Geometric optics: mirrors and lenses

It is mid-October 2026. In Year 4 of an Italian science high school (liceo scientifico), you are somewhere between the last part of mechanics or thermology and geometric optics, which usually takes up October and November. The class test comes four to eight weeks from now, between mid-November and early December. Geometric optics does not mean waves: here light is a ray that travels in a straight line and bends when it meets a mirror or changes medium. Waves, interference and diffraction are another chapter, and you will find them in mechanical waves and sound and in Young’s interference and diffraction. This post stays with rays.

The chapter is short, but it costs marks for one reason only: signs. Once you have fixed a convention, mirrors and lenses are solved with the same equation. The method I use in lessons: draw the rays, write down the sign convention, do the calculation, and use the drawing to check the calculation.

Rays and reflection

In geometric optics, light is described by rays that travel in a straight line in a uniform medium. When a ray hits a smooth surface it is reflected according to two laws:

the incident ray, the reflected ray and the normal to the surface at the point of incidence lie in the same plane;
the angle of reflection equals the angle of incidence: $θ_{r} = θ_{i}$ .

Angles are measured from the normal, not from the surface. This is the most common mistake: if the question gives the angle with the mirror, the angle of incidence is its complement.

Spherical mirrors

A spherical mirror is a cap cut from a sphere of radius $R$ . If the reflecting surface is on the inside of the sphere the mirror is concave; if it is on the outside the mirror is convex. Rays parallel to the axis pass, after reflection, through the focus $F$ , at a distance

f = \frac{R}{2}

from the vertex of the mirror (for a convex mirror it is their extensions, behind the mirror, that pass through it).

The position of the image comes from the conjugate points equation (the mirror equation):

\frac{1}{p} + \frac{1}{q} = \frac{1}{f}

and the linear magnification is

G = - \frac{q}{p}

where $p$ is the distance of the object from the vertex, $q$ is that of the image, and the height of the image is $h^{'} = G h$ .

The sign convention

This is where you win or lose the test. I use the “real is positive” convention, the most common one in high school textbooks:

$p > 0$ if the object is in front of the mirror (on the side the light comes from): this is the normal case.
$q > 0$ if the image is real, that is, in front of the mirror, where the rays really meet. $q < 0$ if it is virtual, behind the mirror.
$f > 0$ for a concave mirror, $f < 0$ for a convex one.
$G < 0$ means the image is inverted, $G > 0$ upright; $∣ G ∣ > 1$ enlarged, $∣ G ∣ < 1$ reduced.

Some books use another convention (for example with the radius of curvature positive “on the side of the incoming rays”, or with the signs of $q$ reversed). Check yours and stay faithful to it from the beginning to the end of the test: the equation is the same, only the signs of the values you substitute change.

The three rays for the ray diagram

To draw the image of a point of the object, two of the three principal rays are enough; the third serves as a check:

the ray parallel to the axis is reflected through the focus;
the ray that passes through the focus is reflected parallel to the axis;
the ray that passes through the centre of curvature $C$ (at a distance $R = 2 f$ from the vertex) arrives perpendicular to the surface and comes back along itself.

The point where the reflected rays meet is the top of the image. If they do not really meet but their extensions behind the mirror do, the image is virtual.

Exercise 1: concave mirror

A concave mirror has radius of curvature $R = 40 cm$ . An object $3.0 cm$ tall is placed $30 cm$ from the vertex. Find the position, magnification and height of the image. Then repeat with the object at $10 cm$ .

Step 1: the focal length. $f = R /2 = 20 cm$ , positive because the mirror is concave.

Step 2: the conjugate points equation with $p = 30 cm$ .

\frac{1}{q} = \frac{1}{f} - \frac{1}{p} = \frac{1}{20} - \frac{1}{30} = \frac{3 - 2}{60} = \frac{1}{60}

So $q = 60 cm$ , positive: the image is real, in front of the mirror.

Step 3: magnification.

G = - \frac{q}{p} = - \frac{60}{30} = - 2

The image is inverted and twice as large: $h^{'} = - 2 \cdot 3.0 = - 6.0 cm$ , that is, $6.0 cm$ tall and upside down.

Check with the drawing. Is the object between $F$ ( $20 cm$ ) and $C$ ( $40 cm$ ), or beyond? At $30 cm$ it is between $F$ and $C$ , and in this zone the image is real, inverted, enlarged, and forms beyond $C$ : $60 cm > 40 cm$ . Consistent.

Second part: $p = 10 cm$ . The object is inside the focal length.

\frac{1}{q} = \frac{1}{20} - \frac{1}{10} = \frac{1 - 2}{20} = - \frac{1}{20}

$q = - 20 cm$ , negative: virtual image, behind the mirror. $G = - (- 20) /10 = + 2$ : upright and twice the size, $6.0 cm$ tall. This is a make-up or shaving mirror: you bring your face inside the focal length and see yourself enlarged and upright. A negative result is not a mistake, it is an answer.

To practise timed questions there are the maths and physics tests for high school: the correct answer is shown straight away, and the step-by-step guided solution unlocks with your first lesson.

Refraction and Snell’s law

When a ray passes from one medium to another it changes direction, because light has different speeds in the two media. Each medium has a refractive index $n = c / v$ , where $c$ is the speed of light in a vacuum and $v$ the speed in the medium: $n \approx 1.00$ for air, $\approx 1.33$ for water, $\approx 1.5$ for ordinary glass. Snell’s law links the angles, measured from the normal:

n_{1} sin θ_{1} = n_{2} sin θ_{2}

A common-sense rule: going into a medium with a larger $n$ the ray bends towards the normal; going into one with a smaller $n$ it bends away. If you got the opposite, you have swapped the two media.

Critical angle and total internal reflection

If the ray goes from a medium with larger $n_{1}$ to one with smaller $n_{2}$ , the angle of refraction is larger than the angle of incidence. As $θ_{1}$ increases, sooner or later $θ_{2}$ reaches $9 0^{\circ}$ : this is the critical angle,

sin θ_{L} = \frac{n _{2}}{n _{1}}

Beyond that angle there is no refracted ray any more: all the light is reflected, and this is total internal reflection. It happens only when going from large $n$ to small $n$ (from glass to air, not the other way round), and it is the principle behind optical fibres and binocular prisms. If you try to calculate $θ_{L}$ with the two indices swapped you get a sine greater than $1$ , which is impossible: a sign that you have turned something upside down.

Exercise 2: Snell and the critical angle

A ray in air ( $n_{1} = 1.00$ ) strikes the surface of a block of glass with $n_{2} = 1.50$ at an angle of $4 0^{\circ}$ . What is the angle of refraction? Then: inside the glass, a ray hits the exit face towards the air at an angle of $4 5^{\circ}$ . Does it come out, or is it totally reflected?

Step 1: the refracted angle.

sin θ_{2} = \frac{n _{1}}{n _{2}} sin θ_{1} = \frac{1.00}{1.50} sin 4 0^{\circ} = \frac{0.6428}{1.50} \approx 0.4285

θ_{2} = arcsin (0.4285) \approx 25. 4^{\circ}

Passing into a medium with larger $n$ the ray has moved closer to the normal, from $4 0^{\circ}$ to $25. 4^{\circ}$ : consistent with the rule.

Step 2: the glass-air critical angle. Here $n_{1} = 1.50$ (glass) and $n_{2} = 1.00$ (air):

sin θ_{L} = \frac{1.00}{1.50} \approx 0.6667 \Rightarrow θ_{L} \approx 41. 8^{\circ}

Step 3: the comparison. An incidence of $4 5^{\circ}$ is larger than $41. 8^{\circ}$ , so the ray does not come out: there is total internal reflection. Check by contradiction: if we tried to apply Snell, $sin θ_{2} = 1.5 \cdot sin 4 5^{\circ} = 1.5 \cdot 0.7071 \approx 1.06$ , a sine greater than $1$ , which does not exist. This check counts as proof: if Snell’s law gives you a sine greater than $1$ , the ray does not get through.

Thin lenses

A thin lens is a piece of glass with two curved surfaces that deflects rays by refraction. There are two families: the converging lens (thicker in the middle) makes parallel rays converge at a real focus; the diverging lens (thinner in the middle) spreads them apart, as if they came from a virtual focus.

The equation is identical to the one for mirrors:

\frac{1}{p} + \frac{1}{q} = \frac{1}{f} G = - \frac{q}{p}

The difference lies in the convention for what “real” means. With a lens the light passes through, so the real image forms on the other side from the object:

$p > 0$ if the object is on the side the light comes from (the normal case);
$q > 0$ if the image is real, that is, on the opposite side of the lens from the object; $q < 0$ if it is virtual, that is, on the same side as the object;
$f > 0$ for the converging lens, $f < 0$ for the diverging one;
$G$ has the same meaning: negative if inverted.

The thing to remember is that, with this convention, “real = positive” is true for both mirrors and lenses. Only the place where the real image forms changes: in front of the mirror, behind the lens.

The three rays for the lens

The ray parallel to the axis comes out passing through the focus (on the other side, for the converging lens).
The ray that passes through the centre of the lens is not deflected.
The ray that passes through the focus (on the object’s side) comes out parallel to the axis.

The image forms where the outgoing rays meet.

Exercise 3: converging lens and diverging lens

A converging lens has focal length $f = 15 cm$ . An object $2.0 cm$ tall is $20 cm$ from the lens. Find $q$ , $G$ and the height of the image. Then move the object to $10 cm$ . Finally, use a diverging lens with focal length $- 10 cm$ and the object at $30 cm$ .

Part a: $p = 20 cm$ , $f = + 15 cm$ .

\frac{1}{q} = \frac{1}{15} - \frac{1}{20} = \frac{4 - 3}{60} = \frac{1}{60} \Rightarrow q = 60 cm

$G = - 60/20 = - 3$ . The image is real, on the other side, inverted and three times larger: $h^{'} = - 3 \cdot 2.0 = - 6.0 cm$ . This is the projector set-up: the object sits between $f$ and $2 f$ and the image forms far away, enlarged.

Part b: $p = 10 cm$ . The object is inside the focal length.

\frac{1}{q} = \frac{1}{15} - \frac{1}{10} = \frac{2 - 3}{30} = - \frac{1}{30} \Rightarrow q = - 30 cm

$G = - (- 30) /10 = + 3$ : virtual image, upright, enlarged three times, $6.0 cm$ tall, on the same side as the object. This is the magnifying glass.

Part c: diverging lens, $f = - 10 cm$ , $p = 30 cm$ .

\frac{1}{q} = - \frac{1}{10} - \frac{1}{30} = \frac{- 3 - 1}{30} = - \frac{4}{30} \Rightarrow q = - 7.5 cm

$G = - (- 7.5) /30 = + 0.25$ : virtual image, upright, reduced to a quarter. A diverging lens always gives a virtual, upright, reduced image, whatever the distance of the object: if you get $q > 0$ with $f < 0$ you have got a sign wrong.

Check with the drawing. In part a the object is between $f$ ( $15 cm$ ) and $2 f$ ( $30 cm$ ): the image is beyond $2 f$ , that is, beyond $30 cm$ . We find $60 cm$ , which fits. In part b the object is inside $f$ : the image is virtual, and indeed $q < 0$ .

The mistakes that cost marks

Sign of $f$ . Concave or converging: $f > 0$ . Convex or diverging: $f < 0$ . This is the mistake that loses the most marks: if $f$ has the wrong sign, everything else is consistent but false.
Sign of $q$ not interpreted. $q < 0$ is not a mistake: it means a virtual image. Always write the words next to the number (“real/virtual, upright/inverted, enlarged/reduced”), because the question almost always asks for them.
$1/ p + 1/ q = 1/ f$ not inverted at the end. You calculate $1/ q$ and then flip the result. On the calculator, $1/ f - 1/ p$ and then the $x^{- 1}$ key, or fractions with a common denominator as above.
Angles measured from the surface. In reflection and refraction you use the angle with the normal. If the question gives the angle with the surface, subtract it from $9 0^{\circ}$ .
Indices swapped in Snell. Check the direction: entering a larger $n$ you move towards the normal. A refracted angle that moves away instead of closer signals the swap.
Total internal reflection in the wrong direction. It exists only from large $n$ to small $n$ . For the opposite passage (air into glass) there is no critical angle and the ray always enters.
The drawing that does not match the calculation. An image that the calculation puts at $60 cm$ but that sits at $12 cm$ on the drawing is not a pencil inaccuracy: one of the two is wrong, and usually it is a sign.

How to prepare

Redo the three exercises without looking, changing the focal length and the distance of the object, and for each result write next to it the three descriptions: real or virtual, upright or inverted, enlarged or reduced. Then draw the three principal rays by hand for each case, on lined paper with a ruler: do the drawing, because in the test it is often worth as much as the calculation. If you want to practise against the clock, try a timed Year 4 test in the practice tests. The overview of the subject is in Physics.

If you want a check before the test

If the formulas make sense but the exercises do not, the problem is almost always in the signs or the drawing, not in the algebra. Write to me from the contact page and we will do a free introductory call: you send me one optics exercise that does not work out and I tell you where it gets stuck.

Geometric optics: mirrors, refraction and lenses