Corrected exercises: refraction and dispersion, with a look ahead to mirrors and lenses (PHYS 101)

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.

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.

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.

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.

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.

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.

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.

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.

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.

Replace each speed by $v=\dfrac{c}{n}$: $\dfrac{c}{n_1}\sin\theta_2=\dfrac{c}{n_2}\sin\theta_1$.

Multiply both sides by $\dfrac{n_1 n_2}{c}$: $n_2\sin\theta_2=n_1\sin\theta_1$, which is Snell's law of a). The two forms are one law: the sheet's form says that the sine of each angle is proportional to the speed on its side.

Pitfall: the indices cross over. On the sheet $v_1$ multiplies $\sin\theta_2$; writing $v_1\sin\theta_1=v_2\sin\theta_2$ puts the larger angle in the slower medium, the opposite of what light does.

Speeds: $v_w=\dfrac{3.00\times 10^{8}}{1.333}=2.25\times 10^{8}$ m/s in the water and $v_a=\dfrac{3.00\times 10^{8}}{1.49}=2.01\times 10^{8}$ m/s in the acrylic.

Water to acrylic: $\sin\theta_a=\dfrac{v_a}{v_w}\sin 30.0°$ with $\dfrac{v_a}{v_w}=\dfrac{2.013}{2.251}=0.894$, so $\sin\theta_a=0.894\times 0.500=0.447$ and $\theta_a=26.6°$.

The light slows down and the angle shrinks: the ray bends toward the normal, as e) says it must.

Acrylic to air: $\sin\theta_{air}=\dfrac{c}{v_a}\sin\theta_a=\dfrac{3.00}{2.013}\times 0.447=0.666$, so $\theta_{air}=41.8°$.

Check: straight from water to air, $\sin\theta=\dfrac{c}{v_w}\sin 30.0°=\dfrac{3.00}{2.251}\times 0.500=0.666$ gives the same $41.8°$. A wall with parallel faces moves the point where the ray leaves, not its final direction, exactly like the glass slab of b).

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.

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.

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.

$NA=\sqrt{1.52^{2}-1.48^{2}}=\sqrt{2.3104-2.1904}=\sqrt{0.1200}=0.346$. 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.

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.

At the critical angle the refracted ray grazes the surface, $\theta_2=90°$, so $v_1\sin 90°=v_2\sin\theta_c$:

A sine cannot exceed $1$, so a solution exists only if $v_1<v_2$: the light must head into the faster medium. This is the rule of b), recovered without a single index.

Dimensions: a speed over a speed, a pure number, as a sine must be.

Limit case: $v_2\to v_1$, two media in which light travels equally fast, gives $\sin\theta_c\to 1$ and $\theta_c\to 90°$: no total reflection at any real angle, which is right, since optically there is no surface left.

Sense of variation: the faster the second medium compared with the first, the smaller $\theta_c$, so the more light is trapped. Diamond, where light is slowest, has the smallest critical angle of a).

Acrylic to air, with the unrounded speed of exercise 2 f): $\sin\theta_c=\dfrac{v_a}{c}=\dfrac{2.013\times 10^{8}}{3.00\times 10^{8}}=0.671$, so $\theta_c=42.2°$ ($42.1°$ with the rounded $2.01$).

From the water: on each face of the wall the speed form keeps $\dfrac{\sin\theta}{v}$ the same, as in exercise 2 f), so $\dfrac{\sin\theta_w}{v_w}=\dfrac{\sin\theta_{air}}{c}$ and the acrylic drops out.

No ray gets out when $\sin\theta_{air}$ would have to exceed $1$, that is $\sin\theta_w\ge\dfrac{v_w}{c}=\dfrac{1}{1.333}$, so $\theta_w\ge 48.6°$, the water to air value of a).

Pitfall: taking $42.2°$ as the limit in the water. It is the limit INSIDE the acrylic; a ray at $42.2°$ in the water reaches the acrylic to air face at only $36.9°$, and gets out.

$f=\dfrac{R}{2}=\dfrac{40.0}{2}=+20.0$ 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 $F$.

$\dfrac{1}{d_i}=\dfrac{1}{f}-\dfrac{1}{d_o}=\dfrac{1}{20.0}-\dfrac{1}{30.0}=\dfrac{3-2}{60.0}=\dfrac{1}{60.0}$, 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.

$\dfrac{1}{d_i}=\dfrac{1}{20.0}-\dfrac{1}{10.0}=\dfrac{1-2}{20.0}=-\dfrac{1}{20.0}$, 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 $F$.

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.

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.

Ray 1, in blue, comes in parallel to the axis and reflects through $F$; ray 2, in green, passes through $F$ and reflects parallel to the axis; ray 3, in violet, passes through $C$, meets the mirror head on and comes straight back. The three reflected rays really cross $60.0$ cm in front of the mirror, below the axis, so BEYOND $C$, which sits at $40.0$ cm. An object between $C$ and $F$ always gives an image beyond $C$, real, inverted and enlarged: it is the case of an object beyond $C$ with object and image swapped. The first tracing below is this construction, and the two others show why c) and d) give nothing a sheet of paper could catch: their reflected rays fan out, and only the dashed extensions meet, behind the mirror.

$\dfrac{1}{d_i}=\dfrac{1}{15.0}-\dfrac{1}{45.0}=\dfrac{3-1}{45.0}=\dfrac{2}{45.0}$, 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.

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

$\dfrac{1}{d_i}=-\dfrac{1}{20.0}-\dfrac{1}{20.0}=-\dfrac{2}{20.0}$, 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.

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.

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.

Ray 1, in blue, enters parallel to the axis and leaves through $F'$, on the far side; ray 2, in green, passes through $F$ on the way in and leaves parallel to the axis; ray 3, in violet, crosses the centre $O$ of the lens without bending. The three really cross $22.5$ cm behind the lens, below the axis, so between $F'$ at $15.0$ cm and $2F'$ at $30.0$ cm: an object beyond $2F$ always gives an image between $F'$ and $2F'$, real, inverted and reduced, which is the camera. The first tracing below is this construction; in the other two the outgoing rays diverge and only their dashed extensions meet, in front of the lens, which is what $d_i<0$ says in b) and c).

$d_o=36.0$: $d_i=\dfrac{36.0\times 12.0}{36.0-12.0}=\dfrac{432}{24.0}=18.0$ 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.

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.

$d_o=8.00$: $d_i=\dfrac{96.0}{8.00-12.0}=\dfrac{96.0}{-4.00}=-24.0$ 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.

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.

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 $f$.