Chapter 01 · Ray Optics · Full book
Ray Optics and Optical Instruments
Light, left to itself, travels in straight lines. That single fact would make optics dull—a subject of rays going nowhere interesting—were it not for surfaces. Everything worth studying in this chapter happens where a ray meets a boundary: it either bounces back into the medium it came from (reflection) or crosses into a new medium and bends (refraction). Two rules, one at each kind of surface, and nothing else. The astonishing part is how far those two rules reach.
From reflection alone come the flat mirror, the shaving mirror, and the car's wide-angle rear-view. From refraction come the broken straw, the shallow-looking pool, the sparkle trapped inside a diamond, the mirage on a hot road, the lens in your eye, and the telescope that resolves the moons of Jupiter. Every one of these is the same pair of laws, read through a single bookkeeping convention for signs—fixed once, at the outset, and never allowed to drift. Master the sign convention and the one keystone result for a curved surface, and ray optics stops being a list of formulas to memorise: it becomes two ideas applied over and over.
As throughout these books, two audiences travel together. Locating images with the mirror and lens equations, and the ray diagrams that go with them, are for everyone (NEET, AP Physics 2); the silvered lenses, achromatic doublets, and Snell's-window geometry are for those going further (JEE Advanced and the Olympiads). Both are the same light, bending at the same surfaces.
Core intuition: Chapter Roadmap
Light, left alone, travels in straight lines. Everything interesting in this
chapter happens at a surface, where a ray either bounces back
(reflection) or bends as it crosses into a new medium (refraction). Astonishingly,
every mirror, lens, prism, microscope, and telescope ever built is assembled from
just those two rules—plus one bookkeeping convention for signs.
- Reflection — the law of equal angles, from the flat mirror to curved ones.
- The mirror formula — \(\dfrac{1}{v}+\dfrac{1}{u}=\dfrac{1}{f}\), derived once from paraxial geometry and governed by a single Cartesian sign convention that we fix at the outset and never abandon.
- Refraction — Snell's law, the refractive index, and why a straw looks broken and a pool looks shallow.
- Total internal reflection — the trapped-light rule behind the sparkle of diamond, the optical fibre, and the desert mirage.
- Refraction at a spherical surface — the keystone result \(\dfrac{n_2}{v}-\dfrac{n_1}{u}=\dfrac{n_2-n_1}{R}\), from which every property of a lens follows.
- Lenses — the lens-maker's equation, the thin-lens formula, power in dioptres, and combinations.
- The prism — deviation, the minimum-deviation condition, and the splitting of white light into colour.
- Optical instruments — the eye, the microscope, and the telescope: lenses arranged to defeat the limits of the unaided eye.
Perplexing Questions
Perplexing questions
Before any formula, let these sit with you. Each is answered—fully—before the
chapter ends.
- The Broken Straw: A straw standing in a glass of water looks snapped at the surface, and the bottom of any pool looks closer than it really is. Are these the same effect, and what exactly is bending—the straw, or the light?
- The Flat Mirror and the Spoon: A flat bathroom mirror shows you life-size no matter how far you stand. Yet the back of a spoon shrinks your face, and the bowl flips it upside-down up close. One rule must cover all three. What is it?
- The Dull Glass Diamond: A real diamond blazes with internal fire; a glass replica cut to the identical shape looks lifeless. The shape is the same—so what property of the material decides whether light escapes or stays trapped?
- The Phantom Lake: On a hot road, the distant tarmac shimmers like a sheet of water reflecting the sky. There is no water. So what is doing the reflecting?
- The Two-Faced Magnifier: Held close to a stamp, a magnifying glass enlarges it. Held at arm's length and aimed across the room, the same lens shows a tiny, inverted world. How can one lens do opposite things?
- The Choosy Window: White light fans into a rainbow through a triangular prism, but passes straight through a flat windowpane with no colour at all. Why does the shape of the glass decide whether colour appears?
Each question is a crack in the everyday picture of light.
By the end of this chapter every one of them will be transparent.
By the end of this chapter every one of them will be transparent.
Reflection and the Plane Mirror
◈ Two inclined mirrors — the 360/θ rule
live benchSet the angle between two mirrors and count the images that ring around the object.
A ray is the idealisation at the heart of this chapter: an infinitely thin line marking the direction in which light energy travels. As long as the obstacles light meets are far larger than its wavelength (\(\sim 500\,\text{nm}\)), light behaves as if it moves along such rays in straight lines. This is geometrical (or ray) optics. The wave nature of light—which dominates when obstacles approach the wavelength—is the subject of the next chapter; here it stays safely hidden.
When a ray strikes a smooth surface, it obeys the two laws of reflection:
- The incident ray, the reflected ray, and the normal to the surface at the point of incidence all lie in one plane.
- The angle of incidence equals the angle of reflection, \(\theta_i = \theta_r\), both measured from the normal.
These laws hold at every point of any surface, flat or curved—the only difference is which way the normal points.
Image in a plane mirror
Trace two rays from a point object \(O\) in front of a plane mirror. Each reflects with \(\theta_i=\theta_r\); the reflected rays diverge as if they came from a single point \(I\) behind the mirror. A little geometry shows that \(I\) lies on the perpendicular from \(O\) to the mirror, as far behind the mirror as \(O\) is in front. The image is therefore:
- virtual — no light actually reaches \(I\); the rays only appear to come from it (you cannot catch it on a screen placed there);
- erect and the same size as the object;
- laterally inverted — left and right are swapped, because front-and-back is reversed by the reflection.
Core intuition: Why “virtual”?
A real image is formed where reflected or refracted rays actually
converge and cross: light is genuinely there, and a screen placed at that
point catches a bright patch. A virtual image is formed where rays only
appear to diverge from: no light reaches that point, so no screen can
catch it—yet your eye, which cannot tell the difference, sees an image. Keeping
these two apart is the single most useful habit in this chapter.
Two mirrors: counting images
Place an object between two plane mirrors inclined at an angle \(\theta\). Repeated reflection produces multiple images arranged on a circle. When \(360^\circ/\theta\) is an even integer, the number of images is \[ N = \frac{360^\circ}{\theta} - 1 . \] For \(\theta = 60^\circ\), \(N = 5\); for \(\theta = 90^\circ\), \(N = 3\). (The case of odd \(360^\circ/\theta\) and the symmetric placement carry small corrections we defer to the worked examples.) Two parallel mirrors, \(\theta\to 0\), give an infinite corridor of images—the effect inside a barber's shop.
Worked Examples
Example 1 · A mirror that turns, a ray that turns twice
A ray of light reflects off a plane mirror. The mirror is then rotated by
\(12^\circ\) about an axis in its own plane, while the incident ray is held fixed.
Through what angle does the reflected ray turn?
Solution.
Given: mirror rotation \(\alpha = 12^\circ\), incident ray fixed.
Find: rotation of the reflected ray.
Setup: rotating the mirror by \(\alpha\) increases the angle of incidence by \(\alpha\); since the reflected ray is measured symmetrically (\(\theta_r=\theta_i\)), the angle between incident and reflected rays changes by \(2\alpha\).
Solve: \[ \Delta(\text{reflected ray}) = 2\alpha = 2\times 12^\circ = 24^\circ . \] Answer: \(\boxed{24^\circ}\) — the reflected ray turns through twice the mirror's rotation.
Check: the factor of two is the working principle of the optical lever, used in galvanometers and torsion balances to amplify a tiny mirror rotation into a large, easily read beam deflection. ✓
Setup: rotating the mirror by \(\alpha\) increases the angle of incidence by \(\alpha\); since the reflected ray is measured symmetrically (\(\theta_r=\theta_i\)), the angle between incident and reflected rays changes by \(2\alpha\).
Solve: \[ \Delta(\text{reflected ray}) = 2\alpha = 2\times 12^\circ = 24^\circ . \] Answer: \(\boxed{24^\circ}\) — the reflected ray turns through twice the mirror's rotation.
Check: the factor of two is the working principle of the optical lever, used in galvanometers and torsion balances to amplify a tiny mirror rotation into a large, easily read beam deflection. ✓
Example 2 · Two mirrors, the odd case
Two plane mirrors are inclined at \(72^\circ\). How many images of a small object
placed between them are seen (i) when the object lies off the angle bisector, and
(ii) when it lies exactly on the bisector?
Solution.
Given: \(\theta = 72^\circ\), so \(360^\circ/\theta = 5\) (an odd
integer).
Find: image count in both placements.
Setup: the counting rule splits on the parity of \(360^\circ/\theta\). When it is even, \(N=360^\circ/\theta-1\) regardless of placement. When it is odd, \(N=360^\circ/\theta-1\) for an object on the bisector but \(N=360^\circ/\theta\) for an object placed asymmetrically.
Solve: \[ \text{(i) off bisector: } N = \frac{360^\circ}{72^\circ} = 5, \qquad \text{(ii) on bisector: } N = \frac{360^\circ}{72^\circ} - 1 = 4 . \] Answer: \(\boxed{N=5 \text{ (asymmetric)},\ N=4 \text{ (symmetric)}}\).
Check: the symmetric case merges the two rearmost images into one on the line of symmetry, dropping the count by exactly one — the “small correction” promised when the rule was stated. For the even case \(\theta=60^\circ\) there is no such ambiguity and \(N=5\) always. ✓
Setup: the counting rule splits on the parity of \(360^\circ/\theta\). When it is even, \(N=360^\circ/\theta-1\) regardless of placement. When it is odd, \(N=360^\circ/\theta-1\) for an object on the bisector but \(N=360^\circ/\theta\) for an object placed asymmetrically.
Solve: \[ \text{(i) off bisector: } N = \frac{360^\circ}{72^\circ} = 5, \qquad \text{(ii) on bisector: } N = \frac{360^\circ}{72^\circ} - 1 = 4 . \] Answer: \(\boxed{N=5 \text{ (asymmetric)},\ N=4 \text{ (symmetric)}}\).
Check: the symmetric case merges the two rearmost images into one on the line of symmetry, dropping the count by exactly one — the “small correction” promised when the rule was stated. For the even case \(\theta=60^\circ\) there is no such ambiguity and \(N=5\) always. ✓
Example 3 · How fast does your reflection rush toward you?
You walk straight toward a fixed plane mirror at \(2.0\,\text{m\,s}^{-1}\). How fast
does your image approach the mirror, and how fast does it approach you?
Solution.
Given: object (you) speed \(v=2.0\,\text{m\,s}^{-1}\) toward a stationary
mirror.
Find: image speed relative to the mirror and relative to you.
Setup: a plane-mirror image is always as far behind the mirror as the object is in front. If the object moves \(\Delta x\) toward the mirror, the image moves \(\Delta x\) toward the mirror on the far side — equal speeds relative to the mirror, oppositely directed in space.
Solve: image speed toward the mirror \(= v = 2.0\,\text{m\,s}^{-1}\). The gap between you and your image closes at \[ v_{\text{rel}} = v + v = 2v = 4.0\,\text{m\,s}^{-1}. \] Answer: \(\boxed{2.0\,\text{m\,s}^{-1}\text{ toward the mirror},\ \ 4.0\,\text{m\,s}^{-1}\text{ toward you}}\).
Check: if the mirror itself moved toward a stationary object at speed \(u\), the image would approach the object at \(2u\) instead — the same doubling, now carried by the mirror. Approach speeds in plane mirrors always come with a factor of two. ✓
Setup: a plane-mirror image is always as far behind the mirror as the object is in front. If the object moves \(\Delta x\) toward the mirror, the image moves \(\Delta x\) toward the mirror on the far side — equal speeds relative to the mirror, oppositely directed in space.
Solve: image speed toward the mirror \(= v = 2.0\,\text{m\,s}^{-1}\). The gap between you and your image closes at \[ v_{\text{rel}} = v + v = 2v = 4.0\,\text{m\,s}^{-1}. \] Answer: \(\boxed{2.0\,\text{m\,s}^{-1}\text{ toward the mirror},\ \ 4.0\,\text{m\,s}^{-1}\text{ toward you}}\).
Check: if the mirror itself moved toward a stationary object at speed \(u\), the image would approach the object at \(2u\) instead — the same doubling, now carried by the mirror. Approach speeds in plane mirrors always come with a factor of two. ✓
Spherical Mirrors and the Mirror Formula
◈ The mirror bench — concave, convex, plane
live benchDrag the object through every case on a live 2D⇄3D bench and read u·v·f·m as the image forms.
A spherical mirror is a small slice of a reflecting sphere. The geometry is named once and used forever:
- the pole \(P\) — the centre of the mirror's surface;
- the centre of curvature \(C\) — the centre of the sphere the mirror is cut from; \(PC = R\), the radius of curvature;
- the principal axis — the line \(PC\);
- the focus \(F\) — where paraxial rays parallel to the axis converge (concave) or appear to diverge from (convex). We will prove \(PF = R/2\).
A concave mirror curves towards the incoming light (reflecting surface on the inside of the sphere); a convex mirror curves away.
The sign convention: fix it now, never drift
Every formula in ray optics is a statement about signed distances. Confuse a sign and a converging lens turns into a diverging one. We adopt the New Cartesian Convention and hold to it for the whole chapter.
Core intuition: The New Cartesian Sign Convention
Place the origin at the pole of the mirror (or the optical centre of a
lens). Measure everything from there.
- Light is taken to travel left to right. Distances measured along the incident light (to the right) are positive; distances measured against it (to the left) are negative.
- Heights above the axis are positive; below, negative.
- A real object sits to the left, so its distance \(u\) is negative.
- For a concave mirror \(f\lt 0\) and \(R\lt 0\); for a convex mirror \(f\gt 0\) and \(R\gt 0\).
- A real image (mirror) forms on the same side as the object, so its \(v\) is negative; a virtual image has \(v\) positive.
Misconception alert: The #1 Optics Trap: Sign-Convention Drift
The single most common error in this entire chapter is to plug magnitudes
into a formula and then “fix the sign at the end by physical reasoning.” Do not.
Assign the sign when you read the data—real object \(\Rightarrow u\lt 0\);
concave mirror \(\Rightarrow f\lt 0\)—and let the algebra return the sign of the
answer. The sign of \(v\) then tells you whether the image is real or
virtual; you should never have to guess it. Examiners build whole questions
around students who carry a stray minus sign.
Deriving the mirror formula
◇ Figure 1
Fig 1.The New Cartesian sign convention. Origin at the pole or optical centre;
light travels left to right. Distances measured along the light are positive,
against it negative; heights above the axis are positive. Every formula in this
chapter assumes these signs.
Birth of the formula
Take a concave mirror with axial object \(O\). A paraxial ray leaves \(O\), strikes
the mirror at a point \(M\) a small height \(h\) above the axis, and reflects to form
the image \(I\) on the axis. The normal at \(M\) passes through the centre \(C\), so the
angle of incidence and reflection are each the angle \(\theta\) between the ray and
\(MC\).
Let \(\alpha\), \(\beta\), \(\gamma\) be the (small) angles that \(MO\), \(MC\), and \(MI\)
make with the axis. By the exterior-angle theorem applied to the two triangles
sharing the vertex \(M\),
\[
\beta = \alpha + \theta, \qquad \gamma = \beta + \theta
\;\;\Longrightarrow\;\; \alpha + \gamma = 2\beta .
\]
For paraxial rays the foot of \(M\) is essentially at the pole \(P\), so each angle is
its height over its distance:
\[
\alpha \approx \frac{h}{PO}, \qquad
\beta \approx \frac{h}{PC}, \qquad
\gamma \approx \frac{h}{PI}.
\]
Substituting and cancelling \(h\),
\[
\frac{1}{PO} + \frac{1}{PI} = \frac{2}{PC}.
\]
Now attach Cartesian signs. With the pole as origin and light travelling left to
right, the object, image, and centre of a concave mirror all lie to the left:
\(PO=-u\), \(PI=-v\), \(PC=-R\). The minus signs cancel throughout, leaving
\[
\boxed{\;\frac{1}{v} + \frac{1}{u} = \frac{2}{R} = \frac{1}{f}\;}
\qquad\text{with}\qquad f = \frac{R}{2}.
\]
The relation \(f=R/2\) follows by sending the object to infinity: a parallel ray
(\(u\to\infty\)) converges to \(v=f\), and the equation gives \(1/f = 2/R\). Although
derived for a concave mirror, the boxed result holds for both mirror types
provided the sign convention is obeyed: a convex mirror simply carries
\(R\gt 0\), hence \(f\gt 0\).
Birth of the formula
An off-axis object of height \(h\) sends its tip's ray to the pole \(P\), where it
reflects symmetrically about the axis (\(\theta_i=\theta_r\)). The object and image
tips, the pole, and the axis form two similar right triangles, giving
\(\dfrac{h'}{h} = \dfrac{PI}{PO}\) in magnitude. Attaching signs (a real, inverted
image lies below the axis, \(h'\lt 0\); object and image distances are both negative
for a real image),
\[
\boxed{\; m = \frac{h'}{h} = -\frac{v}{u}\;}.
\]
Read the result: \(m\lt 0\) means inverted, \(m\gt 0\) erect;
\(|m|\gt 1\) enlarged, \(|m|\lt 1\) diminished.
Misconception alert: “\(m=-v/u\) for mirrors but \(m=+v/u\) for lenses”
The magnification carries a minus sign for mirrors and a plus sign
for lenses. This is not a contradiction—it is a consequence of the same sign
convention applied to two different geometries (a mirror folds light back; a lens
passes it through). Memorise the pair together, never one alone, or you will
import the mirror sign into a lens problem and invert every image you compute.
What kind of image? (concave mirror)
The single equation \(1/v+1/u=1/f\) already contains every case; tabulating it is just reading off answers for a concave mirror (\(f\lt 0\)):
| Object position | Image | Use |
| At infinity | at \(F\), real, inverted, tiny | solar furnace |
| Beyond \(C\) | between \(F\) and \(C\), real, inverted, diminished | |
| At \(C\) | at \(C\), real, inverted, same size | |
| Between \(C\) and \(F\) | beyond \(C\), real, inverted, enlarged | projector |
| At \(F\) | at infinity | searchlight |
| Between \(F\) and \(P\) | behind mirror, virtual, erect, enlarged | shaving/make-up mirror |
A convex mirror (\(f\gt 0\)) is simpler: for any real object it forms a virtual, erect, diminished image between \(P\) and \(F\). That always-shrunk, always-upright, wide-angle view is exactly why it is the passenger-side mirror (“objects are closer than they appear”).
◇ Figure 2
Fig 2.Ray diagrams for spherical mirrors (real rays solid blue, virtual
back-extensions grey dashed). Left: a concave mirror images an object
beyond \(C\) as a real, inverted, diminished image between \(F\) and \(C\).
Right: a convex mirror always forms a virtual, erect, diminished image
behind the mirror.
Worked Examples
Example 4 · Concave mirror, object inside the focus (the shaving mirror)
An object is placed \(10\,\text{cm}\) in front of a concave mirror of focal length
\(15\,\text{cm}\). Locate the image, and state its nature and magnification.
Solution.
Given: concave \(\Rightarrow f=-15\,\text{cm}\); real object
\(\Rightarrow u=-10\,\text{cm}\) (note \(|u|\lt |f|\): object is inside the focus).
Find: \(v\), \(m\), nature.
Setup: mirror formula \(\dfrac1v+\dfrac1u=\dfrac1f\), signs assigned first.
Solve: \[ \frac1v = \frac1f - \frac1u = \frac{1}{-15} - \frac{1}{-10} = -\frac{2}{30}+\frac{3}{30} = \frac{1}{30} \;\Rightarrow\; v = +30\,\text{cm}. \] \[ m = -\frac{v}{u} = -\frac{30}{-10} = +3 . \] Answer: \(\boxed{v=+30\,\text{cm},\ m=+3}\) — a virtual (\(v\gt 0\), behind the mirror), erect (\(m\gt 0\)), enlarged (\(3\times\)) image.
Check: an object inside the focus of a concave mirror is exactly the make-up / shaving-mirror configuration; the always-real intuition fails here precisely because \(|u|\lt |f|\). Compare the checkpoint case \(u=-18,\,f=-12\) (\(|u|\gt |f|\)), which gave a real image — same mirror, opposite image type, decided by which side of \(F\) the object sits. ✓
Setup: mirror formula \(\dfrac1v+\dfrac1u=\dfrac1f\), signs assigned first.
Solve: \[ \frac1v = \frac1f - \frac1u = \frac{1}{-15} - \frac{1}{-10} = -\frac{2}{30}+\frac{3}{30} = \frac{1}{30} \;\Rightarrow\; v = +30\,\text{cm}. \] \[ m = -\frac{v}{u} = -\frac{30}{-10} = +3 . \] Answer: \(\boxed{v=+30\,\text{cm},\ m=+3}\) — a virtual (\(v\gt 0\), behind the mirror), erect (\(m\gt 0\)), enlarged (\(3\times\)) image.
Check: an object inside the focus of a concave mirror is exactly the make-up / shaving-mirror configuration; the always-real intuition fails here precisely because \(|u|\lt |f|\). Compare the checkpoint case \(u=-18,\,f=-12\) (\(|u|\gt |f|\)), which gave a real image — same mirror, opposite image type, decided by which side of \(F\) the object sits. ✓
Example 5 · Object at the centre of curvature
An object sits at the centre of curvature of a concave mirror whose radius of
curvature is \(40\,\text{cm}\). Where is the image, and what is its magnification?
Solution.
Given: concave, \(R=40\,\text{cm}\Rightarrow f=-20\,\text{cm}\) (and
\(R=-40\,\text{cm}\) in signed form); object at \(C\), so \(u=-40\,\text{cm}\).
Find: \(v\), \(m\).
Setup: mirror formula, with the object placed exactly at \(C=2f\).
Solve: \[ \frac1v = \frac{1}{-20} - \frac{1}{-40} = -\frac{2}{40}+\frac{1}{40} = -\frac{1}{40} \;\Rightarrow\; v = -40\,\text{cm},\qquad m = -\frac{v}{u} = -\frac{-40}{-40} = -1 . \] Answer: \(\boxed{v=-40\,\text{cm},\ m=-1}\) — the image forms back at \(C\), real, inverted, and the same size as the object.
Check: object and image coincide at \(C\) only when \(u=v=R\); this is the one placement giving unit magnification for a mirror, and it is the basis of the \(u\)–\(v\) method for measuring \(R\) in the laboratory. ✓
Setup: mirror formula, with the object placed exactly at \(C=2f\).
Solve: \[ \frac1v = \frac{1}{-20} - \frac{1}{-40} = -\frac{2}{40}+\frac{1}{40} = -\frac{1}{40} \;\Rightarrow\; v = -40\,\text{cm},\qquad m = -\frac{v}{u} = -\frac{-40}{-40} = -1 . \] Answer: \(\boxed{v=-40\,\text{cm},\ m=-1}\) — the image forms back at \(C\), real, inverted, and the same size as the object.
Check: object and image coincide at \(C\) only when \(u=v=R\); this is the one placement giving unit magnification for a mirror, and it is the basis of the \(u\)–\(v\) method for measuring \(R\) in the laboratory. ✓
Example 6 · Image height of an extended object
A \(5.0\,\text{cm}\) tall candle stands \(30\,\text{cm}\) from a concave mirror of
focal length \(10\,\text{cm}\). How tall is the image, and is it erect or inverted?
Solution.
Given: concave \(\Rightarrow f=-10\,\text{cm}\); \(u=-30\,\text{cm}\);
object height \(h=+5.0\,\text{cm}\).
Find: image height \(h'\) and orientation.
Setup: find \(v\), then \(m=-v/u\), then \(h'=m\,h\).
Solve: \[ \frac1v = \frac{1}{-10} - \frac{1}{-30} = -\frac{3}{30}+\frac{1}{30} = -\frac{1}{15} \;\Rightarrow\; v=-15\,\text{cm},\qquad m=-\frac{v}{u}=-\frac{-15}{-30}=-0.5 . \] \[ h' = m\,h = (-0.5)(5.0) = -2.5\,\text{cm}. \] Answer: \(\boxed{h'=-2.5\,\text{cm}}\) — the image is \(2.5\,\text{cm}\) tall and inverted (\(h'\lt 0\)).
Check: object beyond \(C\) (\(30\gt 20\,\text{cm}\)) must give a real, inverted, diminished image between \(F\) and \(C\) — and indeed \(|v|=15\,\text{cm}\) lies between \(10\) and \(20\,\text{cm}\), with \(|m|\lt 1\). Every qualitative expectation is met. ✓
Setup: find \(v\), then \(m=-v/u\), then \(h'=m\,h\).
Solve: \[ \frac1v = \frac{1}{-10} - \frac{1}{-30} = -\frac{3}{30}+\frac{1}{30} = -\frac{1}{15} \;\Rightarrow\; v=-15\,\text{cm},\qquad m=-\frac{v}{u}=-\frac{-15}{-30}=-0.5 . \] \[ h' = m\,h = (-0.5)(5.0) = -2.5\,\text{cm}. \] Answer: \(\boxed{h'=-2.5\,\text{cm}}\) — the image is \(2.5\,\text{cm}\) tall and inverted (\(h'\lt 0\)).
Check: object beyond \(C\) (\(30\gt 20\,\text{cm}\)) must give a real, inverted, diminished image between \(F\) and \(C\) — and indeed \(|v|=15\,\text{cm}\) lies between \(10\) and \(20\,\text{cm}\), with \(|m|\lt 1\). Every qualitative expectation is met. ✓
✓ Checkpoint · Plane and Spherical Mirrors
- What is the shortest plane mirror, fixed on a wall, in which a person of height \(H\) can see their full length? Does the answer depend on how far they stand?Height \(H/2\), mounted with its top at eye level. The geometry (equal angles, image as far behind as object in front) makes the required mirror exactly half the person's height, independent of distance.
- An object is placed \(18\,\text{cm}\) from a concave mirror of focal length \(12\,\text{cm}\). Find the image distance and magnification.\(f=-12\), \(u=-18\): \(\frac1v=-\frac{1}{12}+\frac{1}{18}=-\frac{1}{36}\Rightarrow v=-36\,\text{cm}\), \(m=-v/u=-2\) (real, inverted, \(2\times\)).
- (Trap) “A concave mirror always forms a real image.” Correct the statement.False. For an object inside the focus (\(|u|\lt |f|\)) a concave mirror forms a virtual, erect, enlarged image—the shaving-mirror case. It forms a real image only for objects beyond \(F\).
Refraction and Snell's Law
◈ Snell's law, by hand
live benchDrag the incident ray and watch the refracted ray bend as n₁ sinθ₁ = n₂ sinθ₂ stays balanced.
Light slows down in matter. The refractive index of a medium is the ratio of its speed to the speed in vacuum, \[ n = \frac{c}{v_{\text{medium}}} \ge 1, \] so \(n=1.00\) for vacuum, \(1.33\) for water, \(\approx 1.5\) for glass, \(2.42\) for diamond. Because the wave is slower inside, a ray crossing a boundary obliquely bends—towards the normal on entering a denser medium, away from it on leaving. The bending obeys Snell's law: \[ \boxed{\; n_1 \sin\theta_1 = n_2 \sin\theta_2 \;} \] where \(\theta_1\), \(\theta_2\) are measured from the normal. Snell's law is here an experimental fact (we will derive it from Fermat's least-time principle in the problem bank). Like reflection, refraction is reversible: reverse a ray's direction and it retraces its path exactly.
Core intuition: Why does bending mean refraction at all?
Picture a wide marching band crossing from pavement onto mud at an angle. The
rank that hits the mud first slows first, so the whole line pivots—the band's
heading changes. A wavefront does the same: the part that enters the slower
medium first lags, swinging the front around. Snell's law is the geometry of that
swing. The ray bends towards the normal entering a slower (denser) medium
because that side of the front is held back.
Apparent depth
Birth of the formula
A point object \(O\) sits at real depth \(d_{\text{real}}\) in a medium of index
\(n_1\), viewed from above through a medium of index \(n_2\) (looking nearly straight
down, paraxial). A ray leaving \(O\) at a small angle \(\theta_1\) to the normal
refracts to \(\theta_2\) at the surface; the emergent ray, traced back, meets the
vertical at the apparent position \(I\) at depth \(d_{\text{app}}\).
For small angles Snell's law reads \(n_1\theta_1 = n_2\theta_2\), and the two
right triangles sharing the surface point give
\(\tan\theta_1 \approx \theta_1 = x/d_{\text{real}}\) and
\(\theta_2 = x/d_{\text{app}}\) for the same horizontal offset \(x\). Hence
\[
n_1\frac{x}{d_{\text{real}}} = n_2\frac{x}{d_{\text{app}}}
\;\Longrightarrow\;
\boxed{\; d_{\text{app}} = \frac{n_2}{n_1}\, d_{\text{real}} \;}.
\]
For the everyday case—object in water (\(n_1=4/3\)) seen from air
(\(n_2=1\))—this gives \(d_{\text{app}} = \tfrac{3}{4}\,d_{\text{real}}\): the pool
looks only three-quarters as deep as it is. The same algebra, run the other way,
makes a coin under a glass slab appear raised by
\(t\!\left(1-\tfrac{1}{n}\right)\).
◇ Figure 3
Fig 3.Left: apparent depth—a ray from O bends away from the normal on
leaving the water, so the eye traces it back to a shallower image I.
Right: a parallel-sided slab shifts a ray sideways by \(d\) but leaves its
direction unchanged.
Misconception alert: Which index goes “on top”?
Apparent depth shrinks by the ratio (index you look into) over (index the
object sits in). For the usual “object in water, eye in air” picture the
factor is \(n_{\text{air}}/n_{\text{water}} = 3/4 \lt 1\), so the object rises. Swap
the media (a fish looking up at a bird) and the factor inverts, so the bird
appears higher than it is. Always write the ratio as
\(n_{\text{observer side}}/n_{\text{object side}}\) and the direction takes care of
itself.
Worked Examples
Example 7 · Speed, wavelength, and colour in glass
Yellow light of vacuum wavelength \(600\,\text{nm}\) enters glass of refractive
index \(1.5\). Find its speed and wavelength inside the glass, and confirm that its
colour (frequency) is unchanged.
Solution.
Given: \(n=1.5\), vacuum wavelength \(\lambda_0=600\,\text{nm}\),
\(c=3.0\times10^{8}\,\text{m\,s}^{-1}\).
Find: \(v_{\text{glass}}\), \(\lambda_{\text{glass}}\), frequency \(f\).
Setup: by definition \(n=c/v\), so \(v=c/n\). Frequency is fixed at a boundary (the wave cannot pile up), so \(\lambda=v/f=\lambda_0/n\).
Solve: \[ v = \frac{c}{n} = \frac{3.0\times10^{8}}{1.5} = 2.0\times10^{8}\,\text{m\,s}^{-1}, \qquad \lambda_{\text{glass}} = \frac{\lambda_0}{n} = \frac{600}{1.5} = 400\,\text{nm}. \] \[ f = \frac{c}{\lambda_0} = \frac{3.0\times10^{8}}{600\times10^{-9}} = 5.0\times10^{14}\,\text{Hz} = \frac{v}{\lambda_{\text{glass}}} = \frac{2.0\times10^{8}}{400\times10^{-9}} . \] Answer: \(\boxed{v=2.0\times10^{8}\,\text{m\,s}^{-1},\ \lambda=400\,\text{nm},\ f=5.0\times10^{14}\,\text{Hz}}\).
Check: computing \(f\) two ways — from \(c/\lambda_0\) and from \(v/\lambda_{\text{glass}}\) — gives the same \(5.0\times10^{14}\,\text{Hz}\), which is why the light stays yellow: it is the frequency, not the wavelength, that the eye reads as colour. ✓
Setup: by definition \(n=c/v\), so \(v=c/n\). Frequency is fixed at a boundary (the wave cannot pile up), so \(\lambda=v/f=\lambda_0/n\).
Solve: \[ v = \frac{c}{n} = \frac{3.0\times10^{8}}{1.5} = 2.0\times10^{8}\,\text{m\,s}^{-1}, \qquad \lambda_{\text{glass}} = \frac{\lambda_0}{n} = \frac{600}{1.5} = 400\,\text{nm}. \] \[ f = \frac{c}{\lambda_0} = \frac{3.0\times10^{8}}{600\times10^{-9}} = 5.0\times10^{14}\,\text{Hz} = \frac{v}{\lambda_{\text{glass}}} = \frac{2.0\times10^{8}}{400\times10^{-9}} . \] Answer: \(\boxed{v=2.0\times10^{8}\,\text{m\,s}^{-1},\ \lambda=400\,\text{nm},\ f=5.0\times10^{14}\,\text{Hz}}\).
Check: computing \(f\) two ways — from \(c/\lambda_0\) and from \(v/\lambda_{\text{glass}}\) — gives the same \(5.0\times10^{14}\,\text{Hz}\), which is why the light stays yellow: it is the frequency, not the wavelength, that the eye reads as colour. ✓
Example 8 · A coin raised by a glass slab
A coin rests on a table and is viewed from directly above through a
\(6.0\,\text{cm}\) thick glass slab (\(n=1.5\)) laid over it. By how much does the coin
appear raised, and at what apparent depth below the top surface does it seem to
lie?
Solution.
Given: slab thickness \(t=6.0\,\text{cm}\), \(n=1.5\), viewed from air.
Find: apparent raising and apparent depth.
Setup: looking from the rarer side into the denser slab, the apparent depth is \(t/n\); the coin is therefore raised by \(t\!\left(1-\tfrac1n\right)\).
Solve: \[ d_{\text{app}} = \frac{t}{n} = \frac{6.0}{1.5} = 4.0\,\text{cm}, \qquad \text{raised by } t\!\left(1-\frac1n\right) = 6.0\!\left(1-\frac{1}{1.5}\right) = 6.0\times\frac13 = 2.0\,\text{cm}. \] Answer: \(\boxed{\text{raised } 2.0\,\text{cm};\ \text{apparent depth } 4.0\,\text{cm}}\).
Check: \(d_{\text{app}}+\text{raising}=4.0+2.0=6.0\,\text{cm}=t\), as it must, since the two together span the real thickness. The raising is independent of how thick the air gap above the slab is — only the glass shifts the image. ✓
Setup: looking from the rarer side into the denser slab, the apparent depth is \(t/n\); the coin is therefore raised by \(t\!\left(1-\tfrac1n\right)\).
Solve: \[ d_{\text{app}} = \frac{t}{n} = \frac{6.0}{1.5} = 4.0\,\text{cm}, \qquad \text{raised by } t\!\left(1-\frac1n\right) = 6.0\!\left(1-\frac{1}{1.5}\right) = 6.0\times\frac13 = 2.0\,\text{cm}. \] Answer: \(\boxed{\text{raised } 2.0\,\text{cm};\ \text{apparent depth } 4.0\,\text{cm}}\).
Check: \(d_{\text{app}}+\text{raising}=4.0+2.0=6.0\,\text{cm}=t\), as it must, since the two together span the real thickness. The raising is independent of how thick the air gap above the slab is — only the glass shifts the image. ✓
Example 9 · A fish looks up at a bird
A bird hovers \(3.0\,\text{m}\) directly above the still surface of a pond
(\(n=4/3\)). To a fish looking straight up, at what height does the bird appear?
Solution.
Given: real height \(d_{\text{real}}=3.0\,\text{m}\) in air; observer (fish)
in water, \(n_{\text{water}}=4/3\).
Find: apparent height of the bird.
Setup: the apparent-depth ratio is \(d_{\text{app}}=\dfrac{n_{\text{observer side}}}{n_{\text{object side}}}\, d_{\text{real}}\). Here the observer is in water, the object in air, so the factor is \(n_{\text{water}}/n_{\text{air}}=4/3\gt 1\): the image moves farther, not closer.
Solve: \[ d_{\text{app}} = \frac{4/3}{1}\times 3.0 = 4.0\,\text{m}. \] Answer: \(\boxed{4.0\,\text{m}}\) — the bird looks higher than it is.
Check: this is the mirror-image of the “pool looks shallow” case: swap which medium holds the observer and the factor inverts from \(3/4\) to \(4/3\). Writing the ratio as (observer-side index)/(object-side index) makes the direction automatic. ✓
Setup: the apparent-depth ratio is \(d_{\text{app}}=\dfrac{n_{\text{observer side}}}{n_{\text{object side}}}\, d_{\text{real}}\). Here the observer is in water, the object in air, so the factor is \(n_{\text{water}}/n_{\text{air}}=4/3\gt 1\): the image moves farther, not closer.
Solve: \[ d_{\text{app}} = \frac{4/3}{1}\times 3.0 = 4.0\,\text{m}. \] Answer: \(\boxed{4.0\,\text{m}}\) — the bird looks higher than it is.
Check: this is the mirror-image of the “pool looks shallow” case: swap which medium holds the observer and the factor inverts from \(3/4\) to \(4/3\). Writing the ratio as (observer-side index)/(object-side index) makes the direction automatic. ✓
✓ Checkpoint · Refraction, Snell's Law and Apparent Depth
- In one sentence, explain why a straw standing in water looks broken at the surface.Light from the submerged part bends away from the normal on leaving the water, so the eye traces it back to a shallower, displaced apparent position; the underwater part appears raised and shifted, breaking the line at the surface.
- A coin lies \(15\,\text{cm}\) deep in water (\(n=4/3\)). At what depth does it appear when viewed from directly above?\(d_{\text{app}}=(1)/(4/3)\times15=11.25\,\text{cm}\).
- Light goes from air into glass at incidence \(i\), refracting to \(r\). If the ray is instead sent from glass into air at incidence \(r\), what is the angle of emergence, and what principle guarantees it?It emerges at \(i\). The principle of reversibility: reversing a ray's direction makes it retrace its exact path, so \(n_1\sin i=n_2\sin r\) runs both ways.
Solved examples
Five fully-worked problems from this chapter, free — solution and answer shown in full. The complete set of worked examples is in the full book.
Example 1 · Images between inclined mirrors
Two plane mirrors are set face to face at \(60^\circ\). How many images of a small
object placed between them are seen?
Solution.
Given: mirror angle \(\theta = 60^\circ\).
Find: number of images \(N\).
Setup: repeated reflection arranges images on a circle; when \(360^\circ/\theta\) is an even integer, \(N = 360^\circ/\theta - 1\).
Here \(360^\circ/60^\circ = 6\) (even), so \[ N = 6 - 1 = 5 . \] Five images are seen, evenly spaced around the object. Check. \(360^\circ/60^\circ=6\) is even, so the count carries no on-/off-bisector ambiguity—\(N=5\) for any placement of the object. ✓
Setup: repeated reflection arranges images on a circle; when \(360^\circ/\theta\) is an even integer, \(N = 360^\circ/\theta - 1\).
Here \(360^\circ/60^\circ = 6\) (even), so \[ N = 6 - 1 = 5 . \] Five images are seen, evenly spaced around the object. Check. \(360^\circ/60^\circ=6\) is even, so the count carries no on-/off-bisector ambiguity—\(N=5\) for any placement of the object. ✓
Example 2 · Concave mirror, object beyond the centre
An object is placed \(40\,\text{cm}\) in front of a concave mirror of focal length
\(15\,\text{cm}\). Locate the image and find its magnification.
Solution.
Given: concave mirror \(\Rightarrow f = -15\,\text{cm}\); real object
\(\Rightarrow u = -40\,\text{cm}\).
Find: \(v\), \(m\).
Setup: mirror formula \(\dfrac1v + \dfrac1u = \dfrac1f\), signs assigned.
\[ \frac1v = \frac1f - \frac1u = \frac{1}{-15} - \frac{1}{-40} = -\frac{1}{24}\;\text{cm}^{-1} \;\Rightarrow\; v = -24\,\text{cm}. \] \[ m = -\frac{v}{u} = -\frac{-24}{-40} = -0.6 . \] The image is real (\(v\lt 0\), same side), inverted (\(m\lt 0\)), and diminished (\(|m|\lt 1\)), \(24\,\text{cm}\) in front of the mirror—between \(F\) and \(C\), exactly as expected for an object beyond \(C\). Check. \(\dfrac1v+\dfrac1u=-\dfrac1{24}-\dfrac1{40}=-\dfrac1{15}=\dfrac1f\) ✓; and \(|v|=24\,\text{cm}\) lies between \(F=15\) and \(C=30\,\text{cm}\), the band for a real, diminished image. ✓
Setup: mirror formula \(\dfrac1v + \dfrac1u = \dfrac1f\), signs assigned.
\[ \frac1v = \frac1f - \frac1u = \frac{1}{-15} - \frac{1}{-40} = -\frac{1}{24}\;\text{cm}^{-1} \;\Rightarrow\; v = -24\,\text{cm}. \] \[ m = -\frac{v}{u} = -\frac{-24}{-40} = -0.6 . \] The image is real (\(v\lt 0\), same side), inverted (\(m\lt 0\)), and diminished (\(|m|\lt 1\)), \(24\,\text{cm}\) in front of the mirror—between \(F\) and \(C\), exactly as expected for an object beyond \(C\). Check. \(\dfrac1v+\dfrac1u=-\dfrac1{24}-\dfrac1{40}=-\dfrac1{15}=\dfrac1f\) ✓; and \(|v|=24\,\text{cm}\) lies between \(F=15\) and \(C=30\,\text{cm}\), the band for a real, diminished image. ✓
Example 3 · Convex rear-view mirror
A car's convex side mirror has focal length \(15\,\text{cm}\). A motorcycle is
\(20\,\text{cm}\) from it. Where is the image, and how large?
Solution.
Given: convex mirror \(\Rightarrow f = +15\,\text{cm}\);
\(u = -20\,\text{cm}\).
Find: \(v\), \(m\).
Setup: mirror formula with \(f\gt 0\) for a convex mirror.
\[ \frac1v = \frac1f - \frac1u = \frac{1}{15} - \frac{1}{-20} = \frac{7}{60} \;\Rightarrow\; v = \frac{60}{7} \approx +8.6\,\text{cm}. \] \[ m = -\frac{v}{u} = -\frac{60/7}{-20} = +\frac{3}{7} \approx +0.43 . \] The image is virtual (\(v\gt 0\), behind the mirror), erect (\(m\gt 0\)), and diminished. A convex mirror always does this—hence the wide, upright field of view that warns “objects are closer than they appear.” Check. \(\dfrac1v+\dfrac1u=\dfrac{7}{60}-\dfrac{1}{20}=\dfrac{1}{15}=\dfrac1f\) ✓; \(0\lt v\lt f\) and \(0\lt m\lt 1\)—the always-virtual, always-shrunk convex-mirror signature. ✓
Setup: mirror formula with \(f\gt 0\) for a convex mirror.
\[ \frac1v = \frac1f - \frac1u = \frac{1}{15} - \frac{1}{-20} = \frac{7}{60} \;\Rightarrow\; v = \frac{60}{7} \approx +8.6\,\text{cm}. \] \[ m = -\frac{v}{u} = -\frac{60/7}{-20} = +\frac{3}{7} \approx +0.43 . \] The image is virtual (\(v\gt 0\), behind the mirror), erect (\(m\gt 0\)), and diminished. A convex mirror always does this—hence the wide, upright field of view that warns “objects are closer than they appear.” Check. \(\dfrac1v+\dfrac1u=\dfrac{7}{60}-\dfrac{1}{20}=\dfrac{1}{15}=\dfrac1f\) ✓; \(0\lt v\lt f\) and \(0\lt m\lt 1\)—the always-virtual, always-shrunk convex-mirror signature. ✓
Example 4 · Refraction into glass
A ray of light in air strikes a flat glass surface (\(n = 1.5\)) at \(60^\circ\) to
the normal. Find the angle of refraction.
Solution.
Given: \(n_1 = 1.00\), \(\theta_1 = 60^\circ\), \(n_2 = 1.5\).
Find: \(\theta_2\).
Setup: Snell's law \(n_1\sin\theta_1 = n_2\sin\theta_2\).
\[ \sin\theta_2 = \frac{n_1\sin\theta_1}{n_2} = \frac{1.00 \times \sin 60^\circ}{1.5} = \frac{0.866}{1.5} = 0.577 \;\Rightarrow\; \theta_2 = 35.3^\circ . \] The ray bends towards the normal on entering the denser medium, as it must. Check. \(n_2\sin\theta_2=1.5\times\sin35.3^\circ=1.5\times0.578=0.866=n_1\sin\theta_1\) ✓; and \(\theta_2\lt \theta_1\), exactly the bend-toward-normal into the denser glass. ✓
Setup: Snell's law \(n_1\sin\theta_1 = n_2\sin\theta_2\).
\[ \sin\theta_2 = \frac{n_1\sin\theta_1}{n_2} = \frac{1.00 \times \sin 60^\circ}{1.5} = \frac{0.866}{1.5} = 0.577 \;\Rightarrow\; \theta_2 = 35.3^\circ . \] The ray bends towards the normal on entering the denser medium, as it must. Check. \(n_2\sin\theta_2=1.5\times\sin35.3^\circ=1.5\times0.578=0.866=n_1\sin\theta_1\) ✓; and \(\theta_2\lt \theta_1\), exactly the bend-toward-normal into the denser glass. ✓
Example 5 · Apparent depth of a coin
A coin lies at the bottom of a tank holding \(12\,\text{cm}\) of water
(\(n = 4/3\)). How deep does it appear to an observer looking straight down, and by
how much does it seem raised?
Solution.
Given: \(d_{\text{real}} = 12\,\text{cm}\), water \(n_1 = 4/3\), air
\(n_2 = 1\).
Find: \(d_{\text{app}}\) and the shift.
Setup: apparent depth \(d_{\text{app}} = (n_2/n_1)\,d_{\text{real}}\).
\[ d_{\text{app}} = \frac{1}{4/3}\times 12 = \frac{3}{4}\times 12 = 9\,\text{cm}, \qquad \text{shift} = 12 - 9 = 3\,\text{cm}. \] The coin appears \(9\,\text{cm}\) down and seems raised by \(3\,\text{cm}\). Check. \(d_{\text{app}}+\text{shift}=9+3=12\,\text{cm}=d_{\text{real}}\) ✓; the factor \(n_2/n_1=3/4\lt 1\) guarantees the coin rises, never sinks. ✓
Setup: apparent depth \(d_{\text{app}} = (n_2/n_1)\,d_{\text{real}}\).
\[ d_{\text{app}} = \frac{1}{4/3}\times 12 = \frac{3}{4}\times 12 = 9\,\text{cm}, \qquad \text{shift} = 12 - 9 = 3\,\text{cm}. \] The coin appears \(9\,\text{cm}\) down and seems raised by \(3\,\text{cm}\). Check. \(d_{\text{app}}+\text{shift}=9+3=12\,\text{cm}=d_{\text{real}}\) ✓; the factor \(n_2/n_1=3/4\lt 1\) guarantees the coin rises, never sinks. ✓
Problem bank
Five questions from this chapter’s 50-question bank, free — attempt each one before you reveal the answer. The rest of the bank, and the timed test that draws on all of it, are in the full book.
✎ Free practice questions
- Plane Mirror Geometry
You stand \(2.0\,\text{m}\) in front of a plane mirror. How far is your image from you, and if you walk toward the mirror at \(1.5\,\text{m s}^{-1}\), how fast does your image approach you?Image is \(2.0\,\text{m}\) behind the mirror, so \(4.0\,\text{m}\) from you. As you close in at \(1.5\,\text{m s}^{-1}\) the image also approaches the mirror at \(1.5\,\text{m s}^{-1}\), so it approaches you at \(3.0\,\text{m s}^{-1}\). - Concave Mirror, Direct
An object is \(30\,\text{cm}\) in front of a concave mirror of focal length \(10\,\text{cm}\). Find the image distance.\(f=-10\), \(u=-30\): \(\frac1v=\frac1f-\frac1u=-\frac{1}{15}\Rightarrow v=-15\,\text{cm}\) (real, in front). - Image at the Centre of Curvature
An object sits \(20\,\text{cm}\) from a concave mirror of focal length \(10\,\text{cm}\). Find the image distance and magnification, and describe the image.\(f=-10\), \(u=-20\) (object at \(C\)): \(v=-20\,\text{cm}\), \(m=-v/u=-1\). Real, inverted, same size, at \(C\). - Concave Mirror from Magnification
A concave mirror of focal length \(15\,\text{cm}\) forms a real image three times the size of the object. Find the object and image distances.Real image \(\Rightarrow m=-3=-v/u\Rightarrow v=3u\). \(\frac{1}{3u}+\frac1u=\frac1f\Rightarrow u=-20\,\text{cm}\), \(v=-60\,\text{cm}\). - Half a Lens
A convex lens forms a sharp image of a candle on a screen. The upper half of the lens is then covered with opaque card. Describe what happens to the image, and contrast this with cutting the lens in half along its principal axis.Covering half the aperture leaves the whole image (every object point still sends rays through the open half) but at reduced brightness. Cutting the lens along the axis and separating the halves can give two displaced images; each half alone still images the full object, since focal length is set by curvature, not by which part of the aperture is used.
Chapter test
A paper drawn at random from this chapter's bank. Choose the exam you are training for — the marking scheme, pace and difficulty mix follow the real pattern. Work on paper; when you finish (or the clock runs out), the answers are revealed and you mark yourself honestly.
Pattern
Questions
Clock
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- Total Internal Reflection
- Refraction at a Spherical Surface
- Thin Lenses
- The Prism: Deviation and Dispersion
- Optical Instruments
- Common Pitfalls and Exam Strategy
- Summary and Resolution of Perplexing Questions
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