Chapter 1 · Light – Reflection and Refraction

1.1 Reflection of Light and Spherical Mirrors

Light bounces off a mirror so that the angle of incidence equals the angle of reflection. Curve the mirror into part of a sphere and it can gather a beam to a focus, with $R=2f$.

Look up from this page. Everything you can see — the wall, your hand, the screen — reaches your eyes because light leaves it and travels to you. Most objects do not make their own light; they send back part of the light that falls on them. When the surface is smooth and shiny, like a mirror, a still pond or a polished steel spoon, the light comes back in a perfectly orderly way, and we see a clear image. That orderly bouncing back is called reflection.

Reflection obeys two very simple laws, and those two laws are enough to explain the bathroom mirror, the periscope in a submarine, the shiny bowl behind a torch bulb and the curved mirror at a hill-road bend. In this section we state the laws, use them to find the image in a plane mirror, and then curve the mirror into part of a sphere. You will see why a curved mirror can squeeze a wide beam of sunlight into one hot spot — the principal focus — and why its distance from the mirror is exactly half the radius of the sphere.

Light Rays and the Laws of Reflection

Light travels in straight lines in a uniform medium — that is why a solid object casts a sharp shadow and why you cannot see round a corner. We draw the path of light as a straight line with an arrow, called a ray; a bundle of rays is a beam. Where a ray strikes a mirror we draw the normal: the line at $90^\circ$ to the mirror surface at that point. Angles in optics are always measured from the normal, never from the mirror surface.

📘 Definition 1.1 — Incident ray, reflected ray and their angles
The ray that arrives at the mirror is the incident ray; the ray that leaves it is the reflected ray. The point where they meet the mirror is the point of incidence. The angle of incidence $\angle i$ is the angle between the incident ray and the normal; the angle of reflection $\angle r$ is the angle between the reflected ray and the normal.
📐 Law 1.1 — The laws of reflection

For light reflected from any smooth surface — plane or curved:

  1. The angle of incidence is equal to the angle of reflection: $\angle i=\angle r$.
  2. The incident ray, the normal at the point of incidence and the reflected ray all lie in the same plane.

For a curved mirror the laws still hold at every single point; you simply use the normal at that point.

✏️ Example 1 — Measuring from the right line
A ray of light strikes a plane mirror making an angle of $30^\circ$ with the mirror surface. Find the angle of incidence, the angle of reflection, and the angle between the incident and reflected rays.
Idea Angles are measured from the normal, which is at $90^\circ$ to the surface.
Angle of incidence $\angle i=90^\circ-30^\circ=60^\circ$.
Law 1.1 $\angle r=\angle i=60^\circ$.
Between the rays The rays lie on opposite sides of the normal, so the angle between them is $\angle i+\angle r=120^\circ$. Check: the reflected ray also makes $30^\circ$ with the surface, and $30^\circ+120^\circ+30^\circ=180^\circ$ along the straight mirror. ✓
✏️ Example 2 — Two mirrors at right angles
Two plane mirrors $M_1$ and $M_2$ are joined at $90^\circ$. A ray strikes $M_1$ with an angle of incidence of $35^\circ$, reflects, and then strikes $M_2$. Find the angle of incidence at $M_2$ and describe the final ray.
At $M_1$ $\angle r_1=35^\circ$, so the reflected ray makes $90^\circ-35^\circ=55^\circ$ with the surface of $M_1$.
Triangle The ray between the mirrors, $M_1$ and $M_2$ form a triangle with angles $90^\circ$ (at the joint), $55^\circ$ and the angle $x$ the ray makes with $M_2$: $x=180^\circ-90^\circ-55^\circ=35^\circ$.
At $M_2$ $\angle i_2=90^\circ-35^\circ=55^\circ$, so $\angle r_2=55^\circ$.
Result Notice $\angle i_1+\angle i_2=90^\circ$. The final ray travels exactly back parallel to the original ray (turned through $180^\circ$), whatever the first angle was. This is how the "cat's-eye" road studs and the reflectors left on the Moon by astronauts send light straight back to where it came from.
⚠️ Common mistakes

1. Measuring from the mirror. If a question gives the angle with the surface, subtract it from $90^\circ$ first.

2. Forgetting the normal on a curved mirror. On a curved mirror the normal changes from point to point. For a spherical mirror it is always the radius drawn from the centre of the sphere — see below.

3. Adding the angles wrongly. The angle between incident and reflected rays is $2\angle i$, not $\angle i$.

Interactive 1 — The Plane-Mirror Lab

A point source of light $S$ sits in front of a plane mirror. Drag the source (orange) or the point of incidence on the mirror (blue), or use the slider. The dashed normal always splits the angle between the rays equally. Switch to a fan of rays to see why you see an image: every reflected ray, traced backwards (dotted), passes through one point $S'$ behind the mirror.

35°
angle of incidence $i$—
angle of reflection $r$—
object distance (in front)—
image distance (behind)—

Images in a Plane Mirror

Interactive 1 shows what happens when you look into a mirror. Rays from a point on your nose spread out, reflect, and enter your eye. Your brain assumes light travels in straight lines, so it traces the rays backwards and "sees" the point where they appear to come from — a point behind the glass. No light actually goes there; the image is virtual. It cannot be caught on a screen.

📐 Result 1.2 — The image in a plane mirror
The image formed by a plane mirror is
  • virtual (it cannot be obtained on a screen) and erect (the right way up);
  • the same size as the object;
  • as far behind the mirror as the object is in front of it;
  • laterally inverted: left and right are swapped — raise your right hand and your image raises its left.
✏️ Example 3 — Walking towards a mirror
Riya stands $2.0\ \text{m}$ in front of a large plane mirror. (a) How far is she from her image? (b) She walks $0.50\ \text{m}$ towards the mirror. How far is she from her image now? (c) If she walks towards the mirror at $0.75\ \text{m/s}$, how fast does the gap between her and her image close?
(a) The image is $2.0\ \text{m}$ behind the mirror, so the gap is $2.0+2.0=4.0\ \text{m}$.
(b) She is now $1.5\ \text{m}$ from the mirror and so is her image: the gap is $3.0\ \text{m}$. Walking $0.50\ \text{m}$ closed the gap by $1.0\ \text{m}$ — twice as much, because the image also came $0.50\ \text{m}$ closer.
(c) Each second she moves $0.75\ \text{m}$ closer to the mirror and the image does the same, so the gap closes at $2\times0.75=1.5\ \text{m/s}$.
✏️ Example 4 — How long must a mirror be?
Arjun is $1.60\ \text{m}$ tall and his eyes are $10\ \text{cm}$ below the top of his head. What is the shortest vertical plane mirror in which he can see his whole body, and where must its edges be?
Top edge Light from the top of his head must reflect into his eyes. By $\angle i=\angle r$ the reflection point is level with the point halfway between head-top and eyes: $5\ \text{cm}$ below the top of his head, at height $1.60-0.05=1.55\ \text{m}$.
Bottom edge Light from his feet reflects into his eyes from a point halfway between feet and eyes. His eyes are at $1.50\ \text{m}$, so this point is at $0.75\ \text{m}$.
Length $1.55-0.75=0.80\ \text{m}$ — exactly half his height. The distance from the mirror never entered the working, so stepping back does not help!

Spherical Mirrors and Their Vocabulary

Cut a piece from a hollow, shiny sphere and you get a spherical mirror. If the reflecting surface is the inside of the sphere (it curves in, like a cave), the mirror is concave. If the reflecting surface is the outside (it bulges out towards you), it is convex. The inside of a steel tablespoon is a rough concave mirror; the back of the spoon is a convex one — look at your face in each.

PFC R Concave mirrorparallel rays meet at a real focus F PFC Convex mirrorrays spread out as if from a virtual focus F
Pole $P$, principal focus $F$ and centre of curvature $C$ on the principal axis. The radius $CP=R$; the focal length $PF=f$.
📘 Definition 1.2 — Parts of a spherical mirror
  • Pole $P$: the centre of the reflecting surface.
  • Centre of curvature $C$: the centre of the sphere of which the mirror is a part. It lies in front of a concave mirror and behind a convex mirror.
  • Radius of curvature $R$: the radius of that sphere, i.e. the distance $PC$.
  • Principal axis: the straight line through $P$ and $C$.
  • Aperture: the width (diameter) of the reflecting surface.
  • Principal focus $F$: the point on the principal axis where rays parallel to the axis meet after reflection (concave mirror), or from which they appear to spread out after reflection (convex mirror). The focus of a concave mirror is real; that of a convex mirror is virtual.
  • Focal length $f$: the distance $PF$.

Because $C$ is the centre of the sphere, the line from $C$ to any point on the mirror is a radius, and a radius is perpendicular to the sphere's surface. So the normal at any point of a spherical mirror is the line joining that point to $C$. With that one fact and Law 1.1 we can work out where every ray goes.

📐 Result 1.3 — Radius of curvature and focal length
For a spherical mirror of small aperture (rays close to the principal axis), the principal focus lies midway between the pole and the centre of curvature: $$R=2f\qquad\text{or}\qquad f=\frac R2 .$$
✏️ Example 5 — Finding $f$ and $R$
(a) A concave mirror focuses sunlight to a bright spot $18\ \text{cm}$ in front of it. Find its focal length and radius of curvature. (b) A convex mirror is cut from a sphere of radius $50\ \text{cm}$. Where is its focus?
(a) Idea The Sun is so far away that its rays arrive parallel; they meet at the principal focus. So $f=18\ \text{cm}$.
(a) Radius $R=2f=36\ \text{cm}$: the centre of curvature is $36\ \text{cm}$ in front of the mirror.
(b) $f=\dfrac R2=\dfrac{50}{2}=25\ \text{cm}$. For a convex mirror the focus is virtual, $25\ \text{cm}$ behind the mirror.
Interactive 2 — Where Does a Spherical Mirror Focus?

A beam of rays parallel to the principal axis hits a true spherical mirror. Every reflection is computed exactly with $\angle i=\angle r$, using the radius to $C$ as the normal. Make the beam narrow: the rays meet at $F$, halfway between $P$ and $C$. Make it wide: the outer rays cross nearer the mirror and the focus smears — which is why $R=2f$ is a small-aperture rule.

4035%
focal length $f=R/2$—
near-axis rays cross at—
outermost rays cross at—
the reflected beam is—

Four Rays You Can Always Draw

To find an image we need only two rays from the same point of the object; where they meet (or appear to meet) after reflection, the image of that point is formed. Four rays are especially easy, because each follows directly from Law 1.1 and the definitions above.

🧭 The standard rays for spherical mirrors
  1. Parallel to the principal axis → after reflection it passes through $F$ (concave), or appears to come from $F$ behind the mirror (convex).
  2. Through $F$ (concave), or heading towards $F$ behind the mirror (convex) → after reflection it travels parallel to the principal axis. (This is ray 1 run backwards: light paths are reversible.)
  3. Through $C$ (concave), or heading towards $C$ (convex) → it strikes the mirror along the normal ($\angle i=0$) and comes straight back along itself.
  4. At the pole $P$ → it reflects at the same angle on the other side of the principal axis, because the principal axis is the normal at $P$.
✏️ Example 6 — Following individual rays
A concave mirror has $R=30\ \text{cm}$. Describe the reflected ray for: (a) a ray that passes through a point on the principal axis $15\ \text{cm}$ in front of the mirror; (b) a ray through the point $30\ \text{cm}$ in front; (c) a ray striking the pole at $25^\circ$ to the principal axis.
Locate $f=\tfrac{30}{2}=15\ \text{cm}$, so the point $15\ \text{cm}$ in front is $F$ and the point $30\ \text{cm}$ in front is $C$.
(a) A ray through $F$ reflects parallel to the principal axis (standard ray 2).
(b) A ray through $C$ meets the mirror along the normal and returns along its own path (standard ray 3).
(c) At $P$ the principal axis is the normal, so $\angle i=25^\circ$ and the ray leaves at $25^\circ$ on the other side of the axis (standard ray 4).
🚀 Beyond the board syllabus (JEE Main · NEET · JEE Advanced) — optional

In JEE Main, NEET and JEE Advanced 2026; not in the CBSE Class 10 syllabus.

Turning a mirror turns the reflected ray twice as much. Keep the incident ray fixed and rotate a plane mirror through an angle $\theta$. The normal turns through $\theta$, so $\angle i$ changes by $\theta$ and so does $\angle r$; the reflected ray therefore turns through $2\theta$. This is used in instruments that measure tiny rotations: a spot of light reflected from a small mirror on a rotating coil moves twice as far across the scale.

Exercises

I
The laws of reflection.
  • 1 The angle of incidence is $40^\circ$. Find the angle of reflection.
  • 2 A ray makes $25^\circ$ with the surface of a plane mirror. Find $\angle i$ and the angle between the incident and reflected rays.
  • 3 The incident and reflected rays are at $90^\circ$ to each other. Find the angle of incidence.
  • 4 A ray strikes a mirror along the normal. What are $\angle i$ and $\angle r$, and where does the ray go?
  • 5 The reflected ray makes $50^\circ$ with the incident ray. What angle does the reflected ray make with the mirror surface?
  • 6 Describe a simple activity with a sheet of paper and a torch that shows the second law of reflection.
  • 7 Two plane mirrors are at $90^\circ$. A ray strikes the first with $\angle i=50^\circ$. Find the angle of incidence at the second mirror.
  • 8 Do the laws of reflection apply to a spherical mirror? If so, which line is the normal at a point of the mirror?
1 $\angle r=\angle i=40^\circ$.
2 $\angle i=90^\circ-25^\circ=65^\circ$; angle between the rays $=2\times65^\circ=130^\circ$.
3 $2\angle i=90^\circ$, so $\angle i=45^\circ$.
4 $\angle i=\angle r=0^\circ$; the ray retraces its path back along the normal.
5 $2\angle i=50^\circ$ gives $\angle r=25^\circ$ from the normal, so the reflected ray makes $90^\circ-25^\circ=65^\circ$ with the surface.
6 Lay a sheet of paper flat on a table with a plane mirror standing upright on it, and shine a narrow torch beam along the paper at the mirror: the reflected beam also runs along the paper. If you fold the part of the paper carrying the reflected beam downwards, the beam is no longer seen on it — incident ray, normal and reflected ray lie in one plane.
7 As in Example 2, $\angle i_1+\angle i_2=90^\circ$, so $\angle i_2=40^\circ$.
8 Yes, at every point. The normal at a point is the radius joining that point to the centre of curvature $C$.
II
Images in plane mirrors.
  • 1 An object is $25\ \text{cm}$ in front of a plane mirror. Where is its image, and how far is the image from the object?
  • 2 A plane mirror is moved $10\ \text{cm}$ further away from a fixed object. How far does the image move?
  • 3 Why is the word AMBULANCE often painted in mirror writing on the front of an ambulance?
  • 4 List four properties of the image formed by a plane mirror.
  • 5 A wall clock without numbers, seen in a plane mirror, appears to show 3:40. What is the actual time?
  • 6 What is the shortest plane mirror in which a person $1.70\ \text{m}$ tall can see a full-length image?
  • 7 A girl walks straight towards a plane mirror at $1.0\ \text{m/s}$. At what rate does the distance between the girl and her image decrease?
1 $25\ \text{cm}$ behind the mirror; $25+25=50\ \text{cm}$ from the object.
2 If the object is at distance $d$, the image is $2d$ from it. Increasing $d$ by $10\ \text{cm}$ increases $2d$ by $20\ \text{cm}$: the image moves $20\ \text{cm}$ (away from the object).
3 A driver ahead sees the ambulance in the rear-view mirror, which laterally inverts the writing; mirror writing then reads correctly as AMBULANCE.
4 Virtual; erect; same size as the object; as far behind the mirror as the object is in front; laterally inverted (any four).
5 For a clock, actual time + mirror time $=12{:}00$. So the actual time is $12{:}00-3{:}40=8{:}20$. (At 8:20 the hour hand is at $250^\circ$ and the minute hand at $120^\circ$ from 12; reflected about the 12–6 line they sit at $110^\circ$ and $240^\circ$, which is how 3:40 looks.)
6 Half the height: $\tfrac12\times1.70=0.85\ \text{m}$, placed correctly (Example 4).
7 Both she and her image approach the mirror at $1.0\ \text{m/s}$, so the gap closes at $2.0\ \text{m/s}$.
III
Spherical mirrors: words and $R=2f$.
  • 1 A spherical mirror has $R=32\ \text{cm}$. Find $f$.
  • 2 A concave mirror has $f=12.5\ \text{cm}$. Find $R$.
  • 3 A shiny bowl focuses sunlight to a point $20\ \text{cm}$ in front of it. What kind of mirror is it? Find $f$ and $R$.
  • 4 A convex mirror is cut from a shiny ball of diameter $50\ \text{cm}$. Find $R$ and $f$.
  • 5 Why is the focus of a convex mirror called virtual?
  • 6 Name the type of mirror in: (a) a torch reflector, (b) a car's side mirror, (c) a dentist's mouth mirror, (d) a shop's security mirror.
  • 7 Why does a ray through the centre of curvature return along its own path?
  • 8 A concave mirror has $f=15\ \text{cm}$. How far apart are $F$ and $C$?
1 $f=\tfrac{32}{2}=16\ \text{cm}$.
2 $R=2\times12.5=25\ \text{cm}$.
3 It makes parallel rays meet in front of it, so it is concave; $f=20\ \text{cm}$, $R=40\ \text{cm}$.
4 $R=\tfrac{50}{2}=25\ \text{cm}$ (the radius, not the diameter!), $f=12.5\ \text{cm}$ behind the mirror.
5 After reflection the rays spread out; they never actually pass through $F$. They only appear to come from it when traced backwards behind the mirror.
6 (a) concave, (b) convex, (c) concave, (d) convex.
7 The line through $C$ is a radius, which is the normal to the mirror. So $\angle i=0$ and $\angle r=0$.
8 $F$ is $15\ \text{cm}$ and $C$ is $30\ \text{cm}$ from $P$, so they are $15\ \text{cm}$ apart.
IV
Applications and challenges.
  • 1 Solar cooker. A concave reflector has $R=1.2\ \text{m}$. Where should the cooking pot be placed?
  • 2 Headlight. A car headlight needs a parallel beam. Its concave reflector has $R=24\ \text{cm}$. Where must the bulb go?
  • 3 Lunar laser ranging. Corner reflectors (three mirrors at right angles) were left on the Moon. Use the idea of Example 2 to explain why a laser beam fired from Earth comes back towards Earth.
  • 4 Use Interactive 2 to explain why a wide spherical mirror does not focus sunlight to a sharp point, and suggest what shape large solar reflectors use instead.
  • 5 Periscope. Two plane mirrors are fixed parallel to each other, each at $45^\circ$ to the vertical tube. Show that a horizontal ray entering the top leaves the bottom horizontal again, travelling in its original direction.
  • 6 Challenge (uses trigonometric ratios). A ray parallel to the principal axis strikes a concave mirror of $R=20\ \text{cm}$ at a point where the radius to $C$ makes $30^\circ$ with the axis. Show that the reflected ray crosses the axis $\dfrac{R}{2\cos30^\circ}$ from $C$, and find how far from $P$ that is. Compare with $f$.
1 Sunlight is parallel, so it meets at $F$: $f=\tfrac{1.2}{2}=0.60\ \text{m}$ in front of the pole.
2 At the focus, $f=12\ \text{cm}$ from the pole: rays from $F$ reflect parallel to the axis (standard ray 2).
3 In two dimensions two perpendicular mirrors turn any ray through $180^\circ$ (Example 2). Three mutually perpendicular mirrors do the same in three dimensions, so the beam returns parallel to itself, back towards the telescope that sent it. Timing the round trip gives the Earth–Moon distance.
4 Rays far from the axis cross the axis closer to the mirror than $F$, so a wide beam gives a blurred patch instead of a point (spherical aberration). Large solar dishes and searchlights use parabolic reflectors, which send every parallel ray through one focus.
5 At the top mirror the horizontal ray has $\angle i=45^\circ$ and is turned through $90^\circ$, so it travels straight down. At the lower, parallel mirror it again has $\angle i=45^\circ$ and is turned through $90^\circ$ the other way, so it leaves horizontal and in its original direction, just lower down.
6 Let the ray hit the mirror at $M$ and cross the axis at $X$. The incident ray is parallel to $CP$, so $\angle i=\angle MCX=30^\circ$ (alternate angles), and $\angle r=\angle CMX=30^\circ$. Triangle $CMX$ is isosceles with $CX=XM$; dropping a perpendicular from $X$ to $CM$ bisects $CM=R$, so $CX=\dfrac{R/2}{\cos30^\circ}=\dfrac{10}{0.866}=11.5\ \text{cm}$. Then $PX=20-11.5=8.5\ \text{cm}$, noticeably less than $f=10\ \text{cm}$ — the ray misses $F$, exactly as Interactive 2 shows for wide beams.
🎯 Exam corner — board and entrance-exam practice BoardNEETJEE Main

E1. Board State the two laws of reflection of light. A ray of light falls on a plane mirror making an angle of $30^\circ$ with its surface. What is the angle of reflection? [2 marks]

Laws (i) $\angle i=\angle r$; (ii) the incident ray, the normal at the point of incidence and the reflected ray lie in the same plane.
Working $\angle i=90^\circ-30^\circ=60^\circ$, so $\angle r=60^\circ$.
Answer $60^\circ$. Exam tip: "with the surface" is the classic trap — always convert to the angle with the normal.

E2. Board Define the principal focus of a concave mirror. A concave mirror has a radius of curvature of $24\ \text{cm}$. Draw a ray diagram showing two rays parallel to the principal axis after reflection, and find the focal length. [3 marks]

Definition The point on the principal axis where rays parallel to the principal axis meet after reflection from a concave mirror.
Diagram Two rays parallel to the axis, above and below it, reflect from the mirror and both pass through $F$, which is marked halfway between $P$ and $C$.
Answer $f=\dfrac R2=\dfrac{24}{2}=12\ \text{cm}$. Exam tip: label $P$, $F$, $C$ and put arrows on every ray — marks are lost for missing arrows.

E3. Board Assertion (A): The principal focus of a convex mirror is virtual. Reason (R): Rays parallel to the principal axis diverge after reflection from a convex mirror and only appear to come from a point behind it. (A) Both A and R are true and R is the correct explanation of A. (B) Both A and R are true but R is not the correct explanation of A. (C) A is true but R is false. (D) A is false but R is true.

Idea A focus is virtual when the rays do not really pass through it.
Check R is a true statement and it is exactly why the focus is virtual.
Answer (A). Exam tip: in assertion–reason items, first judge A and R separately, then ask "does R answer the question why A?"

E4. NEET Keeping the incident ray fixed, a plane mirror is rotated through $15^\circ$. The reflected ray turns through (A) $7.5^\circ$ (B) $15^\circ$ (C) $30^\circ$ (D) $60^\circ$

Idea The normal turns with the mirror by $15^\circ$, so $\angle i$ and $\angle r$ each change by $15^\circ$.
Working The reflected ray turns through $2\times15^\circ=30^\circ$.
Answer (C). Exam tip: remember "mirror turns $\theta$, ray turns $2\theta$".

E5. JEE Main Two plane mirrors are inclined at $90^\circ$. A ray in the plane perpendicular to both mirrors is reflected once from each. The total angle through which the ray is turned is (A) $90^\circ$ (B) $180^\circ$ (C) $270^\circ$ (D) it depends on the angle of incidence

Idea A reflection with angle of incidence $i$ turns a ray through $180^\circ-2i$.
Working With $i_1+i_2=90^\circ$ (Example 2): total turn $=(180^\circ-2i_1)+(180^\circ-2i_2)=360^\circ-180^\circ=180^\circ$.
Answer (B), for every angle of incidence. Exam tip: test with one easy case ($i_1=45^\circ$) to confirm in seconds.
🔬 Research–Teaching NexusResearch-tutored · optional, about 40 minutes

Research question. How much sunlight can a curved mirror collect, and how do large solar kitchens and solar furnaces use concave reflectors to cook food or melt metal?

Why it matters. India has used solar steam kitchens with rows of dish-shaped concave reflectors to cook meals for thousands of people a day — the temple kitchen at Shirdi in Maharashtra is a well-known example — and solar cookers are promoted for homes without clean fuel. In the French Pyrenees, the Odeillo solar furnace uses a huge curved mirror fed by many flat mirrors to reach temperatures above $3000\ ^\circ\text{C}$ for testing materials. Both rely on one fact from this section: parallel rays are gathered at the focus.

What to do.

  1. Take a concave shaving or make-up mirror outdoors (never look at the Sun or its focused spot). Catch the image of the Sun — or of a distant building — on a sheet of paper and measure the distance from mirror to paper. This is $f$; work out $R=2f$.
  2. Measure the diameter of the mirror and find the area it presents to the Sun, $A=\pi\left(\tfrac d2\right)^2$.
  3. On a clear day bright sunlight delivers roughly $1000\ \text{W}$ per square metre. Estimate the power your mirror collects.
  4. A large solar-kitchen dish may have an area of about $16\ \text{m}^2$. Estimate the power it collects, and how many of your mirrors would match it.
  5. Use Interactive 2 to explain why such dishes are not made spherical.

Tools: concave mirror, ruler, white card, calculator.

Hand in: A half-page note with your measured $f$ and $R$, the power estimates, and one safety rule for working with focused sunlight.

Going further. Find out how a "Scheffler" reflector tracks the Sun during the day, and why the focus stays in a fixed place inside the kitchen building.

Instructor note: a mirror of diameter $0.30\ \text{m}$ has area $\pi(0.15)^2\approx0.071\ \text{m}^2$ and collects about $71\ \text{W}$; a $16\ \text{m}^2$ dish collects about $16\ \text{kW}$, roughly 230 such mirrors. Typical make-up mirrors give $f$ between $15$ and $50\ \text{cm}$.

🧪 Self-check quiz

0/5 correct

Q1. The angle of incidence on a plane mirror is $35^\circ$. The angle between the incident ray and the reflected ray is

Q2. A concave mirror has a radius of curvature of $30\ \text{cm}$. Its focal length is

Q3. The image formed by a plane mirror is

Q4. Which mirror is used as a vehicle's rear-view mirror to give a wide field of view?

Q5. A ray passing through the principal focus of a concave mirror, after reflection,

← Chapter 1 — Light – Reflection and Refraction1.2 Image Formation, the Mirror Formula and Magnification →