Corrected exercises: geometric optics, mirrors and lenses (PHYS 101)

a) The figure gives $35°$ with the SURFACE. The normal is perpendicular to the surface, so the angle of incidence is $\theta_1=90°-35°=55°$. The law of reflection then gives $\theta_r=\theta_1=55°$, also from the normal. Writing $\theta_r=35°$ is the single most expensive line of the chapter: the number is not wrong by a little, it is the complement, and everything built on it follows the wrong ray.

b) Call $\theta$ the angle of incidence on the first mirror. The ray leaves it at $\theta$ from that normal, so it makes $90°-\theta$ with the first mirror surface. The second mirror is perpendicular to the first, so its normal is parallel to the first surface, and the ray therefore arrives on it at $90°-\theta$ from the second normal. It leaves at the same angle. Each reflection turns the ray through $180°-2\theta$ and $180°-2(90°-\theta)=2\theta$, and the two turns add to $180°$. Check with the figure: $\theta=55°$ gives $70°$ then $110°$, and $70°+110°=180°$. A ray turned through $180°$ goes back the way it came, parallel to itself. This is why a bicycle reflector and a road sign send the light straight back to the headlights, at any angle of approach.

c) A plane mirror puts the image as far behind the glass as the object is in front, so $1.8$ m behind. No light actually reaches that place: the rays only appear to come from it, so the image is VIRTUAL, and it cannot be caught on a screen held there. It is upright and the same size, so $m=+1$. A plane mirror is the limiting case $f\to\infty$ of the mirror equation, which gives $d_i=-d_o$ and $m=-d_i/d_o=+1$: the negative sign of $d_i$ is exactly the statement that the image is virtual.

d) Draw the ray from the top of your head that reaches your eye after one bounce. It hits the mirror at the midpoint between head and eye level, so at $\dfrac{1.70+1.60}{2}=1.65$ m. The ray from your feet hits at the midpoint between floor and eye level, so at $\dfrac{0+1.60}{2}=0.80$ m. Everything between those two points is seen, so the mirror needs to be $1.65-0.80=0.85$ m tall, exactly half your height, with its lower edge $0.80$ m above the floor, at half your eye height. Step back and the triangles get longer but stay similar, so the two midpoints do not move: the answer does not depend on the distance at all. That independence is the counter-intuitive part, and it is worth a full mark on its own.

a) Snell's law is $n_1\sin\theta_1=n_2\sin\theta_2$, both angles from the normal. So $\sin\theta_2=\dfrac{1.000\sin 55°}{1.333}=\dfrac{0.8192}{1.333}=0.6145$, and $\theta_2=37.9°$. Take the sine LAST: students who press the inverse sine on $0.8192$ and then divide get $41.5°$, which is wrong and looks plausible. Check the direction: water is slower than air, so the ray must bend toward the normal, and $37.9°<55°$ confirms it. The speed follows from $n=\dfrac{c}{v}$: $v=\dfrac{3.00\times 10^{8}}{1.333}=2.25\times 10^{8}$ m/s.

b) At the first face, $\sin r=\dfrac{1.000\sin 55°}{1.50}=0.5461$ and $r=33.1°$. At the second face the ray goes from glass back to air. The two faces are parallel, so their normals are parallel, and the angle of incidence inside the glass on the second face is the same $r=33.1°$. Snell's law there reads $1.50\sin 33.1°=1.000\sin\theta$, which is the first equation read backwards, so $\theta=55°$. The emerging ray therefore makes the same angle with the same direction of normal as the incident ray: it is parallel to it. A slab does not deviate light, it only shifts it sideways.

c) With $i=55°$ and $r=33.1°$, $e\dfrac{\sin(i-r)}{\cos r}=1.20\times\dfrac{\sin 21.9°}{\cos 33.1°}=1.20\times\dfrac{0.3728}{0.8377}=0.534$ cm, so $5.34$ mm. Two sanity checks that cost nothing: at normal incidence $i=r=0$ and the shift is zero, which is right, since a ray straight through a window is not displaced; and the shift is always smaller than the thickness, since $\sin(i-r)<\cos r$ here. A window pane shifts what you see by a few millimetres, which is exactly why you never notice it.

d) Take a ray leaving the coin at a very small angle $\theta_1$ to the vertical and refracting into the air at $\theta_2$. With small angles the horizontal offset of the exit point is $d\tan\theta_1$ and the eye traces the emerging ray back to a depth $d'$ with the same offset $d'\tan\theta_2$. So $d'=d\dfrac{\tan\theta_1}{\tan\theta_2}$, and for small angles tangent and sine are interchangeable, so $\dfrac{\tan\theta_1}{\tan\theta_2}\approx\dfrac{\sin\theta_1}{\sin\theta_2}=\dfrac{n_2}{n_1}=\dfrac{1}{1.333}$. Hence $d'=\dfrac{d}{n}=\dfrac{2.10}{1.333}=1.58$ m. The pool looks about three quarters of its real depth, which is why people misjudge the deep end.

e) Think of the beam as a marching rank rather than a line. The edge of the rank that crosses the surface first slows down first, the rest is still going at full speed, and a rank whose one end is slower pivots toward that end. Entering a slower medium the pivot turns the beam toward the normal; leaving it, toward the surface. No derivative is needed anywhere in this chapter, and none is accepted: $n_1\sin\theta_1=n_2\sin\theta_2$ is the whole story.

a) The critical angle is the incidence that puts the refracted ray flat along the surface, $\theta_2=90°$, so $n_1\sin\theta_c=n_2\sin 90°$ and $\sin\theta_c=\dfrac{n_2}{n_1}$. Water to air: $\sin\theta_c=\dfrac{1.000}{1.333}=0.7502$ and $\theta_c=48.6°$. Glass to air: $\sin\theta_c=\dfrac{1}{1.50}=0.6667$ and $\theta_c=41.8°$. The denser the first medium, the smaller the critical angle, which is why a diamond, $n=2.42$, keeps light trapped from $24.4°$ onwards and sparkles.

b) Ray 2 arrives at $60°$, above $48.6°$. Snell's law would demand $\sin\theta_2=1.333\sin 60°=1.154$, and no angle has a sine above $1$. There is no refracted ray, so all the energy goes into the reflected ray: total internal reflection. Going the other way, air into water, Snell gives $\sin\theta_2=\dfrac{\sin\theta_1}{1.333}\le 0.75$, which is always a legal sine, so a refracted ray always exists and there is no critical angle in that direction. Hunting for one is a guaranteed loss of time in an exam. Total internal reflection only happens from the SLOWER medium toward the faster one.

c) At the core to cladding boundary, $\sin\theta_c=\dfrac{1.48}{1.52}=0.9737$, so $\theta_c=76.8°$. Careful with the reference again: that angle is measured from the normal to the boundary, and the normal to the wall of a fibre is PERPENDICULAR to the axis. A ray that makes $76.8°$ with the normal makes $90°-76.8°=13.2°$ with the axis. So a ray must stay within $13.2°$ of the axis to be guided; steeper rays leak into the cladding and are lost within centimetres.

d) $NA=\sqrt{1.52^{2}-1.48^{2}}=\sqrt{2.3104-2.1904}=\sqrt{0.1200}=0.346$. The acceptance cone is defined by $n_{air}\sin\alpha=NA$, so $\sin\alpha=0.346$ and $\alpha=20.3°$. Light entering the flat end within $20.3°$ of the axis is refracted to within $13.2°$ inside and is then guided; light entering more steeply is not. That single number is what a catalogue sells, and it is why the end of a fibre has to be polished flat and square.

e) Draw the prism with its right angle at the bottom left and its hypotenuse facing up and to the right. A horizontal ray enters through the vertical face at normal incidence, so it goes straight in with no bending. It then meets the hypotenuse, whose normal is at $45°$ to the horizontal, so the angle of incidence there is $45°$. Since $45°>41.8°$, the ray is totally reflected, turns through $90°$ and leaves through the horizontal face, again at normal incidence, again without bending. Total reflection sends back essentially all the light, where a silvered mirror loses a few per cent at every bounce and tarnishes: that is why binoculars and periscopes use prisms.

a) $f=\dfrac{R}{2}=\dfrac{40.0}{2}=+20.0$ cm, positive because the mirror is concave, that is converging. $C$ is the centre of the sphere the mirror is a piece of, at $40.0$ cm; a ray aimed at $C$ hits the mirror head on and comes straight back. $F$ is halfway, at $20.0$ cm; rays arriving parallel to the axis leave through $F$.

b) $\dfrac{1}{d_i}=\dfrac{1}{f}-\dfrac{1}{d_o}=\dfrac{1}{20.0}-\dfrac{1}{30.0}=\dfrac{3-2}{60.0}=\dfrac{1}{60.0}$, so $d_i=+60.0$ cm. Then $m=-\dfrac{d_i}{d_o}=-\dfrac{60.0}{30.0}=-2.00$. Positive $d_i$ means the outgoing side, which for a mirror is the side the object is on, so the image is REAL, $60.0$ cm in front of the mirror, and the negative $m$ says it is inverted and twice as tall. The three rays of the solution figure meet exactly there, which is the check worth doing before trusting the arithmetic.

c) $\dfrac{1}{d_i}=\dfrac{1}{20.0}-\dfrac{1}{10.0}=\dfrac{1-2}{20.0}=-\dfrac{1}{20.0}$, so $d_i=-20.0$ cm and $m=-\dfrac{-20.0}{10.0}=+2.00$. The minus sign on $d_i$ is not an arithmetic slip, it is the answer: the reflected rays leave diverging, they never cross, and only their backward extensions meet, $20.0$ cm BEHIND the mirror. The image is virtual, upright and twice the size. This is the shaving mirror and the dentist mirror, and it only magnifies while the object stays inside $F$.

d) A convex mirror is diverging, so $f=-\dfrac{R}{2}=-20.0$ cm. Then $\dfrac{1}{d_i}=-\dfrac{1}{20.0}-\dfrac{1}{30.0}=\dfrac{-3-2}{60.0}=-\dfrac{5}{60.0}$, so $d_i=-12.0$ cm and $m=-\dfrac{-12.0}{30.0}=+0.400$. Virtual, upright, reduced to two fifths. A convex mirror does this for every object distance without exception, which is why it is the mirror on a truck and at a blind corner: a small, upright, very wide view.

e) Only b). A sheet of paper catches light that actually arrives on it. In b) the reflected rays genuinely cross $60.0$ cm in front of the mirror, so a paper held there is lit by a sharp inverted picture. In c) and d) the rays leave the mirror diverging and never cross anywhere; what the eye sees comes from extending them backwards, through the glass, where no light goes. That is the whole meaning of the sign of $d_i$, and it is why writing $d_i=20$ cm in c) because a distance cannot be negative destroys the physics while keeping the number.

a) $\dfrac{1}{d_i}=\dfrac{1}{15.0}-\dfrac{1}{45.0}=\dfrac{3-1}{45.0}=\dfrac{2}{45.0}$, so $d_i=+22.5$ cm, and $m=-\dfrac{22.5}{45.0}=-0.500$. Positive $d_i$ for a lens means behind it, so the image is real, $22.5$ cm behind, inverted and half size. The drawing agrees before any arithmetic: the object is beyond $2F$, so the image must fall between $F'$ and $2F'$, inverted and reduced. That is exactly how a camera works.

b) $\dfrac{1}{d_i}=\dfrac{1}{15.0}-\dfrac{1}{10.0}=\dfrac{2-3}{30.0}=-\dfrac{1}{30.0}$, so $d_i=-30.0$ cm and $m=-\dfrac{-30.0}{10.0}=+3.00$. The object is INSIDE the focal length, the emerging rays still diverge, and the image is virtual, $30.0$ cm in front of the lens, upright, three times the size. Hold a magnifying glass closer to the page than its focal length and this is what you see; move it beyond $f$ and the image flips and blurs, which is the moment the sign of $d_i$ changes.

c) $\dfrac{1}{d_i}=-\dfrac{1}{20.0}-\dfrac{1}{20.0}=-\dfrac{2}{20.0}$, so $d_i=-10.0$ cm and $m=-\dfrac{-10.0}{20.0}=+0.500$. Virtual, upright, half size, and a diverging lens gives nothing else whatever you do, which exercise 6 proves in two lines of algebra.

d) Power is $P=\dfrac{1}{f}$ with $f$ in METRES, and the unit, the diopter, is a reciprocal metre. Converging lens: $P=\dfrac{1}{0.150}=+6.67$ D. Diverging lens: $P=\dfrac{1}{-0.200}=-5.00$ D. Forgetting to convert the centimetres is the classic error and it multiplies the answer by one hundred, which no optician would ever write. The sign is kept: an optical prescription reading $-2.00$ D is a diverging lens, and that alone says the eye is short sighted.

e) It is true for a lens and false for a mirror. The convention does not talk about sides of the page, it talks about the OUTGOING side, the side the light goes on after meeting the device. Light passes through a lens, so the outgoing side is the far one; light bounces off a mirror, so the outgoing side is the near one. In both cases $d_i>0$ means a real image and $d_i<0$ a virtual one; only the geography changes. Write the convention at the top of your page, in words, and this question answers itself.

a) $d_o=36.0$: $d_i=\dfrac{36.0\times 12.0}{36.0-12.0}=\dfrac{432}{24.0}=18.0$ cm and $m=-\dfrac{18.0}{36.0}=-0.500$. $d_o=24.0$: $d_i=\dfrac{288}{12.0}=24.0$ cm and $m=-1.00$. $d_o=18.0$: $d_i=\dfrac{216}{6.00}=36.0$ cm and $m=-2.00$. All three have $d_o>f$, so $d_o-f>0$, so $d_i>0$: real and inverted, every time. Notice the middle line: at $d_o=2f$ the image comes back at $2f$ the other side, the same size upside down. That is the only position where a converging lens copies an object at scale $1$, and it is the fastest landmark to memorise.

b) At $d_o=12.0=f$ the denominator $d_o-f$ is zero and the fraction has no value. That is not a failure of the algebra, it is the physics: an object sitting exactly at the focus sends rays that leave the lens perfectly parallel to one another. Parallel rays never meet, so there is no image at any finite distance, and one says the image is at infinity. This is how a headlight and a lighthouse are built, the lamp sitting at the focus to send out a parallel beam.

c) $d_o=8.00$: $d_i=\dfrac{96.0}{8.00-12.0}=\dfrac{96.0}{-4.00}=-24.0$ cm and $m=-\dfrac{-24.0}{8.00}=+3.00$. $d_o=6.00$: $d_i=\dfrac{72.0}{-6.00}=-12.0$ cm and $m=+2.00$. Both have $d_o<f$, so $d_o-f<0$ and $d_i<0$: virtual and upright. And notice that moving the object closer, from $8.00$ to $6.00$ cm, makes the magnification SMALLER, from $3$ to $2$. A magnifier does not magnify more when you push it against the page; it magnifies most just inside the focus.

d) Write $f=-|f|$ with $|f|>0$, and keep $d_o>0$ for a real object. Then $d_i=\dfrac{d_o(-|f|)}{d_o+|f|}$. The numerator is negative and the denominator is a sum of two positive numbers, so $d_i<0$ for every $d_o$: the image is always virtual. The magnification is $m=-\dfrac{d_i}{d_o}=\dfrac{|f|}{d_o+|f|}$, which is positive, so upright, and which has a denominator strictly bigger than its numerator, so $0<m<1$: always reduced. Three conclusions, no numbers, no calculus, and it is worth writing them out because it turns four exam questions into one.

e) Now take a concave mirror, so $f>0$, with the object inside the focus, $0<d_o<f$. The denominator $d_o-f$ is negative while the numerator $d_o f$ is positive, so $d_i<0$: virtual, behind the mirror. And $m=-\dfrac{d_i}{d_o}=-\dfrac{f}{d_o-f}=\dfrac{f}{f-d_o}$, where the denominator $f-d_o$ is positive and SMALLER than $f$, so $m>1$: upright and enlarged. Check on the numbers of exercise 4: $f=20.0$ and $d_o=10.0$ give $m=\dfrac{20.0}{10.0}=2.00$, the value found there. This is the make-up mirror, and it explains why it stops magnifying and flips the moment you hold it further away than $f$.

a) First face: $\sin r_1=\dfrac{\sin 50.0°}{1.532}=\dfrac{0.7660}{1.532}=0.5000$, so $r_1=30.0°$. Then $r_2=A-r_1=60.0°-30.0°=30.0°$. Second face, glass to air: $\sin i_2=1.532\sin 30.0°=0.7660$, so $i_2=50.0°$. Deviation $D=i_1+i_2-A=50.0°+50.0°-60.0°=40.0°$. The path came out perfectly symmetric, $i_1=i_2$ and $r_1=r_2$, which is the position of minimum deviation: for the violet, this incidence happens to be exactly it.

b) Same three steps with $n=1.513$: $\sin r_1=\dfrac{0.7660}{1.513}=0.5063$ and $r_1=30.4°$; $r_2=60.0°-30.4°=29.6°$; $\sin i_2=1.513\sin 29.6°=0.7469$ and $i_2=48.3°$; $D=50.0°+48.3°-60.0°=38.3°$. Do NOT reuse the violet value of $r_1$ for the red, and do not reuse $r_2$ either: the whole point of dispersion is that every step depends on $n$.

c) The two beams leave at $i_2=50.0°$ and $i_2=48.3°$ from the same normal, and their deviations differ by $40.0°-38.3°=1.68°$. That is the entire spread of the visible spectrum from one prism, less than two degrees, which is why the screen has to be far away before the colours separate visibly. Violet is deviated MORE because the glass is slower for it, so its index is larger, so it bends more at each face.

d) At the second face the light goes from glass to air, the direction in which total internal reflection is possible. The critical angles are $\arcsin\dfrac{1}{1.532}=40.7°$ for the violet and $\arcsin\dfrac{1}{1.513}=41.4°$ for the red. Both $r_2$ values are close to $30°$, comfortably below, so both colours emerge. Worth checking every time: increase $i_1$ far enough and $r_2=A-r_1$ grows past the critical angle, the prism goes dark on that face, and a student who skips this check reports an exit angle for a ray that never came out.

e) Red is on the outside, at the top of the primary bow. In a drop, light refracts in, reflects once off the back, and refracts out. Violet, having the larger index, is bent more at each refraction, so it comes back at about $40°$ from the direction of the incoming sunlight while red comes back at about $42°$. The colour you receive from a given drop depends on the angle between that drop and the point opposite the Sun, so the red reaches you from drops HIGHER in the sky than the violet does. The bow is a set of nested cones, and the red one is the widest.

a) FALSE, and only the second half is wrong. A virtual image cannot be caught on a screen, because no light actually reaches the place where it seems to be; the rays only appear to come from there. But it is seen very well: your own reflection, the page under a magnifying glass and the traffic in a convex mirror are all virtual images. The correct version: a virtual image cannot be caught on a screen, but the eye sees it without difficulty, because the eye accepts diverging rays and traces them back for you.

b) FALSE, and this is the single most expensive habit of the chapter. The minus sign is not a mistake to be cleaned up, it is the content of the answer. With the convention written at the top of the page, $d_i=-15$ cm says the image lands on the INCOMING side, that is behind a mirror or in front of a lens, and therefore that it is virtual. Strip the sign and you also strip the physics: $m=-d_i/d_o$ then comes out negative instead of positive, and you report an inverted image where the object is upright. One sign, two wrong answers.

c) FALSE, the direction is backwards. Going from air, $n=1.000$, into water, $n=1.333$, Snell's law gives $\sin\theta_2=\dfrac{\sin\theta_1}{1.333}\le 0.750$, always a legal sine, so a refracted ray always exists and there is no critical angle. Total internal reflection needs the light to start in the SLOWER medium and head for the faster one. The correct version: total internal reflection happens when light goes from water into air at an angle of incidence greater than the critical angle $\theta_c=48.6°$.

d) FALSE, and the figure of exercise 5 shows why in one glance. Every point of the object sends light to the WHOLE surface of the lens, and every part of the lens sends some of it to the matching image point. Cover half the lens and each image point still receives light, just from half as much area: the image keeps its full shape and loses about half its brightness. The correct version: masking half a lens leaves the image complete and makes it about twice as dim. The three principal rays are a drawing convention, not the only rays that travel.

e) TRUE, and it is worth seeing where it comes from rather than learning it. A plane surface is a sphere of infinite radius, so $f=\dfrac{R}{2}\to\infty$ and $\dfrac{1}{f}=0$. The mirror equation becomes $\dfrac{1}{d_o}+\dfrac{1}{d_i}=0$, so $d_i=-d_o$, negative, therefore virtual, and at the same distance on the other side. Then $m=-\dfrac{d_i}{d_o}=+1$: upright, same size. Every statement in the sentence is one consequence of the same line.

a) The object is far away, so $d_o\to\infty$ and $\dfrac{1}{d_o}=0$. The image must land at the far point, on the same side as the object, so it is virtual and $d_i=-0.500$ m. The lens equation gives $\dfrac{1}{f}=0+\dfrac{1}{-0.500}=-2.00$, so $f=-0.500$ m and $P=\dfrac{1}{f}=-2.00$ D. A diverging lens, which is what the negative sign says. Note the shortcut hidden in the algebra: for short sight the focal length is just minus the far point distance.

b) The image is $50.0$ cm in front of the lens, on the same side as the incoming light, and it is virtual: no light gathers there, the rays merely leave the lens as if they came from there. But that is enough, because the eye behind the lens now receives light diverging from a point $50.0$ cm away, which is precisely its far point, the furthest thing it can still focus on the retina. The lens has not repaired the eye, it has moved the world inside the range the eye still has.

c) Now the object is at the reading distance, $d_o=+0.250$ m, and the image must be built at the near point, virtual and on the same side, so $d_i=-0.750$ m. Then $\dfrac{1}{f}=\dfrac{1}{0.250}+\dfrac{1}{-0.750}=4.00-1.333=2.667$, so $f=+0.375$ m and $P=+2.67$ D, a converging lens. The trap is the sign of $d_i$: write $+0.750$ and you get $P=5.33$ D, a number twice too large that an optician would refuse. The image is on the reader's side of the glasses, so $d_i$ is negative, always.

d) $f=\dfrac{1}{P}=\dfrac{1}{20.0}=0.0500$ m, that is $5.00$ cm. With the object at the focus the rays leave parallel, the image is at infinity, the eye is relaxed, and $M=\dfrac{25.0}{5.00}=5.00$. Bring the object slightly closer so that the virtual image lands at the near point and the eye works a little harder for one extra unit: $M=1+\dfrac{25.0}{5.00}=6.00$. A twenty per cent gain paid for with eye strain, which is why catalogue figures quote the relaxed value.

e) The sign IS the diagnosis. A negative power is a diverging lens, which pushes the image of a distant object nearer, to the far point: that is short sight, or myopia. A positive power is a converging lens, which pushes the image of a near object further away, to the near point: that is long sight, or hyperopia. A prescription reading $-2.00$ D and one reading $+2.67$ D therefore describe opposite problems, and the two numbers computed above are exactly those two cases.

a) The lens equation rearranged gives $f=\dfrac{d_o d_i}{d_o+d_i}$. Pair one: $\dfrac{30.0\times 59.2}{89.2}=19.91$ cm. Pair two: $\dfrac{40.0\times 40.6}{80.6}=20.15$ cm. Pair three: $\dfrac{60.0\times 29.5}{89.5}=19.78$ cm. The mean is $\dfrac{19.91+20.15+19.78}{3}=19.9$ cm. The spread is about $0.4$ cm, that is two per cent, which is what you expect when the sharp position of a screen is judged by eye: the picture stays acceptable over a centimetre or so. Reporting $19.945$ cm would claim a precision the bench does not have, so $f=19.9$ cm, three significant figures at most.

b) $m=-\dfrac{d_i}{d_o}=-\dfrac{40.6}{40.0}=-1.015$. So the image should be $1.015\times 2.00=2.03$ cm tall and inverted, and $2.03$ cm is exactly what the ruler read on the screen. That agreement is the real check of the experiment: the two independent roads to the same number, one through the distances and one through the heights, meet. And $d_o$ and $d_i$ being almost equal here is not a coincidence, it is the $2f$ and $2f$ position of exercise 6, which for $f=19.9$ cm sits at about $39.8$ cm.

c) $15.0$ cm is smaller than $f=19.9$ cm, so the object is inside the focal length. Then $d_i=\dfrac{d_o f}{d_o-f}=\dfrac{15.0\times 19.9}{15.0-19.9}=\dfrac{298.9}{-4.9}=-60.5$ cm. The negative sign says the image is on the incoming side, so it is virtual: the rays leave the lens still diverging, they never cross, and there is nowhere along the rail to put a screen. To see this image you look THROUGH the lens, the way you use a magnifying glass. Sliding the screen back and forth for ten minutes is the standard waste of a lab period, and one line of arithmetic predicts it.

d) Let $D=d_o+d_i$ be the fixed object to screen distance. The lens equation $\dfrac{1}{d_o}+\dfrac{1}{d_i}=\dfrac{1}{f}$ can be written $\dfrac{d_o+d_i}{d_o d_i}=\dfrac{1}{f}$, so $d_o d_i=fD$. Now $d_o$ and $d_i$ are two numbers whose sum is $D$ and whose product is $fD$: they are the roots of $x^{2}-Dx+fD=0$. Real positions exist only if the discriminant is not negative, $D^{2}-4fD\ge 0$, that is $D\ge 4f$. With $f=19.9$ cm this gives $D\ge 79.8$ cm: bring the screen closer than about $80$ cm from the object and no lens position gives a sharp picture. At $D=4f$ exactly the discriminant is zero, the two roots merge at $d_o=d_i=\dfrac{D}{2}=2f$, and $m=-\dfrac{d_i}{d_o}=-1$: the one image the same size as the object, upside down. Note that this whole proof is algebra, a sum, a product and a discriminant, and needs no derivative at any point.