PHYS 101 Introductory Physics, Mechanics • McGill University, Montreal

Revision sheet: geometric optics, mirrors and lenses (PHYS 101)

This sheet covers the geometric optics chapter of PHYS 101, Introductory Physics, the algebra based first year course at McGill University in Montreal. It assumes you already have the course notes: what follows is the list of places where marks are actually lost, and the exact gesture that prevents each loss.

One idea runs through all of it. Optics has no hard calculations, it has one convention, and almost every wrong answer on a midterm is a sign that was dropped, inverted or measured from the wrong reference. Write the convention at the top of your page, draw the rays first, and use the numbers to confirm the drawing.

The thread of the chapter

Every question in this chapter is decided by a sign convention you adopt on the first line and never abandon. A negative $d_i$ is not a slip to be tidied away, it is the answer: the image is virtual, on the incoming side. A negative $m$ says inverted. So draw the rays BEFORE computing, to know which sign to expect, and let the arithmetic confirm the drawing.

This chapter is part of PHYS 101, Introductory Physics - Mechanics (McGill)

Before this chapter, you need

• Waves and sound

What is on this sheet

1. The essentials
2. The rules in table form
3. The mistakes that cost marks
4. Which method to choose
5. How the answer is expected to be written
6. Check before you hand in
7. The typical problem, taken apart
8. Learn by heart
9. Frequently asked questions

Topics covered

Sign convention in optics
Law of reflection
Snell's law of refraction
Index of refraction
Critical angle
Total internal reflection
Real and virtual images
Mirror and lens equation
Linear magnification
Principal rays
Optical power in diopters
Dispersion

The essentials

The convention, written once, never abandoned

• Distances are measured from the centre of the lens or from the vertex of the mirror. A real object in front of the device has $d_o>0$ .
• $d_i>0$ when the image lands on the OUTGOING side, and the image is then REAL. The outgoing side is BEHIND a lens, because light goes through it, and IN FRONT of a mirror, because light bounces back.
• $d_i<0$ when the image lands on the incoming side. The image is then VIRTUAL: no light reaches that place, only the backward extensions of the rays meet there, and no screen will ever show it.
• $f>0$ for a converging device, a convex lens or a concave mirror; $f<0$ for a diverging one, a concave lens or a convex mirror. For a mirror, $f=\dfrac{R}{2}$ with the same sign.
• $m=-\dfrac{d_i}{d_o}=\dfrac{h_i}{h_o}$ . Positive means upright, negative means inverted, and $|m|>1$ means enlarged.

Same rule, two geographies: the outgoing side is behind the lens and in front of the mirror, so

d_i>0

points right on the left half and LEFT on the right half.

For a single lens or a single mirror, real and inverted always travel together, and so do virtual and upright. If your answer pairs them the other way, a sign is wrong and you can find it without redoing the problem.

The three equations, and nothing else

• Reflection: $\theta_{\text{reflected}}=\theta_{\text{incident}}$ , both from the normal. It happens at every surface, even when refraction happens too.
• Refraction: $n_1\sin\theta_1=n_2\sin\theta_2$ , both from the normal, with $n=\dfrac{c}{v}\ge 1$ . Toward a slower medium the ray bends TOWARD the normal.
• Critical angle, from the slower medium toward the faster one only: $\sin\theta_c=\dfrac{n_2}{n_1}$ . Past $\theta_c$ there is no refracted ray at all.
• Image: $\dfrac{1}{d_o}+\dfrac{1}{d_i}=\dfrac{1}{f}$ , the same letters for mirrors and for lenses, often more useful as $d_i=\dfrac{d_o f}{d_o-f}$ , where the sign of $d_o-f$ decides everything.
• Power: $P=\dfrac{1}{f}$ in diopters, with $f$ in METRES. A prescription of $-2.00$ D is a diverging lens, so a short sighted eye.

Nothing on this page needs a derivative or an integral, which is the whole point of PHYS 101 against PHYS 131. A solution that starts by differentiating is not a PHYS 101 solution even when the final number is right.

The rules in table form

Each row reads from left to right: the assumptions, then the result. A red cell is not an answer, it is the finding that the form settles nothing and the instruction to rewrite it. Every case is followed by a worked example.

One converging lens of focal length 12 cm, read off the object position

Object position	Where the image lands	Nature, orientation, size
$d_o>2f$	$f<d_i<2f$	real, inverted, reduced
Example: $d_o=36$ cm gives $d_i=18$ cm and $m=-0.5$
$d_o=2f$	$d_i=2f$	real, inverted, same size
Example: $d_o=24$ cm gives $d_i=24$ cm and $m=-1$
$f<d_o<2f$	$d_i>2f$	real, inverted, enlarged
Example: $d_o=18$ cm gives $d_i=36$ cm and $m=-2$
$d_o=f$	nowhere	the rays leave parallel no image at all
Example: $d_o=12$ cm gives $d_o-f=0$ What to do: Say that the image is at infinity and stop. Do not force a number out of a zero denominator, and do not write a huge one either.
$d_o<f$	$d_i<0$	virtual, upright, enlarged
Example: $d_o=8$ cm gives $d_i=-24$ cm and $m=+3$
any $d_o$ , with $f<0$	$d_i<0$ always	virtual, upright, reduced
Example: $d_o=20$ cm with $f=-20$ cm gives $d_i=-10$ cm and $m=+0.5$

The mistakes that cost marks

These are the errors I correct most often in session. Each one costs marks on a paper, even when the reasoning behind it is right.

1. The angle is given from the surface, not from the normal

the whole question, and usually the whole problem built on it

What not to write

“The ray makes $35°$ with the mirror, so it reflects at $35°$ .”

What to write

“The ray makes $35°$ with the surface, so $\theta_1=90°-35°=55°$ from the normal, and it reflects at $55°$ .”

Why: An exam writes it that way on purpose, because that is how a protractor laid flat on a bench measures an angle. The laws accept only the angle from the normal, so the conversion is the first line you write, before any equation.

2. The minus sign is erased from an image distance

two marks, because the sign is wrong twice: on the position and again on the orientation

What not to write

“I get $d_i=-30$ cm, but a distance cannot be negative, so $d_i=30$ cm, image behind the lens.”

What to write

“ $d_i=-30$ cm, so the image is on the incoming side, $30$ cm in front of the lens: it is VIRTUAL, and $m=-\dfrac{-30}{10}=+3$ , so upright and three times the size.”

Object inside the focal length: the rays leave DIVERGING and never cross. Only their backward extensions, dashed, meet at the image, which is why

d_i<0

and why no screen can catch it.

Why: The convention is exactly what turns a number into a physical statement. Strip the sign and $m$ flips with it, so you report an inverted image of an upright object. The figure below shows what the minus sign is describing.

3. The critical angle is hunted in the direction where it does not exist

the whole question, plus the time spent looking for an angle that does not exist

What not to write

“Light goes from air into water at $70°$ , which is past the critical angle, so it is totally reflected.”

What to write

“Air to water is slow-down, so a refracted ray always exists: $\sin\theta_2=\dfrac{\sin 70°}{1.333}=0.705$ , that is $\theta_2=44.8°$ . Total internal reflection is only possible from water toward air, past $\theta_c=48.6°$ .”

Why: $\sin\theta_c=\dfrac{n_2}{n_1}$ only has a solution when $n_2<n_1$ . Going toward the denser medium the required sine is always below one, so there is nothing to check. One glance at which index is bigger settles it.

4. The outgoing side of a mirror is taken to be that of a lens

one to two marks, and any later question that places a screen

What not to write

“ $d_i=+60$ cm on a concave mirror, so the image is $60$ cm BEHIND the mirror.”

What to write

“ $d_i=+60$ cm on a concave mirror, so the image is on the outgoing side, which for a mirror is the side the light came from: $60$ cm IN FRONT of the mirror, and it is real.”

Why: The convention never names a side of the page, it names the side the light travels on after meeting the device. Light passes through a lens and bounces off a mirror, so the same positive sign points in opposite directions. In both cases $d_i>0$ means real, and that is the part that never changes.

5. The inverse sine is taken before dividing by the index

one mark, and the error is invisible because the wrong answer looks completely reasonable

What not to write

“ $\sin\theta_2=\dfrac{\sin 55°}{1.333}$ , so $\theta_2=\dfrac{55°}{1.333}=41.3°$ .”

What to write

“ $\sin\theta_2=\dfrac{\sin 55°}{1.333}=\dfrac{0.8192}{1.333}=0.6145$ , and only then $\theta_2=37.9°$ .”

Why: Dividing the angle is not dividing its sine: the sine is not a proportional function. The two answers here differ by three and a half degrees, which no marker will read as a rounding difference. The rule is mechanical: compute the sine, divide, take the inverse sine LAST.

6. The focal length stays in centimetres inside a power

one mark, and it is the kind of answer that gets no partial credit because the unit is in the definition

What not to write

“ $f=15$ cm, so $P=\dfrac{1}{15}=0.067$ D.”

What to write

“ $f=15$ cm $=0.150$ m, so $P=\dfrac{1}{0.150}=+6.67$ D.”

Why: The diopter IS the reciprocal metre, so the conversion is not a detail of presentation, it is part of the formula. A factor of one hundred also puts the answer outside every range that exists: real eyewear runs from about $-8$ to $+4$ D, and a hand magnifier from $+5$ to $+30$ D.

7. The magnifier is assumed to magnify more when it is pushed closer to the page

the reasoning mark on any question that asks where to hold the lens

What not to write

“Bring the lens nearer the page and the image grows, so the closer the better.”

What to write

“With $f=12$ cm, $d_o=8$ cm gives $m=+3$ but $d_o=6$ cm gives only $m=+2$ : the magnification is $m=\dfrac{f}{f-d_o}$ , which grows as $d_o$ approaches $f$ , not as $d_o$ approaches zero.”

Why: The formula settles it in one line, and the limit cases confirm it: pressed flat on the page ( $d_o\to 0$ ) the lens does nothing at all and $m\to 1$ , while just inside the focus $m$ grows without bound. It is the standard intuition trap of the chapter.

Which method to choose

Where the image lands, decided by the object position alone

Compare $d_o$ with $f$ first, and only then with $2f$ . Everything follows from the SIGN of $d_o-f$ in $d_i=\dfrac{d_o f}{d_o-f}$ , so that single subtraction is the whole decision.

The six object distances of the table, read against

F

12

cm and

2F

24

cm: which side of

F

the mark falls on is the entire decision.

If $f<0$ , the device is diverging → $d_i<0$ whatever $d_o$ : virtual, upright, reduced, no exception to look for

Example: $f=-20$ cm and $d_o=20$ cm give $d_i=-10$ cm, $m=+0.5$

The numerator $d_o f$ is negative and the denominator $d_o-f=d_o+|f|$ is positive, so the quotient cannot be positive. Two lines of algebra replace four exam questions.
If $f>0$ and $d_o>2f$ → real, inverted and reduced, with the image between $f$ and $2f$

Example: $f=12$ cm and $d_o=36$ cm give $d_i=18$ cm, $m=-0.5$

This is the camera case. The further the object, the closer the image creeps to the focal plane.
If $f>0$ and $f<d_o<2f$ → real, inverted and enlarged, with the image beyond $2f$

Example: $f=12$ cm and $d_o=18$ cm give $d_i=36$ cm, $m=-2$

This is the projector case, and it is the mirror image of the previous line: object and image swap roles.
If $f>0$ and $d_o=f$ exactly → no image at any finite distance, the rays leave parallel

Example: $f=12$ cm and $d_o=12$ cm give $d_o-f=0$

Say that the image is at infinity. This is the headlight and the lighthouse, the lamp sitting at the focus.
If $f>0$ and $d_o<f$ → virtual, upright and enlarged, on the incoming side

Example: $f=12$ cm and $d_o=8$ cm give $d_i=-24$ cm, $m=+3$

The magnifying glass and the shaving mirror. Cross $f$ going outward and the image flips, which is the moment the sign of $d_i$ changes.

What happens when a ray meets a surface

Compare the two indices first, and test the angle only in the case where a critical angle can exist. Reversing the two tests wastes time and invents answers.

If $n_2>n_1$ , the second medium is slower → refraction always happens, the ray bends toward the normal, and there is no critical angle to look for

Example: air to glass at $55°$ gives $33.1°$

$\sin\theta_2=\dfrac{n_1}{n_2}\sin\theta_1$ with a factor below one is always a legal sine.
If $n_2<n_1$ and $\theta_1<\theta_c$ → refraction happens, the ray bends away from the normal, and part of the light is also reflected

Example: water to air at $30°$ gives $41.8°$

Compute $\theta_c=\arcsin\dfrac{n_2}{n_1}$ BEFORE the refraction angle, so you know the answer exists before you produce it.
If $n_2<n_1$ and $\theta_1\ge\theta_c$ → no refracted ray at all: total internal reflection, and all the energy leaves at $\theta_1$ on the same side

Example: water to air at $60°$ needs $\sin\theta_2=1.154$ , which is impossible

The optical fibre, the endoscope and the prism in a pair of binoculars are all this line.

How the answer is expected to be written

A marker ticks steps. Here they are in order, with the concluding sentence expected word for word.

An image problem, written the way the marker reads it

When to use it: Every question that ends with find the image, and every question that gives you a device, an object and a distance.

1 State the convention in words, once: “Distances from the lens; $d_i>0$ behind it, so real; $f>0$ converging; $m=-d_i/d_o$ .” One line, and it is usually worth a mark on its own.
2 List the data WITH their signs: “ $f=+15.0$ cm (converging), $d_o=+10.0$ cm, $h_o=2.00$ cm.” A sign written here is a sign you will not lose later.
3 Place the object against $F$ and $2F$ and predict: “ $d_o<f$ , so I expect a virtual, upright, enlarged image, that is $d_i<0$ and $m>1$ .” This is the sentence that catches an arithmetic slip before it costs anything.
4 Isolate before substituting: “ $\dfrac{1}{d_i}=\dfrac{1}{f}-\dfrac{1}{d_o}$ , so $d_i=\dfrac{d_o f}{d_o-f}$ ”, then put the numbers in. Substituting first turns one algebra error into three.
5 Turn the signs back into words, explicitly: “ $d_i=-30.0$ cm, so the image is virtual and $30.0$ cm in front of the lens; $m=+3.00$ , so it is upright and three times the size, that is $6.00$ cm tall.”

Concluding sentence

“The image is virtual, $30.0$ cm in front of the lens, upright, and $6.00$ cm tall.”

Check before you hand in

Five minutes of checking recover more marks than one more problem started in a hurry.

Do the sign of $d_i$ and the drawing agree?

Look at your ray diagram. If the outgoing rays actually cross somewhere, $d_i$ must be positive; if they leave diverging and only their dashed extensions meet, $d_i$ must be negative. Any disagreement is an arithmetic slip, and you know which sign is right before hunting for it.

Object inside the focus of a converging lens: the rays diverge on the way out, so a positive $d_i$ is impossible.
Do the signs of $d_i$ and $m$ agree with each other?

For a single lens or a single mirror, real goes with inverted and virtual goes with upright, never the other way. So $d_i>0$ demands $m<0$ , and $d_i<0$ demands $m>0$ . This is a one second check and it catches the most common sign error of the chapter.

$d_i=-20$ cm with $m=-2$ is impossible: one of the two minus signs has been invented.
Did the ray bend the right way at the surface?

Toward a slower medium, that is a larger index, the angle must DECREASE; toward a faster one, it must increase. Compare $\theta_2$ with $\theta_1$ and you know in one look whether the two indices were swapped in Snell's law.

Air to glass with $\theta_1=55°$ : a $\theta_2=72°$ is certainly wrong, since $33.1°$ is the only possible shape of answer.
Is the order of magnitude physically possible?

An index below $1$ does not exist. A magnification of $50$ from a single hand lens does not exist. A spectacle power of $200$ D does not exist. Most unit errors in this chapter announce themselves by producing a number no catalogue would print.

$P=\dfrac{1}{15}=0.067$ D means the centimetres were never converted: the real value is $+6.67$ D.
Does the answer contain a derivative?

If it does, it is not a PHYS 101 answer. Everything in this chapter is algebra and right angle trigonometry, and a solution copied from a calculus based course will be marked as not following the method taught, even when the final number matches.

The proof that a screen needs $D\ge 4f$ is a discriminant, $D^{2}-4fD\ge 0$ , not a minimum found by differentiating.

The typical problem, taken apart

A converging lens used as a magnifier, from the convention to the sentence

A converging lens has a focal length of $15.0$ cm. An arrow $2.00$ cm tall stands upright on the axis, $10.0$ cm from the lens.

Locate the image, say what kind it is, give its orientation and its height, and state the power of the lens in diopters.

Step 1

Convention: distances from the centre of the lens, $d_i>0$ behind it and therefore real, $f>0$ for a converging lens, $m=-\dfrac{d_i}{d_o}$ . Data: $f=+15.0$ cm, $d_o=+10.0$ cm, $h_o=2.00$ cm.

Why

Written before anything else, this line fixes the meaning of every sign that follows. Markers give a mark for it, and more importantly it is what stops you from deleting a minus sign three steps later.

Step 2

$d_o=10.0$ cm is smaller than $f=15.0$ cm, so the object sits INSIDE the focal length. Expect a virtual image, hence $d_i<0$ , upright, hence $m>0$ , and enlarged, hence $|m|>1$ .

Why

The prediction is made from the drawing, not from the arithmetic. It costs five seconds and it turns the computation that follows into a check rather than a leap of faith.

Step 3

$\dfrac{1}{d_i}=\dfrac{1}{f}-\dfrac{1}{d_o}=\dfrac{1}{15.0}-\dfrac{1}{10.0}=\dfrac{2-3}{30.0}=-\dfrac{1}{30.0}$ , so $d_i=-30.0$ cm.

Why

Isolate $\dfrac{1}{d_i}$ first and put the numbers over a common denominator: the sign then comes out of the subtraction by itself, instead of being decided at the end by a student who does not like negative distances.

Step 4

$m=-\dfrac{d_i}{d_o}=-\dfrac{-30.0}{10.0}=+3.00$ , so $h_i=m\,h_o=3.00\times 2.00=6.00$ cm, upright.

Why

The two minus signs cancel, and that cancellation IS the physics: a virtual image is an upright image for a single lens. If your $m$ had come out negative here, the error would be in step 3, not in this one.

Step 5

$f=15.0$ cm $=0.150$ m, so $P=\dfrac{1}{0.150}=+6.67$ D.

Why

The conversion to metres belongs to the definition of the diopter, not to the presentation. The positive sign confirms a converging lens, which is the one consistency check available on this last line.

The conclusion, written out

The image is virtual, $30.0$ cm in front of the lens, upright and $6.00$ cm tall, so three times the object; the lens has a power of $+6.67$ D. Every one of those five statements is a sign read out loud.

Learn by heart

• Every angle in this chapter is measured from the NORMAL, never from the surface. An angle given with the surface is $90°$ minus the angle you need.
• $d_i>0$ means real and on the outgoing side; $d_i<0$ means virtual and on the incoming side. The outgoing side is behind a lens and in front of a mirror.
• $m=-\dfrac{d_i}{d_o}$ : positive upright, negative inverted, $|m|>1$ enlarged. For one lens or one mirror, real and inverted always go together.
• $f=\dfrac{R}{2}$ , positive for a concave mirror and a convex lens, negative for a convex mirror and a concave lens.
• $d_i=\dfrac{d_o f}{d_o-f}$ : the sign of $d_o-f$ decides the whole answer, so compute that subtraction first.
• $n=\dfrac{c}{v}\ge 1$ ; toward a larger index the ray bends toward the normal and the angle decreases.
• $\sin\theta_c=\dfrac{n_2}{n_1}$ , and it only exists going from the slower medium toward the faster one. Water to air: $48.6°$ . Glass to air: $41.8°$ .
• $P=\dfrac{1}{f}$ with $f$ in METRES. Negative power corrects short sight, positive power corrects long sight.
• A screen can only show a real image, and only if the object to screen distance satisfies $D\ge 4f$ .
• No derivative and no integral anywhere: the whole chapter is algebra and right angle trigonometry.

Frequently asked questions

What does a negative image distance mean in optics?

It is not a mistake, it is the answer. A negative image distance says the image sits on the incoming side of the device, in front of a lens or behind a mirror, and therefore that the image is virtual. No light actually reaches that place: the rays leave diverging and only their backward extensions meet there, so no screen will ever show it, although your eye sees it perfectly well. Erase the minus sign and you also flip the magnification, so you end up reporting an inverted image of an upright object.

How do I know if an image is real or virtual?

Look at whether the outgoing rays really cross. If they do, light genuinely arrives where they meet, the image is real, and a screen held there shows it. If they leave diverging and never cross, only their backward extensions meet, and the image is virtual. In the equation this is the sign of the image distance: positive means real, negative means virtual. For a single lens or a single mirror, real images are always inverted and virtual images are always upright.

Do mirrors and lenses use the same equation?

Yes, letter for letter: one over the object distance plus one over the image distance equals one over the focal length, and the magnification is minus the image distance divided by the object distance. Only the geography changes. Light passes through a lens, so a positive image distance puts the image behind it; light bounces off a mirror, so a positive image distance puts the image in front of it. In both cases positive means real, and that is the part worth remembering.

When does total internal reflection happen?

Only when light travels from the slower medium toward the faster one, that is from the larger index toward the smaller one, and only when the angle of incidence measured from the normal exceeds the critical angle. From water to air the critical angle is about forty nine degrees, from ordinary glass to air about forty two. Going the other way, from air into water or into glass, a refracted ray always exists and there is no critical angle at all, so looking for one is wasted time.

Does PHYS 101 optics at McGill use calculus?

No, and that is the point of the course. PHYS 101 is the algebra based introduction, closed to students who already have PHYS 131 or the equivalent cegep objective, so every result in the optics chapter is obtained with algebra, proportions and right angle trigonometry. Most solutions found online are written for a calculus based course and are therefore not usable: a solution that begins by differentiating is not a PHYS 101 solution even when its final number is correct.

Practise it

Corrected exercises: Geometric optics, mirrors and lenses, PHYS 101 at McGill

A method is proved on a paper, not on a sheet. The set for the same chapter takes each of these traps into a problem, with the solution written out step by step.

10 corrected exercises
100 points
150 minutes

Do the exercises

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Revision sheet: geometric optics, mirrors and lenses (PHYS 101)

The essentials

The convention, written once, never abandoned

The three equations, and nothing else

The rules in table form

One converging lens of focal length 12 cm, read off the object position

The mistakes that cost marks

1. The angle is given from the surface, not from the normal

2. The minus sign is erased from an image distance

3. The critical angle is hunted in the direction where it does not exist

4. The outgoing side of a mirror is taken to be that of a lens

5. The inverse sine is taken before dividing by the index

6. The focal length stays in centimetres inside a power

7. The magnifier is assumed to magnify more when it is pushed closer to the page

Which method to choose

Where the image lands, decided by the object position alone

What happens when a ray meets a surface

How the answer is expected to be written

An image problem, written the way the marker reads it

Check before you hand in

The typical problem, taken apart

A converging lens used as a magnifier, from the convention to the sentence

Learn by heart

Frequently asked questions

What does a negative image distance mean in optics?

How do I know if an image is real or virtual?

Do mirrors and lenses use the same equation?

When does total internal reflection happen?

Does PHYS 101 optics at McGill use calculus?

Corrected exercises: Geometric optics, mirrors and lenses, PHYS 101 at McGill

See also

Looking for a tutor in Montreal for this chapter?