Part 4 · Chapter 34

Images

One goal of physics is to discover the laws that govern light; a broader goal is to put those laws to work, and no use is more important than the production of images. In this chapter we take the law of reflection and the law of refraction from Chapter 33 and let them build pictures of the world. We first sort every image into one of two kinds — real or virtual — then learn to locate the image a plane mirror, a curved mirror, a single curved refracting surface, or a thin lens produces, using one small family of equations and a handful of special rays. Finally we assemble lenses into the instruments that have extended human vision from the bacterium to the distant galaxy.

Fundamentals of Physics Prof. Mithun Mondal Reading time ≈ 70 min
i What you'll learn
  • The two types of image: a real image, where rays actually meet and which can fall on a screen (image distance \(i > 0\)), and a virtual image, where only the backward extensions of the rays meet (\(i < 0\)).
  • Plane mirrors: the image lies as far behind as the object is in front, \(i = -p\), virtual, erect, and the same size (\(m = +1\)).
  • Spherical mirrors: focal length \(f = \tfrac{1}{2}r\), the mirror equation \(\dfrac{1}{p}+\dfrac{1}{i}=\dfrac{1}{f}\), and the lateral magnification \(m = -\dfrac{i}{p}\), located graphically with the four special rays.
  • Spherical refracting surfaces: \(\dfrac{n_1}{p}+\dfrac{n_2}{i}=\dfrac{n_2-n_1}{r}\), with a sign rule for \(r\) that is the reverse of the one for mirrors.
  • Thin lenses: \(\dfrac{1}{p}+\dfrac{1}{i}=\dfrac{1}{f}\) and the lens-maker's equation \(\dfrac{1}{f}=(n-1)\!\left(\dfrac{1}{r_1}-\dfrac{1}{r_2}\right)\); converging lenses (\(f>0\)) make real or virtual images, diverging lenses (\(f<0\)) only virtual.
  • Optical instruments: the simple magnifier \(m_\theta = \dfrac{25\,\mathrm{cm}}{f}\), the compound microscope \(M = -\dfrac{s}{f_{\text{ob}}}\,\dfrac{25\,\mathrm{cm}}{f_{\text{ey}}}\), and the refracting telescope \(m_\theta = -\dfrac{f_{\text{ob}}}{f_{\text{ey}}}\).
Section 34-1

What Is Physics?

The first photographic images, made in the 1820s, were little more than novelties; today our world thrives on images. Whole industries rest on producing them on television, computer, and theater screens. Satellite images guide both military planners and environmental scientists, and surveillance cameras can make a transit system safer even as they raise hard questions about privacy. Researchers have even produced crude visual sensations in some blind people by stimulating the brain's visual cortex directly. Our first task here is to define and classify images; then we examine the handful of basic ways — reflection at mirrors and refraction at surfaces and lenses — by which they are produced.

Section 34-2

Two Types of Image

To see an object, your eye intercepts some of the light rays spreading from it and your visual system — retina to visual cortex — automatically infers where those rays came from. It assumes light travels in straight lines, so it traces the intercepted rays straight back to where they appear to originate and places the image there: a reproduction of the object derived from light. Two cases arise, and the distinction runs through the whole chapter.

A real image forms where the rays themselves actually converge and cross; it exists whether or not anyone is watching, and it can be caught on a screen — the image a projector throws on a wall, or the picture on your retina. A virtual image forms where only the backward extensions of the diverging rays meet; no light energy is present there, the image cannot be projected onto a screen, and it exists only for an observer whose eye gathers the rays — the "you" standing behind a bathroom mirror.

📐
A mirage is a virtual image
The shimmering "pool" on a hot road is a virtual image of the sky, formed when light bends in air whose index of refraction changes with temperature.

Near a sun-warmed road the air is hotter and slightly less dense below than above, so its index of refraction increases with height. Rays from the low sky descending toward the road bend gradually until they sweep back upward, and your visual system, assuming straight-line travel, traces them back to the road surface — where it paints a bluish patch that looks like water but recedes as you approach. Heat shimmer above a fire or a hot car hood is the same effect on a smaller scale.

Section 34-3

Plane Mirrors

A mirror is a surface that reflects a beam in a single direction rather than scattering or absorbing it; a polished metal surface qualifies, a concrete wall does not. Put a point source — the object \(O\) — a perpendicular distance \(p\) in front of a flat (plane) mirror. The rays that strike the mirror reflect according to the law of reflection, and they spread apart after reflecting, so they never truly meet. Extend the reflected rays straight back behind the mirror, however, and the extensions all cross at one point a distance \(i\) behind it. Your eye, intercepting the reflected light, places the image \(I\) there.

The plane-mirror relation
\[ i = -p \]
The image sits as far behind the mirror as the object is in front. The minus sign is a convention that flags a virtual image: the rays do not pass through it, so its image distance is taken as negative. A plane mirror is the limiting case of a spherical mirror whose radius of curvature is infinite.

An extended object is just a collection of point sources, so the mirror builds up a composite virtual image, point by point. Drawing rays from the top and bottom of an upright arrow shows the image is the same height, the same way up, and the same distance behind — yet it has undergone a curious switch we call left–right reversal (more precisely, a front–back reversal that makes your right hand look like the image's left).

Why "left–right," not "up–down"? A mirror does not actually swap left and right — it reverses front and back, flipping the axis pointing into the glass. Your raised right hand maps onto the image's hand on the same side of the room, but because the image faces you, you read that hand as its "left." Up and down lie in the mirror plane and are untouched, which is why the puzzle feels lopsided.
Section 34-4

Spherical Mirrors

Bend a plane mirror into a small piece of a sphere and you get a spherical mirror. Curve it so the reflecting side caves in toward the object and it is concave; bulge it the other way and it is convex. Bending the mirror moves the image: a concave mirror brings the image closer and can enlarge it, while a convex mirror pushes the image farther away and shrinks it, widening the field of view (the reason for passenger-side and store-security mirrors).

The key new quantity is the focal point \(F\). Rays arriving parallel to the central axis reflect through \(F\) for a concave mirror, or appear to diverge from \(F\) behind a convex mirror. Its distance from the mirror is the focal length \(f\), fixed by the radius of curvature alone.

Focal length of a spherical mirror
\[ f = \tfrac{1}{2}r \]
The focal length is half the radius of curvature. By convention \(f\) and \(r\) are positive for a concave mirror (real focus, on the object's side) and negative for a convex mirror (virtual focus, behind it). A plane mirror has \(r=\infty\) and therefore \(f=\infty\).
Section 34-5

Images from Spherical Mirrors

For rays that make only small angles with the central axis (the paraxial approximation), one compact equation ties together the object distance, the image distance, and the focal length — and it holds for concave, convex, and plane mirrors alike.

The mirror equation
\[ \frac{1}{p}+\frac{1}{i}=\frac{1}{f} \]
The object distance \(p\) is positive. The image distance \(i\) is positive for a real image (formed in front of the mirror, on the object's side) and negative for a virtual image (behind the mirror). A convex or plane mirror can form only a virtual image, whatever the object's position.
Lateral magnification
\[ m = -\frac{i}{p}, \qquad |m| = \frac{h'}{h} \]
Here \(h\) and \(h'\) are the object and image heights. A positive \(m\) means the image is erect (same orientation as the object); a negative \(m\) means it is inverted. For a plane mirror \(i=-p\), so \(m=+1\) — same size, erect, as expected.
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Locating an image with the four special rays
Pick any two of these four rays from an off-axis point; where they cross (or their extensions cross) is the image of that point.

1. A ray parallel to the central axis reflects through the focal point \(F\). 2. A ray that passes through \(F\) reflects back parallel to the axis. 3. A ray through the center of curvature \(C\) retraces its own path. 4. A ray striking the mirror at its central point reflects symmetrically about the axis. For a convex mirror, read "toward \(F\)" and "toward \(C\)" for rays 2 and 3, since those points lie behind the mirror.

Section 34-6

Spherical Refracting Surfaces

We now turn from reflection to refraction at a single spherical surface separating two transparent media of indices \(n_1\) (where the object sits) and \(n_2\). As a ray crosses, it bends toward the normal if it enters a higher-index medium and away if it enters a lower-index one. Whether the bent rays converge to a real image or diverge into a virtual one depends on the indices, the curvature, and how far the object lies from the surface — but again a single paraxial equation captures every case.

Single refracting surface
\[ \frac{n_1}{p}+\frac{n_2}{i}=\frac{n_2-n_1}{r} \]
As with mirrors, \(p\) is positive and \(i\) is positive for a real image, negative for a virtual one. The crucial twist is the sign rule for the radius: when the object faces a convex surface, \(r\) is positive; when it faces a concave surface, \(r\) is negative — the reverse of the mirror convention, so handle it with care.
A bug in amber, magnified. Refraction at a curved surface explains why an insect trapped in a polished, convex amber bead looks larger and nearer than it really is: rays leaving the bug refract at the amber–air surface so their backward extensions form an enlarged virtual image. Note the contrast with mirrors — for a refracting surface, real images form on the side opposite the object, because the light passes through.
Section 34-7

Thin Lenses

A lens is a transparent body with two refracting surfaces sharing a common central axis. Light refracts on entering the lens and again on leaving it; a thin lens is one whose thickness is negligible next to \(p\), \(i\), and the radii, so we can treat all the bending as happening at one central plane. A lens that makes parallel rays converge is a converging lens (thicker in the middle, \(f>0\)); one that makes them diverge is a diverging lens (thinner in the middle, \(f<0\)).

Thin-lens equation
\[ \frac{1}{p}+\frac{1}{i}=\frac{1}{f} \]
Identical in form to the mirror equation, with the same sign meanings: \(i>0\) for a real image (on the far side of the lens from the object), \(i<0\) for a virtual image (on the object's side). The magnification is again \(m=-i/p\).
The lens-maker's equation
\[ \frac{1}{f}=(n-1)\!\left(\frac{1}{r_1}-\frac{1}{r_2}\right) \]
For a lens of index \(n\) in air, with \(r_1\) the radius of the surface nearer the object and \(r_2\) that of the far surface, signed by the refracting-surface rule above. A lens bends light only because its index differs from the surroundings — submerge it in a medium of index \(n_{\text{med}}\) and replace \(n\) by \(n/n_{\text{med}}\); matched indices give no bending at all.
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Which image does a lens make?
A converging lens makes a real, inverted image of an object outside its focal point, and a virtual, erect, enlarged image of an object inside it; a diverging lens always makes a virtual, erect, reduced image.

Locate it with three special rays from an off-axis point: a ray parallel to the axis passes through the far focal point; a ray through the near focal point emerges parallel; and a ray straight through the lens center is undeviated. Slide an object toward a converging lens and watch the image flip from small-and-distant, to enormous at the focal point, to the erect magnified image of a handheld lens — the whole behavior of a camera, a projector, and a magnifier in one motion.

Section 34-8

Optical Instruments

The eye focuses sharply only on objects from infinity down to a near point \(P_n\), taken here as 25 cm. Bring an object closer and it blurs; the detail you can resolve is limited by the angle the object subtends. Optical instruments work by enlarging that angle — by increasing the angular magnification \(m_\theta=\theta'/\theta\), the ratio of the angle the final image subtends to the angle the object subtends at the unaided eye.

Simple magnifying lens
\[ m_\theta = \frac{25\,\mathrm{cm}}{f} \]
Place the object just inside the focal point of a converging lens. The lens forms a distant, enlarged virtual image the eye can focus comfortably, subtending a larger angle than the object could at the near point — a short focal length giving the most power.
Compound microscope
\[ M = m\,m_\theta = -\frac{s}{f_{\text{ob}}}\,\frac{25\,\mathrm{cm}}{f_{\text{ey}}} \]
The objective (focal length \(f_{\text{ob}}\)) forms a real, inverted, enlarged image with lateral magnification \(m=-s/f_{\text{ob}}\), where \(s\) is the tube length. The eyepiece (focal length \(f_{\text{ey}}\)) then views that image as a simple magnifier, and the total magnification is the product.
Refracting telescope
\[ m_\theta = -\frac{f_{\text{ob}}}{f_{\text{ey}}} \]
In a telescope the objective's rear focal point coincides with the eyepiece's front focal point. A long objective focal length and a short eyepiece focal length give large angular magnification — the opposite balance from a microscope. The minus sign marks the inverted final image. For astronomy, light-gathering power (a large objective) usually matters more than raw magnification.
Worked Examples

Putting It to Work

1 A concave mirror forms a real image

Problem. A concave mirror has radius of curvature \(r = 24\,\mathrm{cm}\). An object 1.0 cm tall is placed \(p = 36\,\mathrm{cm}\) in front of it. Find the image distance, the magnification, and describe the image.

Solution. The focal length is \(f = r/2 = +12\,\mathrm{cm}\) (positive, concave). Solve the mirror equation for \(i\).

1/i = 1/f − 1/p
\[ \frac{1}{i} = \frac{1}{12} - \frac{1}{36} = \frac{3-1}{36} = \frac{1}{18} \;\Rightarrow\; i = +18\,\mathrm{cm} \]
m = −i/p
\[ m = -\frac{18}{36} = -0.50, \qquad h' = |m|h = 0.50\,\mathrm{cm} \]

Since \(i>0\) the image is real and lies 18 cm in front of the mirror; \(m<0\) means it is inverted and, at half size, smaller than the object — exactly what you see in the bowl of a shiny spoon held at arm's length.

2 A convex mirror's virtual image

Problem. A convex security mirror has \(|r| = 40\,\mathrm{cm}\). A shopper stands \(p = 2.0\,\mathrm{m}\) away. Locate the image and give its magnification.

Solution. For a convex mirror \(f = r/2 = -20\,\mathrm{cm} = -0.20\,\mathrm{m}\) (negative). With \(p = 2.0\,\mathrm{m}\):

1/i = 1/f − 1/p
\[ \frac{1}{i} = \frac{1}{-0.20} - \frac{1}{2.0} = -5.0 - 0.5 = -5.5\,\mathrm{m^{-1}} \;\Rightarrow\; i \approx -0.18\,\mathrm{m} \]
m = −i/p
\[ m = -\frac{-0.18}{2.0} \approx +0.09 \]

The negative \(i\) marks a virtual image about 18 cm behind the mirror; \(m \approx +0.09\) says it is erect and roughly one-tenth size. Shrinking the world this way is precisely how a convex mirror packs a wide aisle into a small reflecting disk.

3 Designing a converging lens

Problem. A double-convex lens of glass (\(n = 1.50\)) has surface radii of magnitude 20 cm and 30 cm. (a) Find its focal length. (b) An object is placed 60 cm in front of it; find the image distance and magnification.

Solution (a). The object faces a convex first surface, so \(r_1 = +20\,\mathrm{cm}\); it faces a concave second surface, so \(r_2 = -30\,\mathrm{cm}\). Apply the lens-maker's equation.

1/f = (n−1)(1/r₁ − 1/r₂)
\[ \frac{1}{f} = (1.50-1)\!\left(\frac{1}{20} - \frac{1}{-30}\right) = 0.50\!\left(\frac{1}{20} + \frac{1}{30}\right) = 0.50\,(0.0833) \;\Rightarrow\; f \approx +24\,\mathrm{cm} \]

Solution (b). With \(p = 60\,\mathrm{cm}\) and \(f = 24\,\mathrm{cm}\):

1/i = 1/f − 1/p
\[ \frac{1}{i} = \frac{1}{24} - \frac{1}{60} = \frac{5-2}{120} = \frac{1}{40} \;\Rightarrow\; i = +40\,\mathrm{cm}, \quad m = -\frac{40}{60} \approx -0.67 \]

The image is real, inverted, and about two-thirds size — the object lies outside the focal point, so a converging lens behaves like a camera lens here.

4 A telescope's magnification

Problem. A refracting telescope has an objective of focal length \(f_{\text{ob}} = 100\,\mathrm{cm}\) and an eyepiece of focal length \(f_{\text{ey}} = 5.0\,\mathrm{cm}\). (a) Find its angular magnification. (b) How long is the telescope tube?

Solution. Use the telescope relation, then note that the two focal points coincide so the tube length is the sum of the focal lengths.

mθ = −f_ob / f_ey
\[ m_\theta = -\frac{100}{5.0} = -20 \]
L = f_ob + f_ey
\[ L = 100 + 5.0 = 105\,\mathrm{cm} \]

The image is magnified 20-fold and inverted (hence the minus sign) — fine for stars and planets, where "upside down" hardly matters. The barrel is just over a meter long, set by the long objective that gives both the magnification and the light-gathering reach.

Review

Chapter Summary

✓ Two types of image

Real images form where rays cross (\(i>0\), can be projected); virtual images form where ray extensions cross (\(i<0\), cannot).

✓ Plane mirror

\(i = -p\); virtual, erect, same size, \(m = +1\); front–back reversal.

✓ Spherical mirror

\(f = \tfrac{1}{2}r\); \(\dfrac{1}{p}+\dfrac{1}{i}=\dfrac{1}{f}\); \(f>0\) concave, \(f<0\) convex.

✓ Magnification

\(m = -\dfrac{i}{p}\), \(|m| = \dfrac{h'}{h}\); \(m>0\) erect, \(m<0\) inverted.

✓ Refracting surface

\(\dfrac{n_1}{p}+\dfrac{n_2}{i}=\dfrac{n_2-n_1}{r}\); \(r>0\) for a convex face — reverse of mirrors.

✓ Thin lens

\(\dfrac{1}{p}+\dfrac{1}{i}=\dfrac{1}{f}\); lens-maker \(\dfrac{1}{f}=(n-1)\!\left(\dfrac{1}{r_1}-\dfrac{1}{r_2}\right)\).

✓ Magnifier & microscope

\(m_\theta = \dfrac{25\,\mathrm{cm}}{f}\); \(M = -\dfrac{s}{f_{\text{ob}}}\dfrac{25\,\mathrm{cm}}{f_{\text{ey}}}\).

✓ Telescope

\(m_\theta = -\dfrac{f_{\text{ob}}}{f_{\text{ey}}}\); long objective, short eyepiece; image inverted.

Practice

Problems

Take the near point at 25 cm. For mirrors, \(f\) and \(r\) are positive for concave, negative for convex; for refracting surfaces and lenses, sign each radius by the surface the object faces (convex \(\Rightarrow r>0\)). In every case \(p>0\), while \(i>0\) is a real image and \(i<0\) a virtual one.

  1. You stand 1.5 m in front of a plane mirror. (a) How far behind the mirror is your image? (b) If you walk 0.5 m toward the mirror, by how much does the distance between you and your image change?
  2. A concave mirror has a focal length of 15 cm. An object is placed 45 cm in front of it. Find (a) the image distance, (b) the magnification, and (c) whether the image is real or virtual, erect or inverted.
  3. An object is placed 10 cm in front of a concave mirror of focal length 15 cm. Locate the image and find its magnification. What kind of image is it?
  4. A convex mirror has a radius of curvature of 30 cm. An object 2.0 cm tall stands 20 cm in front of it. Find the image distance, the image height, and describe the image.
  5. A makeup mirror is to produce an erect image magnified 1.5 times when a face is held 20 cm from it. (a) Should the mirror be concave or convex? (b) What focal length is required?
  6. A point object sits in air 30 cm from the convex end (\(r = 10\,\mathrm{cm}\)) of a long glass rod (\(n = 1.50\)). Find the image distance inside the glass.
  7. A small fish is 20 cm below the flat surface of a pond (\(n = 1.33\)). To a bird looking straight down, how deep does the fish appear to be? (Use the single-surface formula with \(r = \infty\).)
  8. A double-convex lens (\(n = 1.52\)) has surfaces of radii 25 cm and 25 cm. Find its focal length, and the image of an object placed 50 cm in front of it.
  9. A thin lens has focal length \(f = -20\,\mathrm{cm}\). An object is placed 30 cm in front of it. Find the image distance and magnification, and identify the lens type and image type.
  10. A converging lens of focal length 10 cm is used as a simple magnifier. (a) Find its angular magnification. (b) What focal length would double that magnification?
  11. A compound microscope has an objective of focal length 0.80 cm and an eyepiece of focal length 2.5 cm separated so the tube length is \(s = 16\,\mathrm{cm}\). Find the overall magnification.
  12. A refracting telescope is to have an angular magnification of 40. If the eyepiece focal length is 2.0 cm, find (a) the objective focal length and (b) the length of the telescope tube.
  13. An object is placed in turn at 8 cm, 12 cm (the focal point), and 24 cm in front of a converging lens of focal length 12 cm. For each, find the image distance and state whether the image is real or virtual — and explain what happens at the focal point.
  14. Two thin lenses of focal lengths \(+10\,\mathrm{cm}\) and \(+15\,\mathrm{cm}\) are placed in contact. Show that the combination behaves like a single lens and find its focal length. (Hint: add the powers \(1/f\).)
Tip: three habits keep this chapter clean. First, fix your signs before plugging in numbers — for mirrors \(f>0\) means concave, but for lenses and refracting surfaces the radius is signed by the surface the object faces, and that rule runs opposite to the mirror's, so write down each sign explicitly. Second, let the answer's sign read back the physics: a positive image distance is a real image you could catch on a screen, a negative one a virtual image behind the optic, while a positive magnification is erect and a negative one inverted — if a "real" image comes out with \(i<0\), you have a sign slip, not a new effect. Third, before reaching for an equation, sketch two of the special rays; a quick ray diagram tells you at a glance whether the image is real or virtual, enlarged or reduced, and on which side it sits, so the algebra only has to confirm what the picture already showed.