.Construction of radio telescopes is cheaper and simpler because a wire mesh is used instead of a mirror . As long as mesh size < λ/20, radio waves will be reflected.

The parabolic dish of a radio telescope is extremely similar to the objective mirror of a reflecting optical telescope.A radio telescope must move across an area to build up an image , unlike optical telescopes.

Both optical and radio telescopes can be built on the ground ( ground-based ) since both radio waves and optical light can pass easily through the atmosphere .Radio telescopes experience man-made interference from radio transmissions, phones, microwave ovens etc. Optical telescopes experience interference from weather conditions, light pollution, stray radiation .

Worked Example02

Compare the angular resolution of: (a) A 10 m optical telescope using light of wavelength 500 nm, (b) A 76 m radio telescope using waves of wavelength 21 cm. By what factor is optical resolution better?

1

Optical: θ₁ = λ/D = (500 × 10⁻⁹)/10 = 5 × 10⁻⁸ rad

2

Radio: θ₂ = λ/D = 0.21/76 = 2.76 × 10⁻³ rad

3

Factor = θ₂/θ₁ = (2.76 × 10⁻³)/(5 × 10⁻⁸) = 55,200

Answer: The optical telescope has ~55,000× better resolution (smaller θ = better).

Radio Telescope Arrays

Since single radio dishes have poor resolution, astronomers connect multiple dishes as an interferometer. The effective diameter becomes the maximum separation (baseline).

Interferometry

When signals from multiple telescopes are combined, the array acts like a single telescope with diameter equal to the baseline.θ = λ / D_baseline

Examples

VLA (New Mexico): 27 dishes, 36 km baseline. Event Horizon Telescope: Earth-diameter baseline - imaged a black hole!

Collecting Power

A telescope's ability to detect faint objects depends on how much radiation it can collect, determined by the area of the primary mirror or lens.

Collecting Power

Collecting Power ∝ D²The collecting power is proportional to the area of the objective: Area = πD²/4, so collecting power ∝ (diameter)²

Why This Matters

Greater collecting power = brighter images. Doubling the diameter increases collecting power by a factor of 4 (not 2), allowing detection of fainter objects.

Worked Example03

The Hubble Space Telescope has a primary mirror of diameter 2.4 m. The James Webb Space Telescope has a mirror diameter of 6.5 m. By what factor is JWST's collecting power greater?

1

Collecting power ∝ D²

2

Ratio = (D_JWST/D_Hubble)² = (6.5/2.4)²

3

Ratio = (2.708)² = 7.34

Answer: JWST collects approximately 7.3× more light than Hubble.

Infrared Telescopes

Infrared telescopes use infrared radiation to create images of astronomical objects. They reveal cool objects invisible to optical telescopes.

Telescope Design

Infrared telescopes consist of large concave mirrors which focus radiation onto a detector.

What IR Reveals

Cooler regions in space

Dust clouds and star-forming regions

Cool stars and brown dwarfs

Distant galaxies (redshifted light)

Critical Challenges

Atmosphere absorbs most IR → must be launched into space and accessed remotely from ground

Must be cooled using cryogenic fluids (liquid nitrogen or hydrogen) to almost absolute zero

Must be well shielded to avoid thermal contamination from nearby objects and its own IR emissions

Why Cooling to Near Absolute Zero?

All warm objects emit IR radiation as heat. If the telescope is warm, its own emission swamps faint astronomical signals. Cooling eliminates this thermal noise.

Ultraviolet Telescopes

Ultraviolet telescopes use ultraviolet radiation to create images of astronomical objects.

Why Space-Based?

The ozone layer blocks all ultraviolet rays that have a wavelength of less than 300 nm. This means UV telescopes also need to be positioned in space.

UV Telescope Design

Use Cassegrain configuration to bring UV rays to a focus

Detected by solid state devices

Use photoelectric effect to convert UV photons into electrons

Electrons then pass around a circuit to create the signal

What UV Reveals

Interstellar medium

Star formation regions

Hot young stars

Active galactic nuclei

X-ray Telescopes

X-ray telescopes use X-rays to create images of astronomical objects.

Why Space-Based?

Since all X-rays are absorbed by the atmosphere, these telescopes need to be in space to collect data.

Special Mirror Design Required

X-rays have such high energy that using mirrors like an ordinary optical telescope would not work - they would just pass straight through.

X-ray telescopes must use a combination of parabolic and hyperbolic mirrors (grazing incidence), all of which must be extremely smooth. The rays enter the telescope, skim off the mirrors at shallow angles, and are brought to focus on CCDs which convert light into electrical pulses.

What X-rays Reveal

Since X-rays are high-energy, they can be used to observe high-energy events and areas of space such as active galaxies, black holes and neutron stars.

Gamma-ray Telescopes

Gamma telescopes use gamma radiation to create images of astronomical objects.

No Mirrors At All

These telescopes do not use mirrors at all as gamma rays have so much energy they would just pass straight through.

Instead, they use a detector made of layers of pixels. As the gamma photons pass through, they cause a signal in each pixel they come into contact with.

What Gamma Telescopes Observe

Gamma ray bursts (GRBs), quasars, black holes and solar flares

Short-lived GRBs

Last between 0.01 and 1 second

Associated with merging neutron stars (forming a black hole), or a neutron star falling into a black hole

Long-lived GRBs

Last between 10 and 1000 seconds

Associated with a Type II supernova (the death of a massive star)

Observatory Sites

Ground-based optical telescopes are best placed on mountain tops. Space telescopes avoid atmospheric problems entirely.

Why Mountain Tops?

Above dust and atmospheric pollution

Above low cloud, mist and fog

Far from light pollution from cities

Thinner air means less atmospheric absorption

Fewer convection currents so less image shake (better "seeing")

Ground-Based

Advantages: Larger mirrors possible, cheaper, easier to repair

Examples: Mauna Kea (Hawaii), Atacama (Chile)

Space-Based

Advantages: No atmospheric absorption/distortion, can observe UV/X-ray/gamma, 24/7 observation

Examples: Hubble, JWST, Chandra

CCDs: Charge-Coupled Devices

Modern telescopes use CCDs as detectors. CCDs are arrays of light-sensitive pixels that become charged when exposed to light by the photoelectric effect.

How CCDs Work

1. Photons hit a silicon chip divided into pixels

2. Each photon releases an electron (photoelectric effect)

3. Charge builds up - depends on intensity and exposure time

4. Cells are scanned electronically to build a digital image

Quantum Efficiency

A key measure of detector performance - how good is it at converting photons into detected signals?

Quantum Efficiency

QE = (number of electrons detected per second) / (number of photons incident per second) × 100%A device with high QE produces many electrons for a given number of incident photons.

Typical Quantum Efficiency Values

DetectorQuantum EfficiencyHuman eye1 - 4%Photographic film4 - 10%CCD70 - 90%

Key Point

A CCD detects up to 90% of incident photons vs only 1-4% for the eye. CCDs can detect objects far too faint for the eye to see.

CCDs vs The Human Eye

CCDs have largely replaced the human eye and photographic plates due to several key advantages.

PropertyHuman EyeCCD

Quantum efficiency1-4%70-90%

Spectral rangeVisible onlyIR, UV, visible

Spatial resolution~100 μm~10 μm

Integration time~0.1 s (limited)Hours possible

StorageMemory onlyPermanent digital

ConvenienceSimple, no equipmentNeeds setup, but digital

Summary

CCDs are better for detecting finer details and producing images that can be shared and stored. The eye's only advantage is convenience.

Worked Example04

What aperture would be needed to give a resolving power of 10⁻³ radians using: (a) a radio telescope at 20 cm wavelength, (b) an optical telescope at 600 nm wavelength?

1

Rearrange θ = λ/D to give D = λ/θ

2

(a) Radio: D = 0.20/(10⁻³) = 200 m

3

(b) Optical: D = (600 × 10⁻⁹)/(10⁻³) = 6 × 10⁻⁴ m = 0.6 mm

Answer: (a) 200 m for radio (b) 0.6 mm for optical - showing why radio telescopes must be enormous!

Practice

Question 1: A radio telescope array has a baseline of 10 km and operates at wavelength 6 cm. Calculate the angular resolution in arcseconds.

1

θ = λ/D = 0.06/(10,000) = 6 × 10⁻⁶ rad

2

Convert: θ = 6 × 10⁻⁶ × 206,265 = 1.24 arcseconds

Answer: 1.2 arcseconds

Practice

Question 2: Telescope A has diameter 8 m. Telescope B has diameter 2 m. (a) By what factor is A's collecting power greater? (b) By what factor is A's resolution better (same wavelength)?

1

(a) Collecting power ratio = (D_A/D_B)² = (8/2)² = 16

2

(b) Resolution: θ ∝ 1/D, so ratio = D_A/D_B = 8/2 = 4

Answer: (a) 16× more collecting power (b) 4× better resolution

Practice

Question 3: Explain why X-ray telescopes must be placed in space, and describe one design feature that differs from optical telescopes.

1

X-rays are completely absorbed by the Earth's atmosphere, so they cannot reach ground-based detectors.

2

Design feature: Use grazing incidence mirrors (parabolic + hyperbolic). X-rays pass through normal mirrors but reflect at very shallow angles.

Key points: All X-rays absorbed by atmosphere + grazing incidence mirrors needed.

Summary

Key Formulas

QuantityFormulaNotes

Angular resolutionθ = 1.22λ/D ≈ λ/Dθ in radians; smaller = better

Collecting power∝ D²Proportional to area

Quantum efficiency(e⁻/photons) × 100%Eye ~1-4%, CCD ~70-90%

Key Concepts

Radio: Large dishes, wire mesh OK, daytime operation, man-made interference

IR: Cooled with cryogenic fluids, atmosphere absorbs most IR

UV: Ozone blocks λ < 300 nm, must be in space, Cassegrain + solid state detectors

X-ray: All absorbed by atmosphere, grazing incidence (parabolic + hyperbolic) mirrors

Gamma: No mirrors, layered pixel detectors, observe GRBs (short: 0.01-1s, long: 10-1000s)

CCDs: Higher QE, wider spectral range, longer integration, digital storage

Observatory sites: Mountains - above dust/cloud/pollution, thinner air, less convection