The Magnifying Glass and the Compound Microscope – Optical Instruments (Senior 4 Physics) | Ganzaa

Optical Instruments

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Senior 4 · Physics

Optical Instruments

4 lessons

Lesson 2 of 4

The Magnifying Glass and the Compound Microscope

25 min

How big does it look?

A grain of pollen is about 30 micrometres across. Bring it closer to your eye and it looks bigger, but only up to a point: closer than about 25 cm, a normal young eye can no longer focus, and the grain blurs long before you can see any detail. To see cells, bacteria in a water sample or the scales on a butterfly's wing, you need an instrument that makes the object look bigger while giving your eye an image it can focus on. That is the job of the magnifying glass and the compound microscope.

Angular size and angular magnification

How large an object looks depends on the size of its image on the retina, and that depends on the angle the object subtends at the eye. An object of height $h$ at distance $d$ subtends $\alpha \approx \frac{h}{d}$ (small angles, in radians). The closest distance at which an eye can focus comfortably is its near point; for a normal eye we take $D = 25\,\mathrm{cm}$ . So the largest angle you can get without an instrument is $\alpha = \frac{h}{D}$ .

Figure 2.1: A magnifying glass increases the angle the object subtends at the eye, from α to β.

Definition

The angular magnification (or magnifying power) $M$ of an optical instrument is $M = \frac{\beta}{\alpha}$ , where $\beta$ is the angle subtended at the eye by the final image and $\alpha$ is the angle subtended by the object seen without the instrument (at the near point for a microscope or magnifier; at its real distance for a telescope). It has no unit. Example: a lens that makes a stamp look as if it subtended 5 times the angle it subtends at 25 cm has $M = 5$ .

The magnifying glass (simple microscope)

Hold a converging lens close to your eye and place the object just inside its focal length (Figure 2.1b). The lens forms a virtual, upright, magnified image. Two settings are used.

Image at the near point. Put the image at $v = - D$ . The eye is at the lens, so the image subtends $\beta = \frac{h_i}{D}$ , and since $\frac{h_i}{h} = \frac{D}{u}$ , $\beta = \frac{h}{u}$ . Then $M = \frac{\beta}{\alpha} = \frac{h/u}{h/D} = \frac{D}{u}$ . From the lens formula, $\frac{1}{u} = \frac{1}{f} - \frac{1}{-D} = \frac{1}{f} + \frac{1}{D}$ , so

M = \frac{D}{u} = 1 + \frac{D}{f} \qquad\text{(image at the near point)}

Image at infinity. Put the object at F. The rays leave parallel, the relaxed eye sees the image at infinity, and $\beta = \frac{h}{f}$ , so

M = \frac{D}{f} \qquad\text{(image at infinity, relaxed eye)}

The shorter the focal length, the greater the magnification; but a very strong single lens has severe aberrations, which in practice limits a simple magnifier to about $10\times$ .

Worked example 1: a reading lens

A magnifying glass has a focal length of 5.0 cm. Find its angular magnification when the image is (a) at the near point, (b) at infinity, and (c) where the object must be placed in case (a).

\text{(a)}\;\; M = 1 + \frac{25}{5.0} = 6.0 \qquad \text{(b)}\;\; M = \frac{25}{5.0} = 5.0

\text{(c)}\;\; \frac{1}{u} = \frac{1}{5.0} + \frac{1}{25} = \frac{6}{25} \;\Rightarrow\; u = 4.2\,\mathrm{cm}

The extra "1" in case (a) costs the eye some effort: it must focus at its near point, which is tiring for long use.

The compound microscope

A compound microscope uses two converging lenses (Figure 2.2). The objective, of very short focal length $f_{o}$ , is placed so that the object lies just outside its principal focus. It forms a real, inverted, magnified image $I_{1}$ inside the tube. The eyepiece, of focal length $f_{e}$ , is a magnifying glass used to look at $I_{1}$ : it forms the final virtual image $I_{2}$ .

Figure 2.2: The compound microscope. The objective magnifies; the eyepiece magnifies the objective's image again.

The total angular magnification is the product of the two stages:

M = m_o \times M_e = \frac{v_o}{u_o} \times M_e, \qquad M_e = 1 + \frac{D}{f_e}\ \ (\text{final image at } D) \;\text{ or }\; \frac{D}{f_e}\ \ (\text{final image at infinity})

When the final image is at infinity, $I_{1}$ lies at the focal point of the eyepiece. If the distance between the inner focal points of the two lenses is $L$ (the tube length, typically 16 cm), then $v_o = f_o + L \approx L$ and $u_o \approx f_o$ , giving the useful approximation

M \approx \frac{L}{f_o} \times \frac{D}{f_e}

So both lenses should have short focal lengths, and the objective's should be the shorter. A school microscope with a $10\times$ objective and a $10\times$ eyepiece magnifies about $100\times$ . Light microscopes cannot usefully go much beyond about $1000\times$ to $2000\times$ : diffraction blurs detail smaller than about half the wavelength of light (roughly 0.2 µm), so more magnification only enlarges the blur. Viruses need an electron microscope.

Worked example 2: a model microscope

A microscope has an objective of focal length 1.0 cm and an eyepiece of focal length 5.0 cm. An object is 1.1 cm from the objective, and the final image is at the near point (25 cm). Find the magnifying power and the separation of the lenses.

Objective: $\frac{1}{v_o} = \frac{1}{1.0} - \frac{1}{1.1} = \frac{0.1}{1.1}$ , so $v_o = 11.0\,\mathrm{cm}$ and $m_o = \frac{11.0}{1.1} = 10.0$ .

Eyepiece (final image virtual, $v_e = -25\,\mathrm{cm}$ ): $\frac{1}{u_e} = \frac{1}{5.0} + \frac{1}{25} = \frac{6}{25}$ , so $u_e = 4.17\,\mathrm{cm}$ and $M_e = 1 + \frac{25}{5.0} = 6.0$ .

M = m_o M_e = 10.0 \times 6.0 = 60, \qquad \text{separation} = v_o + u_e = 11.0 + 4.17 = 15.2\,\mathrm{cm}

For a relaxed eye (final image at infinity) the eyepiece is moved out to $u_e = 5.0\,\mathrm{cm}$ : $M = 10.0 \times 5.0 = 50$ and the separation becomes 16.0 cm.

Worked example 3: a laboratory microscope

A microscope has $f_o = 0.50\,\mathrm{cm}$ , $f_e = 2.5\,\mathrm{cm}$ and a tube length $L = 16\,\mathrm{cm}$ . Estimate its magnifying power for a relaxed eye.

M \approx \frac{L}{f_o}\times\frac{D}{f_e} = \frac{16}{0.50} \times \frac{25}{2.5} = 32 \times 10 = 320

An exact calculation gives the same: $I_{1}$ is at $v_o = 0.50 + 16 = 16.5\,\mathrm{cm}$ , so $u_o = 0.516\,\mathrm{cm}$ , $m_{o} = 32$ , and $M = 32 \times 10 = 320$ .

Experiment 2.1: building a microscope on a bench

Apparatus: two converging lenses (about 5 cm and 10 cm focal length) in holders, a metre rule, a small lit object (printed millimetre scale on tracing paper over a lamp), a white screen.

Use the 5 cm lens as the objective. Place the object about 6 to 7 cm from it and find the real, magnified image on the screen. Measure $u_{o}$ and $v_{o}$ and calculate $m_{o}$ .
Remove the screen. Place the 10 cm lens beyond the image position as an eyepiece and look through it, moving it until you see a sharp, magnified, inverted image of the scale.
Compare your measured lens separation with the value predicted by the lens formula.

Warning

Safety. When you use a microscope with a mirror, never reflect direct sunlight into it: the objective concentrates it into your eye. Carry a microscope with one hand under the base and one on the arm.

Warning

Common mistakes. (1) Multiplying focal lengths instead of magnifications. (2) Using $D = 25\,\mathrm{cm}$ in a formula but other distances in metres. (3) Saying that the objective forms a virtual image: it forms a real one, which the eyepiece then views. (4) Believing a microscope can show any detail if the magnification is high enough: resolution is limited by the wavelength of light.

Tip

Microscope = objective (real, magnified image) + magnifying glass. Work out the objective with the lens formula first, then treat its image as the object for the eyepiece.

The compound microscope: how to draw ray diagrams (Ben Ryder). Follow how the objective forms a real magnified image and how the eyepiece, used as a magnifying glass, forms the final virtual image.

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Practice

Define angular magnification and explain why it, not linear magnification, describes a magnifying glass.
Find the angular magnification of a magnifying glass of focal length 8.0 cm when the image is (a) at infinity, (b) at the near point.
A magnifying glass gives $M = 4.0$ with the image at infinity. Find its focal length and power.
A microscope has an objective of focal length 2.0 cm and an eyepiece of focal length 6.0 cm. An object is 2.5 cm from the objective, and the final image is at infinity. Find the magnifying power and the distance between the lenses.
A microscope has $f_o = 4.0\,\mathrm{mm}$ , $f_e = 25\,\mathrm{mm}$ and $L = 160\,\mathrm{mm}$ . Estimate its magnifying power for a relaxed eye.
Explain why a microscope's objective should have a shorter focal length than its eyepiece.
A company advertises a light microscope that magnifies 10 000 times. Evaluate this claim.

Show answersHide answers

$M = \frac{\beta}{\alpha}$ , the ratio of the angle subtended by the image to that subtended by the object at the near point. With a magnifier the image can be at any distance (even infinity), where its height means little; what the eye senses is the angle, which sets the size of the retinal image.
(a) $M = \frac{25}{8.0} = 3.1$ ; (b) $M = 1 + \frac{25}{8.0} = 4.1$ .
$f = \frac{25}{4.0} = 6.25\,\mathrm{cm} \approx 6.3\,\mathrm{cm}$ ; $P = \frac{1}{0.0625} = +16\,\mathrm{D}$ .
$v_o = \frac{1}{\frac{1}{2.0} - \frac{1}{2.5}} = 10.0\,\mathrm{cm}$ , $m_{o} = 4.0$ ; $M_e = \frac{25}{6.0} = 4.2$ ; $M = 17$ ; separation $= 10.0 + 6.0 = 16.0\,\mathrm{cm}$ .
$M \approx \frac{160}{4.0} \times \frac{250}{25} = 40 \times 10 = 400$ .
The objective's magnification $\frac{v_o}{u_o}$ is roughly $\frac{L}{f_o}$ ; a tube of fixed length gives a large $m_{o}$ only if $f_{o}$ is small. The eyepiece also benefits from a short $f_{e}$ , but it must be long enough to be comfortable to look through and to view the whole of $I_{1}$ .
Doubtful. Light microscopes cannot resolve detail smaller than about 0.2 µm because of diffraction; beyond about 1000× to 2000× the image grows but shows no new detail ("empty magnification"). The claim is technically possible but useless, or it describes an electron microscope.

Quick check

Key points

Angular magnification $M = \frac{\beta}{\alpha}$ ; without an instrument, $\alpha = \frac{h}{D}$ with $D = 25\,\mathrm{cm}$ .
Magnifying glass: $M = 1 + \frac{D}{f}$ (image at the near point) or $\frac{D}{f}$ (image at infinity).
Compound microscope: $M = m_o M_e \approx \frac{L}{f_o}\cdot\frac{D}{f_e}$ ; the objective forms a real image that the eyepiece magnifies.
Diffraction limits useful light-microscope magnification to about 1000× to 2000×.

Sources and credits

26.4: Microscopes, from College Physics (1st edition) by Paul Peter Urone and Roger Hinrichs, OpenStax, hosted by LibreTexts, licensed under CC BY 4.0. Adapted by Ganzaa.
Photo: Light Optical Microscope.jpg by Jeremyida002, Wikimedia Commons, licensed under CC BY-SA 4.0.
Video: The compound microscope - how to draw ray diagrams, Ben Ryder, embedded from YouTube.

The derivations, the two diagrams, the experiment, the worked examples and the practice exercises are written by Ganzaa.

1How big does it look?

2Angular size and angular magnification

3The magnifying glass (simple microscope)

4The compound microscope

5Experiment 2.1: building a microscope on a bench

How big does it look?

Angular size and angular magnification

The magnifying glass (simple microscope)

The compound microscope

Experiment 2.1: building a microscope on a bench