PHYS 101 Introductory Physics, Mechanics • McGill University, Montreal
Revision sheet: refraction and dispersion, with a look ahead to mirrors and lenses (PHYS 101)
This sheet covers the light 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.
Beyond PHYS 101: mirrors, thin lenses and their equations are taught in PHYS 102, Introductory Physics - Electromagnetism, in the winter, and the PHYS 101 formula sheet does not give them. What IS on the PHYS 101 programme is the speed of light in a medium, v=nc, refraction written with the speeds, v1sinθ2=v2sinθ1, and dispersion: the first block below, the refraction traps and the surface tree. Everything about images is kept as a look ahead to PHYS 102.
One idea runs through all of it. Optics has no hard calculations, it has conventions, 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.
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Every question in this chapter is decided by a sign convention you adopt on the first line and never abandon. A negative di 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.
On your formula sheet: this chapter's lines, and when each one holds
•v=c/n: n≥1, so light is never faster in a material than in vacuum, and its wavelength there is nλ. With c=3.00×108 m/s: water, n=1.333, gives 2.25×108 m/s; acrylic, n=1.49, gives 2.01×108 m/s.
•Refraction, written with the speeds and with the indices CROSSED: v1sinθ2=v2sinθ1. Divide by c and it becomes n1sinθ1=n2sinθ2.
•So vsinθ is the same on both sides of every surface: the slower side holds the smaller angle, and through parallel faces the media in between drop out.
•Critical angle, toward the faster medium only: sinθc=v2v1, a speed over a speed, which tends to 90° when the two speeds become equal.
•Dispersion: one index, so one speed, per colour. Violet is slower than red in glass and is bent more; the frequency never changes at a surface, the wavelength λ=fv does.
Water, acrylic, air: 30.0°, 26.6°, 41.8°. The angle shrinks where light slows down, and the exit angle is the one water to air would give, as if the wall were not there.
The sheet gives nothing else on rays: no mirror equation, no lens equation, no magnification. Those belong to PHYS 102, and the blocks on images below are a look ahead.
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 do>0.
•di>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.
•di<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=2R with the same sign.
•m=−dodi=hohi. 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 di>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: θreflected=θincident, both from the normal. It happens at every surface, even when refraction happens too.
•Refraction: n1sinθ1=n2sinθ2, both from the normal, with n=vc≥1. Toward a slower medium the ray bends TOWARD the normal.
•Critical angle, from the slower medium toward the faster one only: sinθc=n1n2. Past θc there is no refracted ray at all.
•Image, a look ahead to PHYS 102: do1+di1=f1, the same letters for mirrors and for lenses, often more useful as di=do−fdof, where the sign of do−f decides everything.
•Power, a look ahead to PHYS 102: P=f1 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.
Curved mirrors: the three principal rays
•Ray 1 comes in parallel to the axis and reflects through the focus F.
•Ray 2 passes through F on the way in and reflects parallel to the axis.
•Ray 3 passes through the centre of curvature C, meets the mirror head on and comes straight back on itself.
•Where the reflected rays really cross, the image is real; where only their dashed backward extensions meet, behind the mirror, it is virtual. Two rays locate the image, the third checks it.
Concave mirror, f=20 cm, object at 60 cm, beyond C: rays 1 in blue, 2 in green and 3 in violet really cross at di=+30 cm, between C and F, in front of the mirror. Real, inverted and half the size, m=−0.5.
The mirror is drawn as a straight line through its vertex V, hatched on the back. That is the approximation of the course: the rays stay close to the axis, where the sphere and its tangent plane cannot be told apart.
Concave mirror, object inside F: the shaving mirror
•Inside F the reflected rays leave diverging and never cross in front of the mirror.
•Their dashed backward extensions meet BEHIND the mirror: the image is virtual, upright and enlarged, and di<0.
•Move the object outward across F and the image jumps to the front, real and inverted: the sign of do−f decides, exactly as for a lens.
Concave mirror, f=20 cm, object at 10 cm: the three reflected rays spread apart, and their dashed extensions meet 20 cm behind the mirror. So di=−20 cm and m=+2, the make-up and shaving mirror.
Ray 2 can no longer pass through F on the way in, since F is behind the object: it is drawn as if it CAME from F, dashed back to it. Ray 3 likewise lines up with C.
Convex mirror: always virtual, upright and reduced
•A convex mirror is diverging: f=−2R<0, and F and C sit BEHIND it.
•Ray 1 reflects as if it came from F; ray 2 aims at F and reflects parallel; ray 3 aims at C and comes straight back.
•The dashed extensions always meet behind the mirror, between V and F: virtual, upright, reduced, for every object distance, with no exception to look for.
Convex mirror, f=−20 cm, object at 30 cm: the reflected rays fan out, and their dashed extensions meet 12 cm behind the mirror. So di=−12 cm and m=+0.4, the truck mirror and the shop corner mirror.
Because the image always lands between V and F, a convex mirror squeezes a very wide field of view into a small upright picture.
Thin lenses: the three principal rays
•Ray 1 comes in parallel to the axis and leaves through the image focus F′, on the far side.
•Ray 2 passes through the object focus F, on the near side, and leaves parallel to the axis.
•Ray 3 passes through the centre O and goes straight on, without bending.
•For a diverging lens F and F′ swap sides, and rays 1 and 2 only AIM at them or seem to come from them.
Converging lens, f=12 cm, object at 36 cm, beyond 2F: the three rays really cross at di=+18 cm, between F′ and 2F′, so the image is real, inverted and half the size.
The lens is drawn as a straight line with arrowheads: pointing outward for a converging lens, inward for a diverging one. All the bending is done on that line.
Diverging lens: always virtual, upright and reduced
•Ray 1 comes in parallel and leaves as if it came from F′, which sits on the incoming side.
•Ray 2 aims at F, on the far side, and leaves parallel to the axis.
•Ray 3 crosses O without bending. The dashed extensions meet on the incoming side, between F′ and the lens: virtual, upright and reduced, whatever the object distance.
Diverging lens, f=−20 cm, object at 20 cm: the outgoing rays diverge, and their dashed extensions meet 10 cm in front of the lens. So di=−10 cm and m=+0.5, the last row of the object position table.
This is the lens of a short sighted eye: it moves the image of a distant object in, to the far point, where the eye can still focus.
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
do>2f
f<di<2f
real, inverted, reduced
Example: do=36 cm gives di=18 cm and m=−0.5
do=2f
di=2f
real, inverted, same size
Example: do=24 cm gives di=24 cm and m=−1
f<do<2f
di>2f
real, inverted, enlarged
Example: do=18 cm gives di=36 cm and m=−2
do=f
nowhere
the rays leave parallelno image at all
Example: do=12 cm gives do−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.
do<f
di<0
virtual, upright, enlarged
Example: do=8 cm gives di=−24 cm and m=+3
any do, with f<0
di<0 always
virtual, upright, reduced
Example: do=20 cm with f=−20 cm gives di=−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 θ1=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 di=−30 cm, but a distance cannot be negative, so di=30 cm, image behind the lens.”
What to write
“di=−30 cm, so the image is on the incoming side, 30 cm in front of the lens: it is VIRTUAL, and m=−10−30=+3, so upright and three times the size.”
With f=15 cm and the object at 10 cm, inside the focal length, the rays leave DIVERGING and only their dashed extensions meet, 30 cm in front of the lens: di=−30 cm, m=+3, and 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: sinθ2=1.333sin70°=0.705, that is θ2=44.8°. Total internal reflection is only possible from water toward air, past θc=48.6°.”
Why: sinθc=n1n2 only has a solution when n2<n1. 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
“di=+60 cm on a concave mirror, so the image is 60 cm BEHIND the mirror.”
What to write
“di=+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.”
With f=20 cm and the object at 30 cm, the reflected rays come BACK to the object side and really cross 60 cm in front of it: for a mirror, di=+60 cm means a real image on the incoming side.
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 di>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
“sinθ2=1.333sin55°, so θ2=1.33355°=41.3°.”
What to write
“sinθ2=1.333sin55°=1.3330.8192=0.6145, and only then θ2=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 speeds are put on the same side as their angles
the whole question, and every angle built on it
What not to write
“v1sinθ1=v2sinθ2, so from air into water at 40°, sinθ2=2.253.00sin40°=0.857 and θ2=59.0°.”
What to write
“On the sheet the indices cross, v1sinθ2=v2sinθ1, so sinθ2=3.002.25sin40°=0.482 and θ2=28.8°, toward the normal.”
Air into water at 40°: the sheet's form gives 28.8°, toward the normal (green); the speeds swapped give 59.0°, away from it (red), which no ray ever does.
Why: Divided by c, the sheet's form is n1sinθ1=n2sinθ2: the slower side, with the larger index, holds the smaller angle. The swapped version bends light AWAY from the normal as it slows down, and past 48.6° it even asks for a sine above 1.
7.The critical angle of a wall is taken in the wrong medium
the conclusion of the question, often the last two marks
What not to write
“Acrylic to air has θc=42.2°, so a ray in the aquarium water at 45° is trapped.”
What to write
“vsinθ is the same on every face, so the wall drops out: from the water nothing gets out past sinθw=cvw, that is 48.6°, and a ray at 45° gets out.”
Why: 42.2° is the critical angle INSIDE the acrylic. Coming from the water at 45°, a ray reaches the acrylic to air face at only 39.2° and leaves at 70.5°. Through parallel faces only the first and last media count.
8.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=151=0.067 D.”
What to write
“f=15 cm =0.150 m, so P=0.1501=+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.
9.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, do=8 cm gives m=+3 but do=6 cm gives only m=+2: the magnification is m=f−dof, which grows as do approaches f, not as do approaches zero.”
Why: The formula settles it in one line, and the limit cases confirm it: pressed flat on the page (do→0) the lens does nothing at all and m→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 do with f first, and only then with 2f. Everything follows from the SIGN of do−f in di=do−fdof, so that single subtraction is the whole decision.
The six object distances of the table, read against F at 12 cm and 2F at 24 cm: which side of F the mark falls on is the entire decision.
If f<0, the device is diverging → di<0 whatever do: virtual, upright, reduced, no exception to look for
Example: f=−20 cm and do=20 cm give di=−10 cm, m=+0.5
The numerator dof is negative and the denominator do−f=do+∣f∣ is positive, so the quotient cannot be positive. Two lines of algebra replace four exam questions.
If f>0 and do>2f → real, inverted and reduced, with the image between f and 2f
Example: f=12 cm and do=36 cm give di=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<do<2f → real, inverted and enlarged, with the image beyond 2f
Example: f=12 cm and do=18 cm give di=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 do=f exactly → no image at any finite distance, the rays leave parallel
Example: f=12 cm and do=12 cm give do−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 do<f → virtual, upright and enlarged, on the incoming side
Example: f=12 cm and do=8 cm give di=−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 di 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 n2>n1, 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°
sinθ2=n2n1sinθ1 with a factor below one is always a legal sine.
If n2<n1 and θ1<θc → 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 θc=arcsinn1n2 BEFORE the refraction angle, so you know the answer exists before you produce it.
If n2<n1 and θ1≥θc → no refracted ray at all: total internal reflection, and all the energy leaves at θ1 on the same side
Example: water to air at 60° needs sinθ2=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.
1State the convention in words, once: “Distances from the lens; di>0 behind it, so real; f>0 converging; m=−di/do.” One line, and it is usually worth a mark on its own.
2List the data WITH their signs: “f=+15.0 cm (converging), do=+10.0 cm, ho=2.00 cm.” A sign written here is a sign you will not lose later.
3Place the object against F and 2F and predict: “do<f, so I expect a virtual, upright, enlarged image, that is di<0 and m>1.” This is the sentence that catches an arithmetic slip before it costs anything.
4Isolate before substituting: “di1=f1−do1, so di=do−fdof”, then put the numbers in. Substituting first turns one algebra error into three.
5Turn the signs back into words, explicitly: “di=−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 di and the drawing agree?
Look at your ray diagram. If the outgoing rays actually cross somewhere, di must be positive; if they leave diverging and only their dashed extensions meet, di 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 di is impossible.
Do the signs of di 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 di>0 demands m<0, and di<0 demands m>0. This is a one second check and it catches the most common sign error of the chapter.
di=−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 θ2 with θ1 and you know in one look whether the two indices were swapped in Snell's law.
Air to glass with θ1=55°: a θ2=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=151=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≥4f is a discriminant, D2−4fD≥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, di>0 behind it and therefore real, f>0 for a converging lens, m=−dodi. Data: f=+15.0 cm, do=+10.0 cm, ho=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
do=10.0 cm is smaller than f=15.0 cm, so the object sits INSIDE the focal length. Expect a virtual image, hence di<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
di1=f1−do1=15.01−10.01=30.02−3=−30.01, so di=−30.0 cm.
Why
Isolate di1 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=−dodi=−10.0−30.0=+3.00, so hi=mho=3.00×2.00=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=0.1501=+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.
•di>0 means real and on the outgoing side; di<0 means virtual and on the incoming side. The outgoing side is behind a lens and in front of a mirror.
•m=−dodi: positive upright, negative inverted, ∣m∣>1 enlarged. For one lens or one mirror, real and inverted always go together.
•f=2R, positive for a concave mirror and a convex lens, negative for a convex mirror and a concave lens.
•di=do−fdof: the sign of do−f decides the whole answer, so compute that subtraction first.
•On the sheet: v1sinθ2=v2sinθ1, indices crossed, and sinθc=v2v1, toward the faster medium only. The slower side holds the smaller angle.
•n=vc≥1; toward a larger index the ray bends toward the normal and the angle decreases.
•sinθc=n1n2, and it only exists going from the slower medium toward the faster one. Water to air: 48.6°. Glass to air: 41.8°.
•P=f1 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≥4f.
•No derivative and no integral anywhere: the whole chapter is algebra and right angle trigonometry.
•Mirrors, thin lenses, magnification and the eye are a look ahead to PHYS 102: the PHYS 101 formula sheet does not give their equations.
Frequently asked questions
Are mirrors and lenses on the PHYS 101 exam at McGill?
Not according to the 2026 formula sheet, the one students annotate and bring to the exam. For light it gives only the speed of light in a medium, refraction written with the speeds, two slit interference, single slit diffraction and the Rayleigh criterion. Mirrors, thin lenses, magnification and optical instruments are taught in PHYS 102, Introductory Physics - Electromagnetism, in the winter. This sheet keeps them as a look ahead; your own course outline has the final word.
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.
Corrected exercises: Refraction and dispersion, with a look ahead to 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.
Something missing from this sheet?A method, a trap, an exercise to practise it, a mistake you spotted: tell me, I read every request.
Sending goes through your free space: that is how I can answer you and tell you when the exercise is added. An email is enough, no password. Create your space.
Rather not create an account? Write to me directly.
Get in touch for a first session. We go back over the points of method that lose marks in an assessment, then put them to the test on problems set at the real level of the exam.
