1. Magnifying Glass
Angular magnification (m): cm/f < m < 25cm/f + 1
relaxed eye,
image at (normal) far point
image at 25 cm
(= normal near point)
For more magnification, first use a lens to form an enlarged real image, and then use a magnifying glass to make an enlarged virtual image of that:
Compound Microscope

2. For more magnification, first use a lens (the objective lens) to form an enlarged real image, and then use a magnifying glass (the eyepiece) to make an enlarged virtual image of that: Compound Microscope

3. Compound Microscope Objective Lens: MO = -q1/p1 1/p1 = 1/fO – 1/q1
p1 = q1fO/(q1 - fO)
MO = - (q1-fO) / fO = -(q1/f0 -1)
For large MO, need large q1: put object slightly outside the focal point fO.
[Note the microscope length L > q1. Therefore, for a given objective lens (f0), one must increase the length of the microscope to increase M0.]

4. p2 = L-q1
Eyepiece Lens:
The angular magnification
me = 25 cm/fe, for image at infinity.
[To obtain this, need p2  fe.]
Total magnification:
M  Mome = - [(q1-fO) / fO] (25 cm/fe)
M = - [(L – p2 – fO)/fO] (25 cm / fe)
M  - [(L – fe – fO)/fO] (25 cm / fe)
Finally, if p1 is very close to fO,
L > q1 >> fO, fe
M  - (L/fO) (25 cm / fe)
It is inverting because it is a virtual image of a real image.

5. If f0 = 2mm, fe = 1 cm, p1 = 2.05 mm, q1 = p1f0/(p1-f0) = 82 mm.
M  - (L/fO) (25 cm / fe)
For maximum |M|, want to minimize focal lengths. This means want converging lenses with small R’s – i.e. fat lenses. Practical minimum values for these (taking into account aberrations and diffraction, which determine resolution) are fO  2 mm and fe  1 cm.
If f0 = 2mm, fe = 1 cm, p1 = 2.05 mm, q1 = p1f0/(p1-f0) = 82 mm.
Taking L  q1, this gives M  1025.
[Note, that our approximation L >> fe is not great in this case!]

6. If f0 = 2mm, fe = 1 cm, p1 = 2.05 mm, q1 = p1f0/(p1-f0) = 82 mm.
M  - (L/fO) (25 cm / fe)
For maximum |M|, want to minimize focal lengths. This means want converging lenses with small R’s – i.e. fat lenses. Practical minimum values for these are fO  2 mm and fe  1 cm.
If f0 = 2mm, fe = 1 cm, p1 = 2.05 mm, q1 = p1f0/(p1-f0) = 82 mm.
Taking L  q1, this gives M  1025.
Suppose using this same microscope (i.e.  same L), you switched the objective lens to one with fO = 5 mm. The magnification is now |M| = 410.
Since q1  L = 82 mm, this means you use a larger (easier) object distance: p1 = q1f0/(q1-f0) = 5.3 mm.
[Using a longer focal length objective decreases magnification but gives you more”working room” (and depth of focus).]

7. Problem: Suppose your microscope has a magnification M = -100 and eyepiece with focal length fe = 2 cm. How long must the microscope be if the objective lens is in focus when p = 5 mm and what is the focal length of the objective lens?

8. L/fO = -M fe/25 cm = (100) (2 cm)/25 cm = 8
Problem: Suppose your microscope has a magnification M = -100, eyepiece with focal length fe = 2 cm. How long must the microscope be if the objective lens is in focus when p = 1.5 cm?
M  - (L/fO) (25 cm / fe)
L/fO = -M fe/25 cm = (100) (2 cm)/25 cm = 8
1/p1 + 1/q1 = 1/fO, but q1  L  1/p1 + 1/L  1/f0
L/p1 + 1  L/fO = 8
 L  7p1 = 10.5 cm
fO = L/8 = 1.3 cm
[Note, that our approximations L >> fe and L >> fO are both not great in this case!]

9. i.e > 400 nm for visible light.
Besides magnification, another consideration is resolution, x, i.e. “how small an object (or separation) can be observed. In most cases, the resolution is limited by aberrations. More fundamentally, the resolution is limited by diffraction (Physics 232), and, if aberrations are negligible, the smallest distance that can be resolved is on the order of the wavelength,
i.e > 400 nm for visible light.
Much better resolution is possible with an electron-microscope, since electrons have much smaller wavelengths, inversely proportional to V, their accelerating voltage. The best electron microscopes, with accelerating voltages  100 kV, have atomic resolution (x  0.2 nm), limited by spherical aberrations of the electron optics as well as diffraction.
UK electron microscope with resolution x  2 nm

10. Telescope
Unlike the microscope, which forms a large image of very small objects, the telescope forms images of very large objects (e.g. planets) that are very far away: pobject  . Therefore the rays that come from each point in the object are  parallel. The telescope is used to increase the angular spread of rays coming from each point on the object, to make it “look larger”.
Want to create a virtual image in which image > object.

11. Let object = O
and final image = .
For an object at p = , the objective lens forms a real, inverted image at its focal point, fO. The size of its image = h’, where
h’ = -fO tan O  -fO O.
i.e. O  -h’/fO
If the image is at the focal point of the eyepiece, fe, it will form a virtual image at , making an angle
 tan  = h’ / fe.

12. Therefore in the paraxial limit
(tan   ),
O  -h’/fO
 h’/ fe
angular magnification:
m = /O = - fO/ fe
Also, the length of the telescope (distance between lenses) L = fO + fe
[The final image is inverted: it is a virtual (non-inverted) image of a real (inverted) image.]

13. Telescope m = - fO / fe , L = fO + fe
The amount of light/per second that enters the telescope is proportional to the area of the objective lens, so to image faint objects, a large diameter lens is needed. However, it is difficult to support a large lens by its edges (don’t want to block light by building in supports), because gravity will cause it to deform.

14. The largest refracting (i. e
The largest refracting (i.e. lens) telescope is at the Yerkes Observatory in Wisconsin. It’s objective lens has a diameter of 100 cm and a focal length fO = 19.4 m with a maximum magnification |m| = 194; i.e. it’s “eyepiece” has an “effective” focal length of fe = 10 cm. [“Effective”, because modern research telescopes do not have simple eyepieces for viewing.]
Because of atmospheric disturbance, it is actually not useful to have |m| larger than  200 for earth based telescopes. The collecting power (i.e. area of the objective) is a much more important criterion.

15. Galileo’s telescope (1609), with which he first observed the moons of Jupiter and the phases of Venus, was 92 cm long. Its objective lens had a diameter of 3.7 cm and focal length = 98 cm. (Its eyepiece had a diverging lens with a focal length of 2.2 cm, so its total length was less than fO and it was non-inverting.) It had a magnification |m|  40.
Inexpensive home telescope with diameter = 7 cm and fO = 70 cm.
It has two eyepieces:
fe = 20 mm  |m| = 35.
fe = 4 mm  |m| = 175 ( Yerkes!)

16. To overcome the difficulties in constructing a large objective lens, large telescopes use a reflecting, concave mirror. (Since light does not pass through the mirror, it can be supported from behind). Use of a mirror also avoids chromatic aberration. Also, to remove spherical aberration, a parabolic mirror is used: for a parabolic mirror, all rays from  are brought to focus at the focal point, not just paraxial rays.
A disadvantage is that it is necessary to block some of the light in the center, but this fraction of the light can be kept very small.

17. The largest reflecting telescope (for visible light) is Gran Telescopio Canarias in the Canary Islands (Spain), with a diameter of 10.4 m,
fO = 16.5 m. Its maximum magnification |m| = 140, corresponding to an “effective” fe = 11.8 cm.

18. The objective mirror of the main telescope in the MacAdam Student Observatory has a diameter of 50 cm and a focal length 3.45 m. The maximum magnification |m|  200, corresponding to an “effective” fe  1.7 cm.

19. Because of atmospheric disturbance, it is actually not useful to have
|m| larger than  200 for earth based telescopes: so put the telescope in space!
The Hubble Space Telescope has an objective mirror of diameter 2.4 m and focal length fO = 58 m and maximum |m| = 4700, corresponding to an effective fe = 0.5 mm!

20. Most radio telescopes have parabolic mirrors (good metal surfaces are perfect reflectors of radio waves) that can be directed at the point of the sky of interest. A parabolic mirror does not have spherical aberration; all parallel rays (i.e. rays from “” are reflected to focus at the parabola’s focus and a radio detector is placed there. Imaging is done by changing the direction of the parabola’s axis. The sensitivity (radio power detected) is proportional to the area of the dish.
Radio telescopes have been built for wavelengths between  1 mm (300 GHz) and 30 m (10 MHz).
In 1964, Penzias and Wilson were testing a new radio telescope at Bell Labs. They observed a lot of noise, with a peak frequency  160 GHz (  2 mm), no matter how they pointed the telescope. They at first thought it was due to a malfunction of their detector (or bird droppings on the telescope).
-- In fact, they had discovered the Cosmic Microwave Background, residual radiation from the big bang.

21. Prior to 2016, the largest radio telescope in the world was in Aracebo, Puerto Rico, with a diameter of 305 m. It is built into the ground and cannot be moved (or pointed). Therefore, instead of parabolic, it is spherical. Radio waves from different directions in the sky are measured by moving the detector along cables strung above the dish to the point where they are focused. (Because the diameter is so large, spherical aberration is small.)
In 2016, a larger radio telescope opened in China: “FAST”:
Five hundred meter Aperture Spherical Telescope.

22. Note the size of the dish not only affects the collecting power (sensitivity) but also the spatial resolution: because of diffraction of the radio waves, the resolution improves as the size of the detector increases. (Radio telescopes operate at wavelengths between 1 mm and 30 m and, because of diffraction, the angular resolution   /size.)
One can greatly improve spatial resolution (although not increase the collecting power) by using multiple (small) radio telescopes separated by precise distances when their detectors are precisely “sync’d”. Now the resolution   /separation.
National Radio Astronomy Observatory Very Large Array (New Mexico)