Telescopes and resolving power

Four moons of Jupiter

In the winter of 1609 to 1610, Galileo pointed a small telescope at Jupiter. He saw four tiny points of light near the planet, and from night to night they moved. They were moons going round it. A telescope does not make a star bigger. It makes the angle between two objects bigger, and it gathers more light than the eye can. Let us see how.

The astronomical refracting telescope

An astronomical telescope has two convex lenses.

• The objective has a long focal length $f_o$ and a large diameter. It forms a real, inverted image of a far object at its focal plane.
• The eyepiece has a short focal length $f_e$. It works as a magnifying glass on that first image.

A distant object sends parallel rays to the objective. Rays from one point of the object, such as the edge of the Moon, all come in at the same slope, an angle $\alpha$ to the axis. The objective brings them to a point in its focal plane. This is the first image, $I_1$. Its height is $h' = f_o\,\alpha$.

Normal adjustment. We move the eyepiece until $I_1$ is at its focus. Then the rays leave the eyepiece parallel, and the final image is at infinity. A relaxed eye can look at it for a long time. The rays leave at an angle $\beta = \dfrac{h'}{f_e}$. The magnifying power is the ratio of the angle that the image makes to the angle that the object makes:

The minus sign says that the final image is upside down. The distance between the lenses is

Look at the picture. The rays come in with a slope of $0.1$ and leave with a slope of $0.4$ in the other direction. The ratio is $4$, which is $\dfrac{f_o}{f_e} = \dfrac{10}{2.5}$.

The image at the near point. Sometimes the final image is brought to $D = 25\ \text{cm}$. Then $I_1$ is inside the focus of the eyepiece, at $|u_e| = \dfrac{Df_e}{D + f_e}$. The angle is $\beta = \dfrac{h'}{|u_e|}$, so $m = \dfrac{f_o}{|u_e|}$ and, with $\dfrac{1}{|u_e|} = \dfrac{1}{D} + \dfrac{1}{f_e}$,

Example 1. A telescope has $f_o = 100\ \text{cm}$ and $f_e = 2.5\ \text{cm}$. In normal adjustment, $|m| = \dfrac{100\ \text{cm}}{2.5\ \text{cm}} = 40$ and the length is $100\ \text{cm} + 2.5\ \text{cm} = 102.5\ \text{cm}$.

Example 2. We want a magnifying power of $60$ in a tube that is $122\ \text{cm}$ long. Then $f_o + f_e = 122\ \text{cm}$ and $f_o = 60f_e$. So $61f_e = 122\ \text{cm}$. We want $f_e$, so $f_e = 2.0\ \text{cm}$ and $f_o = 120\ \text{cm}$.

Start with the defaults. Read the magnifying power and the length. Now change the eyepiece to $5.0\ \text{cm}$. How does the power change? Then compare the two lengths, for the normal adjustment and for the near point.

Practise with a telescope

Why the objective must be large

A telescope with a larger objective does two things at once.

It gathers more light. The light that enters is proportional to the area of the objective, which is proportional to $D^2$. A mirror of twice the diameter gathers four times as much light. So it shows fainter stars, or the same star in a shorter time.

It resolves finer detail. Light is a wave and spreads when it passes through the circular opening of the objective. Two stars that are too close give two overlapping spots. The smallest angle between two points that the telescope can still separate is

This is the Rayleigh criterion. The wavelength $\lambda$ is in the same unit as $D$, and the angle is in radian. A larger $D$ gives a smaller angle. A shorter wavelength also does.

Example. A space telescope has a mirror of diameter $2.4\ \text{m}$. For green light of $550\ \text{nm}$, $\theta_{\min} = \dfrac{1.22 \times 550 \times 10^{-9}\ \text{m}}{2.4\ \text{m}} = 2.8 \times 10^{-7}\ \text{rad}$. One arcsecond is $4.85 \times 10^{-6}\ \text{rad}$, so this is $0.058\ \text{arcsecond}$.

On the ground, the moving air blurs the picture of a star to about $1$ arcsecond. That is why a telescope above the atmosphere can do so much better. The magnifying power does not change the resolution. Once the two spots overlap, no eyepiece can pull them apart.

Start with the defaults. Now double the aperture. By how much does the smallest angle fall, and by how much does the light gathered rise? Then change the wavelength to $400\ \text{nm}$.

Practise with resolution and light

Reflecting telescopes

A big lens has many problems. It can be held only by its edge, so it sags under its own weight. Glass absorbs some light. A lens bends light of different colours by different amounts, so the image has coloured edges, which is called chromatic aberration. A mirror has none of these problems. A concave mirror reflects every colour in the same way. A mirror can be supported from the whole of its back, so it can be very large. A parabolic mirror brings all parallel rays to one focus, which removes the blur called spherical aberration. That is why the largest telescopes are reflecting telescopes.

The mirror that collects the light is the primary mirror. It replaces the objective, and the magnifying power is $\dfrac{f_o}{f_e}$ with $f_o$ the focal length of the mirror, or the effective focal length of the whole system.

The Newtonian telescope. A small flat mirror, set at $45^\circ$ near the focus, turns the converging beam to the side of the tube. The eyepiece is on the side.

The Cassegrain telescope. A small convex mirror sends the light back through a hole in the primary mirror. The eyepiece is behind the large mirror.

In the picture the primary has $f = 2.0\ \text{m}$. The small mirror forms a final image $4$ times as far from itself as the point where the beam would have focused. So the system acts as one mirror of effective focal length $4 \times 2.0\ \text{m} = 8.0\ \text{m}$, in a tube not much longer than $2\ \text{m}$. An eyepiece of $4.0\ \text{cm}$ then gives $\dfrac{800\ \text{cm}}{4.0\ \text{cm}} = 200$.

Many large telescopes use the Cassegrain idea. The Hubble Space Telescope has a primary mirror of $2.4\ \text{m}$. In India, the Devasthal Optical Telescope near Nainital has a mirror of $3.6\ \text{m}$. The small secondary mirror blocks a little light, but this is a small price.

Two more telescopes

• The terrestrial telescope, for looking at distant things on Earth, must show an upright image. It adds a third lens, of focal length $f$, between the two. Its length is $f_o + 4f + f_e$. Prism binoculars do the same job with prisms.
• The Galilean telescope has a concave lens as the eyepiece. The image is upright. It is short, $f_o - |f_e|$ long, and has a narrow field of view. Opera glasses are Galilean telescopes.

Try it in Python

The program lists five instruments, from the eye to a space telescope. For each it finds the smallest angle that it can separate and the light that it gathers.

Common mistakes

• Putting the eyepiece focal length in the numerator. The objective has the longer focal length, and $|m| = \dfrac{f_o}{f_e}$.
• Thinking that the telescope length is $f_o - f_e$. That belongs to the Galilean telescope. For the astronomical one the length is $f_o + f_e$.
• Believing that more magnification means finer detail. The aperture and the wavelength set the limit, $\theta = \dfrac{1.22\,\lambda}{D}$.
• Saying that the light gathered depends on $D$ and not on $D^2$. It depends on the area.
• Forgetting that the image is upside down. An astronomical telescope inverts the image. That is no problem for stars.
• Using the formula for the normal adjustment when the question says the near point. Read where the final image is.

Remember these

• Normal adjustment: $|m| = \dfrac{f_o}{f_e}$ and length $f_o + f_e$. For the final image at the near point, $|m| = \dfrac{f_o}{f_e}\left(1 + \dfrac{f_e}{D}\right)$.
• A large aperture gathers more light, in proportion to $D^2$, and resolves better, $\theta_{\min} = \dfrac{1.22\,\lambda}{D}$.
• Large telescopes are reflectors: no chromatic aberration, the mirror is supported from behind, and a parabolic mirror removes spherical aberration.
• Newtonian: a flat mirror at $45^\circ$ and a side eyepiece. Cassegrain: a convex secondary mirror and a hole in the primary.
• Terrestrial telescope: erect image, length $f_o + 4f + f_e$. Galilean telescope: concave eyepiece, erect image, length $f_o - |f_e|$.

Check yourself

Try each one on your own first. If you are stuck, open a hint. Then check with Show the answer.

1. A telescope in normal adjustment has an objective of focal length $72\ \text{cm}$ and an eyepiece of $2.4\ \text{cm}$. What is the magnifying power and the length of the tube?

2. Give two reasons why a telescope for looking at faint stars should have a large objective.

3. Give two reasons why the large research telescopes are reflectors and not refractors.

4. What is the smallest angle that a telescope with a mirror of diameter $1.0\ \text{m}$ can separate, for light of $600\ \text{nm}$? Give it in arcsecond.

5. Telescope A has an objective of diameter $10\ \text{cm}$ and telescope B has one of $30\ \text{cm}$. How do the light gathered and the smallest angle compare?

References

• NCERT Physics Part II (Class 12), Chapter 9, the section on the telescope.
• OpenStax University Physics Volume 3, Chapter 2, the section on the telescope, and Chapter 4 on diffraction and resolution.
• OpenStax College Physics 2e, Chapter 26, the sections on telescopes.