Waves and Modern Physics 203-SN3-RE • CEGEP, Montreal
Revision sheet: geometric optics and optical instruments (203-SN3-RE)
CEGEP optics takes up the thin lens equation seen in high school, but applies it to chains of elements: a microscope, a telescope, a corrected eye. The difficulty is no longer the formula, it is knowing where the object of each stage is, and whether it is real or virtual.
This sheet collects what actually costs marks in assessments: the powers, the intermediate image, the two conventions for the real side of a lens and of a mirror, the correction of the eye, and the confusion between the magnification of an instrument and its resolving power, which depends only on the diameter. It also draws, in colour, the ray diagrams of the classic cases: the real image of a lens and of a concave mirror, the virtual images of the concave mirror, the convex mirror, the magnifying glass and the diverging lens, and the microscope to scale.
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The thread of the chapter
A single equation governs lenses AND mirrors. The whole chapter therefore comes down to two habits: respect the sign convention, and treat each instrument as a chain in which the image from the first lens becomes the object of the second.
•Power in dioptres: P=f1 with f in METRES. Two thin lenses in contact have POWERS that add, never their focal lengths.
Same object at 30 cm, same f=10 cm, same di=+15 cm: the lens forms its real image BEHIND it, the mirror IN FRONT of it, on the object side, where the light came back.
The word “real” does not name a side of space but a property: the rays truly cross there. It is because a lens transmits and a mirror sends back that this crossing changes sides.
Spherical mirrors: three rays, five positions
•Ray 1: parallel to the axis, it reflects through F. Ray 2: passing through F, it reflects parallel to the axis. Ray 3: passing through the centre of curvature C, it returns along itself. Two rays place the image, the third checks it.
•CONCAVE mirror, object beyond C: real image between C and F, inverted, reduced. Object on C: real image on C, inverted, same size. Object between C and F: real image beyond C, inverted, enlarged.
•Object on F: the reflected rays are parallel, there is no image, di1=0. Object between F and V: VIRTUAL image behind the mirror, upright, enlarged, the shaving mirror image.
•CONVEX mirror: F and C are behind it, and wherever the object is, the image is virtual, between V and F, upright and reduced: this is the rear-view mirror.
Two virtual images behind the mirror, where the dashed lines cross: the concave one, object between F and V, enlarges it 2.5 times; the convex one reduces it, here to 0.625 times, wherever the object is.
A virtual image is built with the EXTENSIONS of the reflected rays, dashed behind the mirror: no light passes there, it is the eye that places the source there.
Thin lenses: three rays, the same positions
•Ray 1: parallel to the axis, it emerges through F', or seeming to come from F' if the lens diverges. Ray 2: aimed at F, it emerges parallel to the axis. Ray 3: through the centre O, it is not deviated.
•CONVERGING lens, object beyond 2F: real image between F' and 2F', inverted, reduced (camera). Object on 2F: image on 2F', same size. Object between 2F and F: image beyond 2F', enlarged (projector).
•Object on F: emerging rays parallel, image pushed out to infinity. Object between F and O: VIRTUAL image on the object side, upright, enlarged: the magnifying glass.
•DIVERGING lens: its focal points are swapped, F' is on the object side, and the image is always virtual, between F' and O, upright and reduced.
Two virtual images on the object side, where the dashed lines cross: the converging lens, object between F and O, enlarges, this is the magnifying glass; the diverging lens reduces, wherever the object is.
The object focal point F is the boundary: an object beyond it gives a real image, an object inside it a virtual image. For the concave mirror, F plays exactly the same role, and 2F that of C.
The eye and the instruments
•Normal eye: far point at infinity, near point at 25 cm.
•NEARSIGHTEDNESS: the eye converges too much, the far point is at a FINITE distance. It is corrected with a DIVERGING lens whose image focal point coincides with that far point, so f=−dfar.
•FARSIGHTEDNESS: the near point is too far away, and it is corrected with a CONVERGING lens.
•Magnifying glass: angular magnification M=f25 with f in CENTIMETRES. Microscope: M=∣mobjective∣×Meyepiece. Telescope: M=feyepiecefobjective.
•In an instrument with two lenses, the image from the first is the OBJECT of the second: do2=L−di1, and this quantity can be NEGATIVE, which means a virtual object.
Afocal telescope, distances scaled down (f1=4f2): the beam entering at angle θ leaves parallel at θ′, and M=θθ′≈f2f1=4. The telescope multiplies an ANGLE.
A magnification is a ratio of ANGLES, not of lengths: it says how many times larger the object appears, whereas the lateral magnification compares two actual sizes. Many textbooks call both “M”, hence the confusion: here the lateral magnification is m and the angular magnification M.
What limits an instrument
•CHROMATIC aberration: n depends on colour, so blue converges closer than red. It is corrected by an achromatic doublet satisfying ν1P1+ν2P2=0.
•A MIRROR has NO chromatic aberration: reflection does not disperse. This is the reason large telescopes use mirrors.
•SPHERICAL aberration: marginal rays converge too soon. It is corrected by stopping down or by making the surface aspheric.
•Diffraction limit: θ≈D1.22λ. Sharpness depends on the DIAMETER of the instrument, never on its magnification.
•Lensmaker's equation: f1=(n−1)(R11−R21), where n−1 becomes nmediumn−1 if the lens is immersed in something other than air.
Top, the index depends on colour: blue converges before red. Bottom, a single colour but a spherical surface: the marginal rays (pink) cross the axis before the paraxial rays (green).
Magnifying beyond the diffraction limit only enlarges a blurry spot: this is so-called “empty” magnification. A 100 mm telescope resolves 1.4 arcseconds, whatever eyepiece is screwed onto it.
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.Chaining two lenses without going through the intermediate image
the whole question, 8 points, this is the long problem of the chapter
What not to write
“The object is 8.4 mm from the objective, so it is 8.4+193=201.4 mm from the eyepiece, and I apply the equation with that distance.”
What to write
“The image from the objective forms at di1=168 mm; IT is what serves as the object for the eyepiece, at the distance do2=193−168=25 mm.”
Why: Each stage sees only what the previous stage sends it. The real object has left the scene as soon as its rays have crossed the objective: it plays no part in the rest of the calculation.
2.Adding the focal lengths of two lenses in contact
the whole question, and a reversed conclusion
What not to write
“A +20 cm lens in contact with a −50 cm lens gives f=20−50=−30 cm: the combination is diverging.”
What to write
“It is the POWERS that add: P=0.201+−0.501=5.0−2.0=+3.0 D, so f=+33 cm: the combination is CONVERGING.”
Why: The addition of powers comes from the addition of the inverses of the focal lengths, exactly like resistors in parallel. Adding focal lengths has no physical meaning and here gives the opposite of the right result.
3.Looking for the real image of a mirror behind the mirror
3 points, and a wrong figure for the whole question
What not to write
“di=+30 cm, so the image forms 30 cm behind the mirror, as for a lens.”
What to write
“For a MIRROR, di>0 means an image IN FRONT of the mirror, on the object side: that is where the light comes back and where the rays cross.”
f=15 cm, object on C at 30 cm: di=+30 cm, and the real image forms IN FRONT of the mirror, on C, where the reflected rays cross. Behind it, there is only the hatching.
Why: The convention is the same, di>0 for a real image, but the light does not go to the same side. A mirror sends the light back where it came from: nothing real can form behind it.
4.Correcting nearsightedness with a converging lens
the whole question, 5 points
What not to write
“The nearsighted eye sees poorly at a distance, so it does not converge enough: we add a converging lens.”
What to write
“The nearsighted eye converges TOO MUCH: its far point is at a finite distance. We correct with a DIVERGING lens whose image focal point falls on that far point: for a far point at 50 cm, f=−50 cm and P=−2.0 D.”
Distances scaled down. Alone, the nearsighted eye brings the rays from infinity to a focus before the retina; the DIVERGING lens makes them seem to come from the far point, its focal point F', which the eye sees sharply.
Why: The job of the corrective lens is to bring infinity to the far point, where the eye can still accommodate. Only a diverging lens gives, for an object at infinity, a virtual image at a finite distance.
5.Believing that a high magnification gives a finer image
3 points on the discussion question
What not to write
“With a 4 mm eyepiece instead of a 25 mm one, the telescope magnifies six times more, so it separates double stars six times better.”
What to write
“The resolving power depends only on the DIAMETER: θ=D1.22λ. For D=100 mm, it is 6.7×10−6 rad, that is 1.4 arcseconds, whatever the eyepiece.”
The same two stars, seen through two instruments: on the left the diffraction spots are fine enough to be resolved, on the right they overlap. No eyepiece can recreate the two peaks.
Why: Beyond the diffraction limit, magnifying adds no information: you enlarge a blurry spot. Astronomers call this empty magnification, and it is why a telescope is described by its diameter.
6.Attributing chromatic aberration to a mirror
2 points on the theory question
What not to write
“The mirror telescope also suffers from chromatic aberration, like all optics.”
What to write
“Reflection does not DISPERSE: the angle of reflection does not depend on the colour. A mirror therefore has no chromatic aberration, only spherical aberration, corrected by a parabolic shape.”
Top, the lens refracts, hence disperses: blue and red have two focal points. Bottom, the mirror of the same focal length: reflection does not depend on colour, all the white light passes through the same F.
Why: Chromatic aberration comes from the variation of the index with wavelength, and the index only matters in REFRACTION. This was Newton's historical argument in favour of the reflecting telescope.
7.Using the lensmaker's equation without accounting for the medium
4 points, and a conclusion contrary to experiment
What not to write
“This glass lens has a focal length of 20 cm; immersed in water, it still has a focal length of 20 cm.”
What to write
“The factor (n−1)=0.50 becomes (nwatern−1)=1.331.50−1=0.128: the focal length goes to 78 cm, the lens converges almost four times less.”
Why: It is the CONTRAST of index that makes light converge, not the index of the glass alone. A lens of ice in water would deviate almost nothing, and this is also why vision is blurred underwater without a mask.
8.Applying the minimum deviation formula away from the minimum
the whole question, 5 points
What not to write
“The prism deviates by 42∘ at this incidence, so n=sin2Asin2A+42∘.”
What to write
“This formula is valid ONLY at minimum deviation, where the path is symmetric and r1=r2=2A. Away from the minimum, you must go through δ=i1+i2−A and two applications of Snell's law.”
Why: The symmetry of the path is an assumption, not a general property: it holds only for one particular incidence. It is precisely because δmin is a minimum that it can be found experimentally, by rotating the prism until the deviation stops decreasing.
Which method to choose
Chaining two optical elements
The position of the image formed by the first element, relative to the second
If the image from the first lens forms BEFORE the second → real object for the second: do2=L−di1>0
Example: microscope, 193−168=25 mm
If the image from the first would form AFTER the second → VIRTUAL object: do2=L−di1<0, and the equation applies as it stands
Example: a negative do2 is not a mistake, it is a configuration
If the image from the first falls exactly on the focal point of the second → the final image goes to infinity: di21=0
Example: this is the normal setting of a microscope and of a telescope
If the TOTAL magnification is asked for → product of the magnifications, m=m1×m2, signs included
Example: (−20)×(+5)=−100, final image inverted
If the two lenses are IN CONTACT → no chain: add the powers
Example: +5.0 D and −2.0 D give +3.0 D
If the problem gives an angular magnification → it is a ratio of ANGLES: magnifying glass f25, telescope feyefobj
Example: a 5.0 cm magnifier magnifies 5×
The distance between the two lenses is the only piece of data that links the two stages. A problem that gives it announces a chain; a problem that speaks of lenses in contact announces an addition of powers.
Which formula for which situation
The optical object named in the problem
If a thin lens or a spherical mirror alone → do1+di1=f1, then m=−dodi
Example: with f=2R for a mirror
If a single curved surface separating two media → spherical refracting surface: don1+din2=Rn2−n1
Example: the cornea, the bulging face of an aquarium
If the radii of curvature and the index of the glass are given → lensmaker's equation, with the factor nmediumn−1
Example: a lens in water converges four times less
If the problem is about the eye, far or near vision → locate the far or near point, then f=−dfar for nearsightedness
Example: far point at 50 cm: lens of −2.0 D
If the problem is about a magnifier, a microscope, a telescope → angular magnifications, multiplying stage by stage
Example: 20×10=200
If the problem is about sharpness, resolution, a double star → diffraction limit D1.22λ, magnification does not enter
Example: 100 mm of diameter resolve 1.4 arcseconds
A question about the QUALITY of an image is never handled with the thin lens equation: it assumes paraxial rays and monochromatic light. Aberrations and diffraction are precisely what it ignores.
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.
Working through a two-lens instrument
When to use it: The problem gives two focal lengths, a distance between the lenses, and an object position.
1Draw an in-line sketch with the positions marked, and state the convention: light travels from left to right.
2First stage: apply the equation with do1 and f1, obtain di1, then m1=−do1di1, and say where this intermediate image forms.
3Transfer: do2=L−di1, specifying the sign obtained and what it means, real object or virtual object.
4Second stage: same equation with f2, obtain di2 and m2.
5Conclude with the total magnification m=m1m2, with the nature and orientation of the final image.
Concluding sentence
“The objective gives a real image at di1=168 mm, inverted and 20 times larger. This image, located 25 mm from the eyepiece, coincides with its object focal point: the final image therefore forms at infinity, viewable without accommodation, with a total magnification of 200×.”
The trap: Using the object-to-eyepiece distance instead of the intermediate-image-to-eyepiece distance. This is the mistake that loses the entire question, even though the rest of the calculation is correct.
Marking: Often 3 points for the first stage, 2 for the transfer with its commented sign, 3 for the second stage, 2 for the total magnification and the nature of the final image.
Correcting a defect of the eye
When to use it: The problem gives a far point or a near point and asks for the corrective lens.
1Identify the defect from the point given: a far point at a finite distance signals nearsightedness, a near point that is too far away signals farsightedness.
2State what the lens must do: for nearsightedness, give an image of an object AT INFINITY located AT the far point.
3Translate: do=∞ gives di1=f1, with di=−dfar since the image is virtual, hence f=−dfar.
4Convert to power, P=f1 with f in metres, and check the sign: negative for nearsightedness, positive for farsightedness.
Concluding sentence
“The far point is at 50 cm: the lens must give, from infinity, a virtual image at 50 cm in front of the eye, so f=−0.50 m and P=−2.0 dioptres, a diverging lens.”
The trap: Taking di=+dfar. The image formed by the corrective lens is always VIRTUAL, on the same side as the object: its di is therefore negative, and it is this sign that gives the diverging lens.
Marking: 1 point for the defect identified, 2 for stating what the lens must do, 1 for the focal length, 1 for the power with its sign.
Check before you hand in
Five minutes of checking recover more marks than one more problem started in a hurry.
The sign of di translated into a side
Next to each di, write the word “in front” or “behind”, taking the element into account: a lens forms its real image behind, a mirror in front.
A di=+30 cm announced behind a mirror: the side contradicts the sign.
Powers rather than focal lengths
Any addition of elements in contact is done in dioptres. Check that the final focal length is not the sum of the focal lengths, and that its sign matches the sign of the total power.
+20 cm and −50 cm give +33 cm, never −30 cm.
The intermediate image located on the sketch
On a two-lens instrument, physically point to the intermediate image on the drawing before calculating the second stage, and measure its distance from the second lens.
An intermediate image at 168 mm in a tube of 193 mm: it falls 25 mm before the eyepiece, hence at its focal point.
The order of magnitude of a magnification
A school microscope magnifies 40 to 1000 times, a magnifying glass 2 to 10 times, an amateur telescope 30 to 200 times. A value outside these ranges signals a unit conversion error.
A magnification of 2000 for a magnifying glass: centimetres and metres were mixed in f25.
The diameter, sole judge of sharpness
On any question of resolution or sharpness, check that the answer involves D and λ, and never the magnification.
An answer that improves resolution by changing the eyepiece: magnification cannot separate what diffraction has merged.
The typical problem, taken apart
The microscope, stage by stage
A microscope consists of an objective of focal length f1=8.00 mm and an eyepiece of focal length f2=25.0 mm, a distance L=193 mm apart.
The specimen is placed at do1=8.40 mm from the objective. Determine the position and magnification of the intermediate image, then the position of the final image and the total magnification of the instrument.
To scale: the object at 8.40 mm, barely beyond F₁, gives at 168 mm an inverted intermediate image, 20 times larger, which falls on F₂: the rays leave the eyepiece PARALLEL.
Step 1
Objective: di11=f11−do11=8.001−8.401=67.20.40, so di1=168 mm.
Why
The object is placed just beyond the focal point, within 0.40 mm: it is this small margin that produces a very distant and therefore very enlarged image. The whole principle of the microscope is there.
Step 2
Magnification of the objective: m1=−do1di1=−8.40168=−20.0.
Why
The negative sign announces an INVERTED image, which will carry through to the final image. It is also why a slide moves in the opposite direction to what you see in the eyepiece.
Step 3
Transfer: the intermediate image is at do2=L−di1=193−168=25.0 mm from the eyepiece.
Why
This is the decisive step: the eyepiece no longer sees the specimen, it sees this image. The result is positive, so it is a real object for the second stage.
Step 4
Now do2=25.0 mm =f2: the object of the eyepiece is exactly at its focal point, so di21=f21−f21=0 and the final image forms at INFINITY.
Why
This is the normal setting of a microscope: the eye views at infinity, without accommodation, hence without fatigue. The tube length is no accident, it equals di1+f2.
Step 5
Angular magnification of the eyepiece: M2=f225=2.50 cm25.0 cm=10.0.
Why
The 25 in the numerator is the minimum distance of distinct vision, in centimetres: f2 must therefore be converted to centimetres, on pain of a factor of a thousand.
Step 6
Total magnification: M=∣m1∣×M2=20.0×10.0=200, that is 200×.
Why
The two stages multiply because each acts on what the previous one delivers. This is the value engraved on commercial objectives and eyepieces.
Step 7
Check using the tube length: mobjective=f1L=8.00168=21.0, close to the 20.0 obtained, the difference coming from L being counted from the focal point and not from the lens.
Why
This approximate formula is the one used in catalogues. Finding 20 or 21 confirms the order of magnitude; a result of 2 or 200 would signal a unit error on the millimetres.
The conclusion, written out
“The objective forms at 168 mm a real, inverted image, 20.0 times larger. This image falls 25.0 mm from the eyepiece, that is exactly on its focal point: the final image is therefore pushed out to infinity and viewed without accommodation. The total magnification is 20.0×10.0=200×.”
The classic mistake on this problem: Taking the initial object as the object of the eyepiece, at 201.4 mm. The calculation proceeds with no apparent anomaly and gives a magnification of a few units instead of two hundred.
Learn by heart
•do1+di1=f1 and m=−dodi: lenses AND mirrors.
•di>0: real image, BEHIND a lens, IN FRONT of a mirror.
•Ray diagram: ray 1 parallel then through F (F' for a lens), ray 2 through F then parallel, ray 3 through C for a mirror, through O for a lens.
•f>0: converging or concave. f<0: diverging or convex. Mirror: f=2R.
•POWERS add, not focal lengths: P=f1 in dioptres, f in metres.
•Two lenses: do2=L−di1, and a negative result means a virtual object.
•Magnifying glass f25 in cm, microscope ∣mobj∣×Meye, telescope feyefobj.
•A mirror has no chromatic aberration; a lens always has.
•Resolution: θ=D1.22λ, magnification never enters.
Frequently asked questions
Why is the real image of a mirror in front of it and that of a lens behind it?
Because a lens transmits light and a mirror sends it back. In both cases a real image forms where the rays truly cross, but those rays carry on behind the lens, whereas after reflection they head back toward the object side. The sign convention, for its part, is identical.
How do you calculate the focal length of two lenses in contact?
You add the powers, that is, the inverses of the focal lengths expressed in metres, then invert the result. Adding the focal lengths themselves is meaningless and can even give a diverging system where it is in fact converging, as with a twenty-centimetre lens in contact with a minus-fifty one.
How do you handle a microscope or a telescope with two lenses?
You apply the thin lens equation to the first lens, then use its image as the object of the second. The object distance for the second stage is the distance between the lenses minus the image distance of the first. The original object plays no part at all after the first stage.
Which lens corrects nearsightedness?
A diverging lens, of negative power. The nearsighted eye converges too much and sees sharply only out to a finite distance called the far point. The corrective lens must give, for an object at infinity, a virtual image located at that far point, which only a diverging lens can do.
Does more magnification let you see more detail?
No, beyond a certain limit. The sharpness of an instrument is set by diffraction and depends only on its diameter and the wavelength. Magnifying further only enlarges a blurry spot, which astronomers call empty magnification. This is why a telescope is described by its diameter.
Why do large telescopes use mirrors rather than lenses?
Because reflection does not disperse light: a mirror has no chromatic aberration, whereas a lens brings blue to a focus closer than red. A mirror can also be supported from behind, whereas a large lens is held only by its rim and deforms under its own weight.
Practise it
Corrected exercises: Geometric optics and optical instruments, 203-SN3-RE
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