Mr Toogood's Physics

Refracting telescopes

These telescopes use two lenses to collect light and allow astronomical objects to be viewed. These are the easiest to understand, and were the first to be developed. They consist of an objective lens and an eyepiece lens. The objective lens collects the light from stars and brings it to a focus at its focal length, $f_{o}$. This forms an intermediate real image. The eyepiece lens is placed at a distance of 1F from the intermediate image and produces parallel rays of light.

It is important that the eyepiece lens produces parallel rays, otherwise the observer’s eye would have to work harder to view the light and it would cause fatigue. In this arrangement the observer’s eye is relaxed, or unaccommodated. As the emergent rays are parallel, the image created is at infinity. The arrangement is shown below. This is a diagram that you need to learn, and may be expected to reproduce in an exam. In the video below I show you how to draw this important diagram.

Some important things to note about this arrangement, called normal adjustment, are that the two lenses are arranged so that their focal points are in the same place. The objective lens has a much longer focal length that the eyepiece lens and that the angle $β$ is larger than the angle $α$. The focal lengths of the two lenses also define how long the telescope will be.

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Angular magnification

It is these last two points that determine the telescope’s magnification, As the angle $β$ is larger than $α$ the image will have a larger angular size. Angular size is the apparent size that an object appears and depends on both its actual size and its distance from the observer. Two objects can be very different sizes, but if they are different distances from the observed, they may appear the same size. For example the moon has a diameter of $\quantity{3474}{km}$, but as it is $\quantity{384400}{km}$ from Earth, it has an angular size of:

It is unusual to use radians in astronomy; the standard unit of angle is the degree, and its subdivisions, called minutes of arc ($\units{arcmin}$) and seconds of arc ($\units{arcsec}$). There are $\quantity{60}{arcmin}$ in one degree, and $\quantity{60}{arcsec}$ in one $\units{arcmin}$.

It is not unusual in astronomy to measure angular sizes as small as a hundredth of a second of arc! You will not be expected to convert between radians and arcsec, but you may have to convert degrees into minutes or seconds of arc.

The moon’s angular size, is about $\quantity{0.5}{°}$ or $\quantity{30}{arcmin}$

Telescopes magnify the angular size of an object, so that the angle subtended by the object when viewed with the naked eye is much less than the angle subtended by the image when viewed through the telescope.

The angular magnification can, therefore, be calculated by comparing the size of these two angles, $α$ and $β$:

By looking at the diagram below we can also see that the rays of light inside the telescope form two triangle for which the angles $α$ and $β$ can be described as:

$\tan α=\frac{h}{f_{o}}$ and $\tan β=\frac{h}{f_{e}}$

Where $h$ is the height of the intermediate image. In practice, the two angles will be very small, so using the small angle approximation where $\tan θ\approx θ$ when $θ$ is measured in radians:

$α=\frac{h}{f_{o}}$ and $β=\frac{h}{f_{e}}$

So we can show that the ratio of the two angles is also equal to the ratio of the focal lengths of the two lenses:

Therefore telescopes are designed to have very short focal lengths at the eyepiece and a much longer focal length at the objective. This also means that to obtain greater magnifications, much longer telescopes are required. This brings several technical problems, such as the difficulty of producing a structure strong enough to bear the weight of being tilted towards the sky, as well as the need to have the lenses perfectly aligned along the principle axis. Although many very powerful refracting telescopes have been built, these limitations, amongst others, restrict their use in modern day astronomical observations.

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Worked example

An early refracting telescope built by Johannes Hevelius had a length of $\quantity{3.7}{m}$ and a magnification of $50$. Calculate the focal lengths of both the objective and the eye piece lenses.

We assume the telescope is in normal adjustment, so if the total length of the telescope is $\quantity{3.7}{m}$, then:

And if the magnification is $50$ then:

Therefore:

$f_{o}=50 f_{e}$ and $f_{o}=\quantity{3.7}{m}-f_{e}$

So:

The objective lens will, therefore, have a focal length of:

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Reflecting telescopes

Reflecting telescopes use large parabolic mirrors to collect light. Similarly to refracting telescopes, they focus light to an eyepiece, but reflecting telescopes have many advantages over refractors. For astronomical observations, gathering as much light as possible is vital, the majority of objects that are studied are a great distance away, and are dim point objects. Therefore, magnifying a star has little effect, it still appears as a point object, but by gathering more light from it allows us to make brighter images which are more useful to study. The amount of light that can be gathered by a telescope is proportional to the square of the diameter of the telescope.

The pupil in the eye may be up to $\quantity{8}{mm}$ when fully dilated, but relatively cheap reflecting telescope can be $\quantity{15}{cm}$ in diameter, this means that it can collect over 350 times more light:

This allows the observer to see more detail in closer objects, but, more importantly, it allows much dimmer objects to be seen. I would encourage all students to try this out by going outside on a clear evening and looking up at the stars with a pair of binoculars and comparing the difference even a small increase in brightness of around 25 times can make.

Mirrors can be made to very large sizes, so these telescopes are able to see very dim objects indeed.

Reflecting telescopes come in two main forms, Newtonian, and, Cassegrain. The difference between them is in where the eyepiece is situated. Both types consist of one large primary mirror which collects the light and reflects it onto a small secondary mirror. The secondary mirror focuses the light and directs it onto an eyepiece lens.

The Newtonian telescope is designed so that the eyepiece is at the top of the telescope, by the main aperture. This makes it more suited to comfortable observations from a standing or seated position. The Cassegrain arrangement, shown below places the eyepiece behind the primary mirror, and the rays are directed through an aperture in it. This makes it more suitable for placing a camera on the lens, and also is much more useful when scaling up telescopes to large sizes. Looking at the arrangement below, you may think that having the secondary mirror in that position would cause a large obstruction in the image, but it is important to remember that both the rays shown are from the same point on the object, no information is lost by placing the mirror here.

You need to be able to draw this ray diagram for a Cassegrain telescope, so some important things to note are:

• The rays enter the telescope parallel
• The rays do not cross before the secondary mirror, in fact they should not cross until they reach the aperture in the primary mirror.
• The secondary mirror is a convex mirror and must clearly be shown as such.
• Hatching or shading the non-reflective side is a good idea.

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Aberrations

Both refracting and reflecting telescopes are precise scientific instruments which are used for making very careful observations of very small points of light, and therefore have to be made to very high specifications. This means that small imperfections in their design can cause large distortions in the images created. These distortions, or aberrations are usually caused by either the shape of the lens or mirror, or by the fundamental physics of refraction through glass.

If the mirror or lens is made too spherically it can produce a spherical aberration. This causes rays to be brought to different foci depending on their distance from the principle axis. The further from the principle axis the ray is, the shorter its focal length.

These are another set of ray diagrams that you must know how to draw, a useful mnemonic that I use to remember the diagram is that the cLoser the ray is to the principle axis, the Longer its focal length.

The optical effect of a spherical aberration is to cause the image to be very blurred as there are many focal points, instead of just one. The image below was taken by the Hubble Space Telescope, which suffered from spherical aberration when it was first launched due to a manufacturing error which had to be corrected whilst the telescope was in orbit.

The image on the left shows the spherical aberration, and you can easily see how the image is blurred and smeared out into a disc. The image on the right is much more focussed, although some small, unavoidable effects due to diffraction can be seen as rings around the star. Another thing to note about the aberration on the left is that the image has a much wider field of view, $\quantity{2}{arcsec}$ compared to the corrected image which is of only $\quantity{0.05}{arcsec}$, 40 times smaller.

Spherical aberration can be corrected by using a parabolic mirror or lens, as this will bring rays to a single focus no matter their distance from the principle axis, as in the diagram below:

However parabolic mirrors and lenses are much more expensive to produce, so cheaper telescopes suffer more from this aberration more than expensive ones with higher specification optics.

The other type of aberration that telescopes suffer from is chromatic aberration. This is caused by different wavelengths of light being diffracted by a different amount as they pass through a lens. Blue light is brought to a focus closer to the lens and red light will have a longer focal length. This is a fundamental property of refraction, and as such only affects refracting telescopes

The picture of the moon below is an extreme example of chromatic aberration, and usually the effect isn’t so drastic.

Chromatic aberration can be corrected using an achromatic doublet which consists of two lenses, one converging, made of crown glass, which is fixed to a diverging lens made from a different type of glass called flint glass. These lenses refract the light in opposite directions and allows the red light to be brought to a focus at the same point as the blue light.

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Comparison of telescopes

Most modern observatories use reflecting telescopes because they offer many advantages over refractors. The fact that reflectors do not suffer from chromatic aberration is an important factor. However, there are also several other reasons to choose a reflecting telescope over a refractor. Reflecting telescopes can be designed to observe wavelengths of light outside of the visible spectrum, which allows astronomers to view the universe in detail in areas our eyes cannot directly perceive.

Mirrors are much easier to produce than lenses, and as lenses are very heavy, when they get too large, they can distort under their own weight causing a more severe spherical aberration. Large reflecting telescopes can be made from using several smaller mirrors, these are called composite mirrors. This has the advantage that the exact position of each smaller mirror can be controlled which can constantly refine the shape of the overall mirror. The largest refracting telescope ever built was the Yerkes Observatory, built in 1890 and had a diameter on $\quantity{100}{cm}$, which is tiny compared to some of the largest reflecting telescopes currently in use.

The advantages of each type of telescope can be summarised by the points below:

Reflecting telescopes

• The diameter of a mirror can be much larger than that of a lens.
• Mirror surfaces can be made very thin
• Mirrors cannot produce chromatic aberration.
• Reflectors use paraboloidal mirrors and therefore so not produce spherical aberration.
• Reflectors can be used to study the long wavelength >300nm UV that penetrates the Earth's atmosphere.
• Composite mirrors can be made very large.
• Lenses need to be edge mounted so their weight can cause them to deform. Difficult to make glass sufficiently clear for refracting telescopes to see in great detail.
• Lighter and shorter for greater magnifications.

Refracting telescopes

• They are less sensitive to temperature changes than reflectors.
• They require less maintenance than reflectors because mirrors have to be re-aluminised periodically.

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Resolving power

When viewing distant objects along the same line, they can appear to be separated by a small angle, even if they are a considerable distance apart. Their angular separation $θ=\frac{s}{r}$

As well as the aberrations described earlier, if the angular separation between objects is too small it can limit the detail seen by a telescope. In fact this is fundamental limitation on the level of detail they can see stems from the diffraction of light as it enters the telescope. All light is diffracted when it passes through an aperture, the amount of diffraction is greatest when the size of the aperture is the same as the wavelength of the light. When light is diffracted through a slit it produces a series of bright fringes. You will have carried out investigations into diffraction during Year 12, but it is well worth reminding yourself about the physics and the equations that describe the phenomenon.

The equation that describes the position of the first bright fringe, $θ$ is:

Where

• $d$ width of the slit or aperture.
• $θ$ is the angle made by the central maximum.
• $n$ is the order of the maximum.
• $λ$ is the wavelength of the light.

When the aperture or slit is circular, a very similar effect occurs, but in this case the fringes become a circular pattern. About $98\%$ of the light from the source is contained within the bright central maximum, so the subsequent fringes are much dimmer. The central disc is called an airy disc and the overall circular pattern is called Fraunhofer diffraction pattern. Obviously telescopes of all sizes present a circular aperture to a wide range of wavelengths, and although the wavelength of visible light is much smaller than the diameter of a telescope, a small amount of diffraction still occurs. If the light comes from a point source, such as a star, this creates a series of bright circular fringes around the image. This effect doesn’t just affect the images of stars, but also sets a level for the minimum amount of detail that can be obtained when viewing distant objects, even within the solar system.

The angel $θ$ made by the central maximum in this case is very significant as it set the minimum angular resolution of the telescope and the image. It is described by a very similar equation, of which there is a proper form, and a more simplified form which you are able to use. Due to the circular nature of the diffraction pattern, instead of the integer, $n$ being used to determine the order of the maximum, integer multiples of 1.22 are used instead. And as the angle $θ$ in $\units{radians}$ is very small, the small angle approximation where $\sin θ\approx θ$ holds. Therefore the angle made by the central maximum is:

In the AQA specification the 1.22 is dropped to give:

However, you can use either equation , and would not be penalized.

If two objects appear to be too close, their diffraction patterns overlap too much, and it may become impossible to resolve them as separate objects. This is defined by the arbitrary condition called Rayleigh’s criterion, which states that:

Two sources will be (just) resolved if the central maximum of the diffraction pattern of one coincides with the first minimum of the other.

So if the two objects are separated by an angle of $<θ$ their airy discs overlap and they cannot be resolved as separate images. The factors the affect this minimum angular separation are the diameter of the telescope and the wavelength of the light being observed. This gives another advantage to reflecting telescopes as they can be built much larger, so $D$ is bigger and there is less diffraction. In fact the effective diameter of a reflecting telescope is reduced by the spider which holds the secondary mirror, but in exam questions we will neglect this and always perform calculations on the overall diameter of the telescope. However telescopes that observe longer wavelengths of light, such as radio telescopes will suffer from this problems significantly more than optical telescopes.

This problem doesn’t just affect the ability to see stars separately, but also limits the amount of detail that can be seen in closer objects, for example, if a feature on a planet, or moon, or even across a whole galaxy has an angular size of $<θ$ it will not be resolved against nearby objects. A good example of how much detail can be resolved by a telescope is shown in the two photos below. Both have been taken by the same telescope and camera, but at different distances from the object, and you can clearly see how much more detail can be seen in the photo taken from the shorter distance.

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Worked example

The table summarises some of the properties of Vesta, one of the largest objects in the asteroid belt between Mars and Jupiter.

1. Calculate the largest possible distance, in $\units{m}$, between the Earth and Vesta.

Although the unit, $\units{AU}$ hasn’t been covered yet it is worth thinking about the problem in the context of this question. One astronomical unit, $\units{AU}$, is the mean distance between the Earth and the Sun, and is equal to $\quantity{1.50\times 10^{11}}{m}$. It would be easy to rush into this question and convert the largest distance from the Sun into metres, but looking at the diagram below you can see that the greatest distance between the Earth and Vesta is when they are are on opposite sides of the Sun.



So the maximum distance between the two is $\quantity{1}{AU}+\quantity{2.57}{AU}=\quantity{3.57}{AU}$ which in metres is:

$$\quantity{3.57}{AU}\times\quantity{1.50\times 10^{11}}{m}=\quantity{5.36\times 10^{11}}{m}$$



2. Show that when Vesta is at a distance of $\quantity{1.73 \times 10^{11}}{m}$ from Earth, the angle subtended by Vesta to an observer on Earth is about $\quantity{3\times 10^{–6}}{radian}$.

In this question we are given all the data that we need and we have to correctly substitute it into the equation for angle in radians:

$$θ=\frac{s}{r}$$

The examiner will expect to see evidence that you have carried out the calculation, not just written it out with the provided data. The best way to demonstrate this is to give the answer of the calculation to more significant figures than provided, e.g.:

$$θ=\frac{\quantity{5.4\times 10^{5}}{m}}{\quantity{1.73 \times 10^{11}}{m}}=\quantity{3.1}{rad}$$



3. Observations of Vesta have been made by the Infrared Telescope Facility (IRTF) in Hawaii. The IRTF includes a camera capable of detecting infrared radiation with wavelengths in the range $\quantity{1.0}{μm}$ to $\quantity{5.0}{μm}$.
   The smallest angle the telescope can resolve is $\quantity{3.3\times 10^{–7}}{radian}$.
   Calculate the diameter of the objective of the telescope.
   Give your answer to a suitable number of significant figures.

As we have been told that the minimum angular resolution of the telescope is $\quantity{3.3\times 10^{–7}}{radian}$, we can use the equation, $θ\approx\frac{λ}{D}$. In this case the factor of 1.22 can be dropped for simplicity. The key to correctly answering this question is to pick the wavelength that would produce the smallest angle that can be resolved.

The longer the wavelength, the greater the amount of refraction and the larger the smallest resolvable angle. When the limit of resolution is small, more detail can be seen in the image. In this case we take the smallest wavelength of light observable, and divide it by the angle.

$$D=\frac{\quantity{1.0\times 10^{-6}}{m}}{\quantity{3.3\times 10^{–7}}{radian}}=\quantity{3.0}{m}$$

If we performed the same calculation with the other wavelength of light we would get a telescope size of:

$$D=\frac{\quantity{5.0\times 10^{-6}}{m}}{\quantity{3.3\times 10^{–7}}{radian}}=\quantity{15.2}{m}$$

Although this value is a reasonable diameter for a telescope, if we use this diameter to find the smallest resolvable angle for the $\quantity{1.0}{μm}$ wavelength we get:

$$θ=\frac{\quantity{1.0\times 10^{-6}}{m}}{\quantity{15.2}{m}}=\quantity{6.6\times 10^{-8}}{rad}$$

As $\quantity{6.6\times 10^{-8}}{rad}<\quantity{3.3\times 10^{–7}}{rad}$ this cannot be the size of the telescope.

Also note that the answer has been given to two significant figures, as has the data in the question.




4. Discuss the level of detail the IRTF would be able to detect on the surface of Vesta, when Vesta is $\quantity{1.73\times 10^{11}}{m}$ from Earth.

In this question we are being asked to compare the minimum angular resolution of the IRTF to the size of the asteroid Vesta. We have been told that the telescope can resolve angles as small as $\quantity{3.3\times 10^{–7}}{rad}$. In this example Vesta is $\quantity{1.73\times 10^{11}}{m}$ from Earth, and at this distance the IRTF would be able to resolve objects as small as:

$$\quantity{3.3\times 10^{–7}}{rad}\times\quantity{1.73\times 10^{11}}{m}=\quantity{57 090}{m}$$

This means that details which are around $\quantity{50-60}{km}$ can be resolved on the surface of the asteroid.

You could also compare the minimum resolvable angle of the telescope with the angular size of the asteroid at this distance:

$$θ=\frac{\quantity{5.4\times 10^{5}}{m}}{\quantity{1.73\times 10^{11}}{m}}=\quantity{3.12\times 10^{-6}}{rad}$$

This is about ×10 larger than the smallest resolvable angle of the telescope, so features around a 10th of the size of Vesta could be resolved.

To gain full credit for this question you would have to state that the the smallest angular resolution is smaller than the angular size of the asteroid AND support it with a quantitative statement about how much detail can be seen.

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