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Lenses in series, angular magnification, magnifiers, compound microscopes and telescopes, and why aperture matters.
Paper packet. Every task here also exists on screen, where it is checked automatically; answers written on paper are not assessed by Nydus. When you are back at a device, enter your answers there.
By the end of this lesson you will be able to follow images through lenses in series and find the magnification of optical instruments.
From the last lesson you know the thin-lens equation, magnification and lens power, and how a single lens makes real and virtual images. This lesson puts lenses in series, each using the previous image as its object, to build the instruments that extend human sight.
| Term | What it means |
|---|---|
| Near point | The closest distance the eye can focus, taken as $25$ cm. |
| Angular magnification | How many times larger an object's angle appears with the instrument than without. |
| Objective | The lens nearest the object, forming the first real image. |
| Eyepiece | The lens nearest the eye, used as a magnifier on the first image. |
| Telescope magnification | $M = f_o/f_e$. |
| Microscope magnification | $M = m_oM_e \approx -(L/f_o)(25/f_e)$. |
To analyze several lenses, take them one at a time. The first lens forms an image; that image becomes the object for the next lens, at whatever distance it lies from it; and so on. The overall magnification is the product of the individual ones.
For viewing instruments, what matters is how large an angle the final image spans at the eye. A magnifier lets you hold an object closer than your near point: angular magnification $25/f$. A microscope multiplies an objective's magnification by an eyepiece's. A telescope images a distant object at the objective's focus and views it with an eyepiece:
$$M = \frac{f_o}{f_e}.$$
Another way: picture
Picture a relay race. The first runner carries the baton to a handoff point, the second picks it up there and runs on. In an instrument, the first lens carries the light to its image, and the second lens picks up the light from that image as if it were a new object. Only the last runner's finish, the final image, is what the eye sees.
Another way: steps
How big something looks depends on the angle it spans at your eye, not on its actual size. A coin at arm's length can hide the Moon. To see detail, you want a large angle, which means bringing the object close, but the eye cannot focus closer than its near point, about $25$ cm for a young adult.
Instruments get around this. A magnifier lets you put the object much closer; a microscope enlarges a tiny object's image before you look; a telescope makes a distant object's image span a larger angle. In each case the measure of success is angular magnification.
Place an object at the focal point of a converging lens and its rays leave parallel, as if from a distant, enlarged object; the relaxed eye sees it comfortably. The angle it spans is the object's height over $f$ instead of over $25$ cm, so the angular magnification is $25/f$.
Moving the object slightly inside the focal point puts the virtual image at the near point, giving a little more, $1 + 25/f$, at the cost of eye strain. A $5$ cm lens gives about five times; jewelers' loupes and watchmakers' lenses go to ten or twenty.
A microscope's objective has a very short focal length and sits just outside its focal point from the specimen, forming a real, inverted, greatly enlarged image inside the tube, about $16$ cm away. Its magnification is roughly $-L/f_o$.
The eyepiece then acts as a magnifier on that image, adding $25/f_e$. The total is the product: a $40\times$ objective and a $10\times$ eyepiece give $400\times$. Oil-immersion objectives reach $100\times$, for a total of $1000\times$, enough to see bacteria clearly, though finer detail is blurred by diffraction, as the next unit explains.
A telescope's objective, a lens or mirror with a long focal length, forms a real image of a distant object at its focal point. The eyepiece, with a short focal length, is placed so that image sits at its own focal point, and the light leaves parallel for a relaxed eye.
The angular magnification is $f_o/f_e$. With a $1000$ mm objective, a $25$ mm eyepiece gives $40\times$ and a $10$ mm eyepiece $100\times$. Amateur astronomers swap eyepieces to change power. The telescope's length is roughly $f_o + f_e$, which is why long focal lengths once meant very long tubes.
Checking an answer. Telescopes need a long objective and a short eyepiece. Microscope magnifications must multiply, not add. An inverted final image is normal for astronomical telescopes and microscopes.
Each lens bends light according to its own focal length regardless of where the light came from, so the chain can be analyzed one lens at a time. If an image forms beyond the next lens, it becomes a virtual object with a negative object distance, and the same equations still work.
Angular magnification formulas assume small angles, where the angle and its tangent are nearly equal, and the conventional near point of $25$ cm. Individual eyes differ, which is why instruments have focusing knobs.
Early telescopes used lenses. Galileo's, in 1609, had a diverging eyepiece and showed an upright image; he saw Jupiter's moons and the phases of Venus. The largest refractor ever used for research, the $40$-inch lens at Yerkes Observatory in Williams Bay, Wisconsin, was completed in 1897.
Large lenses sag under their own weight and bend colors differently, so every major telescope since has used mirrors. Refracting telescopes remain popular with amateurs for their sharp, high-contrast views of planets, and binoculars are pairs of small refractors with prisms that turn the image upright.
A pair of binoculars labeled $10 \times 50$ has a magnification of $10$ and objectives $50$ mm across. Inside, prisms reflect the light several times by total internal reflection, folding a long light path into a short body and turning the inverted image right side up.
The objective's diameter sets how much light is gathered: a $50$ mm objective collects about fifty times more light than a dark-adapted eye's $7$ mm pupil. Birdwatchers often choose $8\times$ or $10\times$; higher powers magnify hand shake too, which is why astronomers use tripods or image-stabilized binoculars.
Magnification is not a telescope's most important property. Its light-gathering power depends on the area of its objective, which grows as the square of the diameter. The $10$-meter Keck mirrors collect about two million times as much light as the human eye, revealing galaxies far too faint to see.
Larger apertures also resolve finer detail, since diffraction blurs less. That is why astronomers talk about aperture first: a large telescope at modest magnification shows far more than a small one pushed to high power, which only enlarges a blurry, dim image.
Every hospital pathology lab relies on compound microscopes to examine tissue samples, blood smears and bacteria. A pathologist switching between $4\times$, $10\times$, $40\times$ and $100\times$ objectives on a rotating turret moves from an overview of a tissue slice to individual cells.
Surgical microscopes, with long working distances, let surgeons repair blood vessels and nerves thinner than a hair. Digital pathology now scans whole slides into high-resolution images that doctors can read from anywhere, but the optics inside the scanners are still an objective forming a real, magnified image.
Your eye is itself an optical instrument, a camera with a variable lens. When its focusing is off, an extra lens in front corrects it, and the system analysis of this lesson applies: the eyeglass lens forms an image that the eye's lens then images onto the retina.
Astigmatism, where the cornea is curved more in one direction than another, needs lenses with different powers in different directions, cylindrical lenses. An eyeglass prescription lists sphere, cylinder and axis, three numbers describing a single corrective lens that works together with the eye's own optics.
A camera is the simplest instrument of all: one lens, or a group acting as one, forms a real image on a sensor. What changes from camera to camera is the focal length, which sets how large distant objects appear, and the aperture, the diameter of the opening, which sets how much light arrives. Photographers combine them in the f-number, the focal length divided by the aperture diameter.
A lens at f/2 lets in four times the light of the same lens at f/4, because the opening's area goes as the square of its diameter. Wildlife photographers at Yellowstone use long lenses at wide apertures to catch distant animals in dim dawn light; landscape photographers stop down to f/16 so that both the foreground and the far mountains stay in focus. Every setting is a trade among brightness, sharpness and depth, governed by the same lens rules as the telescope.
The $40$-inch refractor at Yerkes Observatory in Williams Bay, Wisconsin, built by the University of Chicago in 1897, has an objective lens a meter across with a focal length of $19.4$ m. Its tube is over $19$ m long, and the observing floor rises and falls like an elevator so astronomers can reach the eyepiece.
With a $5$ cm eyepiece the magnification is $1940/5$, about $390\times$; a $2.5$ cm eyepiece doubles that. Edwin Hubble studied there as a graduate student. Because large lenses sag and bend colors differently, no larger refractor was ever built for research, and every major telescope since has used mirrors. The observatory is now run by a foundation that opens it to the public.
In hospital laboratories across the country, technicians count blood cells and identify bacteria with compound microscopes. A typical instrument has a $10\times$ eyepiece and objectives of $4\times$, $10\times$, $40\times$ and $100\times$, giving totals from $40\times$ to $1000\times$. The $100\times$ objective, with a focal length under $2$ mm, sits almost touching the slide in a drop of immersion oil.
At $1000\times$, an $8$ μm red blood cell looks $8$ mm across, as if held at $25$ cm, easily examined for sickle-cell shapes or parasites. Laboratories certified under federal CLIA rules must check their microscopes regularly, since a smudged objective or misaligned light can hide exactly the details a diagnosis depends on.
It is natural to think a telescope's job is to magnify. Its first job is to gather light: a large objective collects far more light than the eye, revealing faint objects. Magnification comes from the eyepiece and can be changed at will, but pushing it too high on a small telescope only enlarges a dim, blurry image.
A related error is to add the magnifications of lenses in series. They multiply: a $40\times$ objective with a $10\times$ eyepiece gives $400\times$, not $50\times$.
A telescope has a $900$ mm objective and a $25$ mm eyepiece. Find the magnification.
$M = \dfrac{900}{25} = 36$
Ratio of focal lengths.
Switch to a $9$ mm eyepiece. Find the new magnification.
$M = \dfrac{900}{9} = 100$
Shorter eyepiece, more power.
The Moon spans $0.52°$ to the eye. Find its apparent size at $100\times$.
$0.52° \times 100 = 52°$
Filling much of the view.
Find the telescope's length.
$f_o + f_e = 900 + 9 = 909\ \text{mm}$
Just under a meter.
Compare light gathered by a $100$ mm objective with a $7$ mm pupil.
$\left(\dfrac{100}{7}\right)^2 = 204$
Two hundred times more light.
An object is $15$ cm before a lens of $f = 10$ cm; a second lens of $f = 15$ cm is $40$ cm beyond. Find the first image.
$d_{i1} = \dfrac{10 \times 15}{15 - 10} = 30\ \text{cm}$
Real, beyond the first lens.
Find the first magnification.
$m_1 = -\dfrac{30}{15} = -2$
Inverted, doubled.
Find the object distance for the second lens.
$d_{o2} = 40 - 30 = 10\ \text{cm}$
Inside its focal length.
Find the final image.
$d_{i2} = \dfrac{15 \times 10}{10 - 15} = -30\ \text{cm}$
Virtual, on the object side of lens two.
Find the second magnification.
$m_2 = -\dfrac{-30}{10} = 3$
Upright relative to its object.
Find the total magnification.
$M = m_1m_2 = -2 \times 3 = -6$
Inverted overall, six times larger.
An objective of focal length $4.0$ mm and an eyepiece of $2.5$ cm sit in a $16$ cm tube. Find the objective's magnification.
$m_o \approx -\dfrac{16}{0.40} = -40$
Tube length over focal length.
Find the eyepiece's magnification.
$M_e = \dfrac{25}{2.5} = 10$
A magnifier on the first image.
Find the total.
$M = -40 \times 10 = -400$
Inverted, $400\times$.
A red blood cell is $8$ μm across. Find its apparent size.
$400 \times 8\ \mu\text{m} = 3.2\ \text{mm}$
As if viewed at $25$ cm.
Switch to a $100\times$ oil objective. Find the total.
$100 \times 10 = 1000$
The usual maximum.
Explain the limit.
$\text{diffraction blurs details near the wavelength}$
About $0.2$ μm.
Write the magnifier formula.
$M = \dfrac{25}{f}$
With $f$ in centimeters.
Substitute the focal length.
$M = \dfrac{25}{5.0}$
Near point over focal length.
Evaluate the magnification.
A telescope has an objective of focal length $700$ mm and an eyepiece of focal length $28$ mm. What is its angular magnification?
Complete the worked solution: a jeweler's loupe is a converging lens of focal length $12.5$ cm. With a near point of $25$ cm, find its power in diopters, its angular magnification with the image at infinity, and its angular magnification with the image at the near point.
Find the power.
$P = \dfrac{1}{f} =$ p
Focal length in meters.
Find the relaxed-eye magnification.
$M_\infty = \dfrac{25}{f} =$ m
Object at the focal point.
Find the near-point magnification.
$M_N = 1 + \dfrac{25}{f} =$ n
Object a little inside $f$.
Explain the choice of focal length.
$\text{shorter } f, \text{ more magnification}$
But a smaller field of view.
Match each instrument to how it forms its image.
| a long objective images distant objects; the eyepiece magnifies that image | a short objective makes a large real image that the eyepiece magnifies again | one lens forms a real, inverted image on a sensor | one lens forms a virtual, upright, enlarged image | |
|---|---|---|---|---|
| refracting telescope | ||||
| compound microscope | ||||
| camera | ||||
| magnifying glass |
A student microscope has an objective of focal length $2$ mm, an eyepiece of focal length $2.5$ cm, and a tube length of $16$ cm. Taking the near point as $25$ cm, fill in the objective's magnification, the eyepiece's angular magnification, and the total magnification.
| value | |
|---|---|
| objective magnification | |
| eyepiece magnification | |
| total magnification |
A telescope's objective has a focal length of $1000$ mm. Write its angular magnification as a function of the eyepiece's focal length $x$, in mm.
Answer:
Two converging lenses, of focal lengths $10$ cm and $15$ cm, are $40$ cm apart. An object stands $15$ cm in front of the first. Where is the final image, in cm from the second lens? Positive means beyond the second lens.
Answer: cm
The $40$-inch refractor at Yerkes Observatory in Wisconsin, the largest lens telescope ever used for research, has an objective of focal length $19.4$ m. With an eyepiece of focal length $1.6$ cm, what is its angular magnification?
Answer:
Lesson test: one question per skill, one attempt each, no hints. Your answers are checked when you submit.
A telescope's objective has a focal length of $1200$ mm. Write its angular magnification as a function of the eyepiece's focal length $x$, in mm.
Answer:
You can analyze optical instruments. Explain to someone why a telescope's aperture matters more than its magnification.
23. Your turn: a magnifier has a focal length of $5.0$ cm. What is its angular magnification with the image at infinity?, step 3
$M = 5.0$
Five times.