chapter_26

# Ch.26: Geometrical optics (Apr 28,30)

## Units Ch.26

### Non-SI units Ch.26

• Centimeter (cm) - no need to convert to SI in this chapter (as long as all lengths are in the same units)

## Do-it-yourself videos

Video 4.1: Step-by-step instructions on drawing ray diagrams:
Concave mirror
Video 4.2: Concave lens and convex mirror

## Equation Sheet Ch.26

The Law of Reflection

$\theta_r=\theta_i$

$\theta_r=$ angle of reflection
$\theta_i=$ angle of incidence
Focal Length for a Convex Mirror of Radius R

$f=-\frac{1}{2}R$

SI unit: meter (m)
Focal Length for a Concave Mirror of Radius R

$f=\frac{1}{2}R$

SI unit: meter (m)
The Mirror Equation

$\frac{1}{d_o}+\frac{1}{d_i}=\frac{1}{f}$

$f$ is positive for concave mirrors.
$f$ is negative for convex mirrors.
Image Distance $d_i$ is positive for images in front of a mirror (real images).
$d_i$ is negative for images behind a mirror (virtual images).
$d_o$ is positive for objects in front of a mirror (real objects).
$d_o$ is negative for objects behind a mirror (virtual objects).

$\frac{h_o}{-h_i}=\frac{d_o}{d_i}$
$\frac{h_o}{-h_i}=\frac{d_o-R}{R-d_i}$

Magnification, m
$m=\frac{h_i}{h_o}=-\frac{d_i}{d_o}$ Magnification $m$ is positive for upright images.
$m$ is negative for inverted images.
The height of an image
$h_i= -\left(\frac{d_i}{d_o}\right)h_o$
The Refraction of Light
a wave propagates from a medium in which its speed is $v_1$ to another in which its speed is $v_2<v_1$

$\frac{\sin \theta_1}{v_1}=\frac{\sin \theta_2}{v_2}$
Definition of the Index of Refraction, $n$
$v=\frac{c}{n}$ $v=$ speed of light in a given medium
Snell's Law
$n_1\sin \theta_1=n_2 \sin \theta_2$
Critical Angle for Total Internal Reflection, $\theta_c$
$\sin \theta \leq 1$, $n_1 \geq n_2$
$\sin \theta_c=\frac{n_2}{n_1}$
Total Polarization, Brewster's Angle, $\theta_B$
Reflected light is completely polarized when the reflected and refracted beams are at right angles to one another.
The direction of polarization is parallel to the reflecting surface.

$\tan \theta_B=\frac{n_2}{n_1}$
The Thin-Lens Equation

$\frac{1}{d_o}+\frac{1}{d_i}=\frac{1}{f}$

Equations to the right are only used to derive the thin-lens equation above in this particular case. The thin-lens equation above is true for all thin lenses, however.
$\frac{h_o}{f}=\frac{-h_i}{d_i-f}$
$\frac{h_o}{d_o}=\frac{-h_i}{d_i}$
Magnification, $m$

$m=-\frac{d_i}{d_o}$
$f$ is positive for converging (convex) lenses.
$f$ is negative for diverging (concave) lenses.
$m$ is positive for upright images (same orientation as object).
$m$ is negative for inverted images (opposite orientation of object).
$d_i$ is positive for real images (images on the opposite side of the lens from the object).
$d_i$ is negative for virtual images (images on the same side of the lens as the object).
Object Distance $d_o$ is positive for real objects (from which light diverges).
$d_o$ is negative for virtual objects (toward which light converges).

## Examples Ch.26

### Problem 4.1

A fireman moving at a constant speed starting at point A (1 minute shortest distance from the river) needs to get water at point M and bring it to the house at point B (2 minutes shortest distance from the river, horizontal distance between A and B is 3 minutes).

1. Tabulate the fireman's travel time for the following choices of x (the horizontal displacement between A and M):
1. 0.8 min
2. 0.9 min
3. 1 min
4. 1.1 min
5. 1.2 min
2. Plot the fireman's travel time as a function of x

Steps:

1. Noting that the horizontal distance between M and B is $3-x$, the total travel time is (in minutes):
• $\sqrt{1+x^2}+\sqrt{2^2+(3-x)^2}$. Using this formula,
x (min) Total travel time (min)
0.8 4.25
0.9 4.24
1.0 4.24
1.1 4.25
1.2 4.25

### Problem 4.2

A customer, trying a new hat and new shoes, is standing 1 m away from a mirror mounted on a vertical wall. The customer's height (including the hat) is 1.7 m, and the vertical distance from the top of the hat to the eye level is 12 cm.

1. How tall should be the mirror to enable seeing both the shoes and the hat?
2. How far off the floor should the mirror be mounted?

Steps:

1. On the diagram below,
• The triangle $EM_1M_2$ is proportional to the triangle $ET'B'$ with a scaling factor $\frac{1}{2}$, because the distance from the mirror to the object is same as the distance from mirror to the image behind the mirror (the heights of the two triangles along the dashed line form a ratio $\frac{1}{2}$).
• $M_1M_2=\frac{1}{2}T'B'$
• Here $T'B'$ is the height of the object, so the mirror length is $M_1M_2=\frac{1.7\,{\text m}}{2}$ = 0.85 m
2. To find the height to hang (from the floor to the bottom edge of the mirror)
• notice that triangle $EBB'=2\times M_2M_3B'$
• so $M_2M_3=$ $\frac{1}{2}EB=0.5\cdot 1.58$ $=0.79$ m

### Problem 4.3

What is the ideal width of a flat rear-view mirror inside a car, if the distance from the front window to the rear window of the car is L, the width of the rear window is W, and the distance from the driver's eyes to the mirror is d ?

Steps:

1. Notice that this is another similar triangle problem:
• triangle $EM_1M_2$ is approximately similar to triangle $EW_1W_2$
2. For triangle $EM_1M_2$, the distance from eyes to the mirror is $d$.
3. For triangle $EW_1W_2$, the total distance is $d + L$, where $L$ is the distance from the mirror to the window.
4. We generate a ratio:
• $\frac{d}{d+L}=\frac{\text{mirror width}}{\text{window width}}$

### Problem 4.4

The object of height $h_o$ is at a distance $\;d_o\!=\!\frac{1}{2}f\;$ from the concave mirror of focal distance $f$. Find the image distance, size and other properties.

Steps:

1. Solve the thin lens equation for $d_i$, the sign of $d_i$ determines if image is real or virtual
• $\frac{1}{f}=$ $\frac{1}{d_o}+\frac{1}{d_i}$
• After plugging in known values:
• $\frac{1}{f}=\frac{1}{0.5\,f}+\frac{1}{d_i}$
• Multiply both the numerator and denominator in $\frac{1}{0.5f}$ both by 2 to isolate $f$ as the common denominator:
• $\frac{1}{f}=\frac{2}{2\cdot 0.5\,f}+\frac{1}{d_i}$
• $\frac{1}{f}=\frac{2}{f}+\frac{1}{d_i}$
• Isolate $\frac{1}{d_i}$:
• $\frac{1}{f}-\frac{2}{f}=\frac{1}{d_i}$
• $\frac{1}{d_i}=-\frac{1}{f}$
• Take the inverse to solve for $\frac{1}{d_i}$:
• $d_i= -f$
• Because $\frac{1}{d_i}$ is a negative number (here $f$ is the positive focal length of a concave mirror), we know the image produced is virtual, on the back side of the mirror
2. Solve magnification equation for $m$, the sign of $m$ determines if image is enlarged or reduced
• $m=-\frac{d_i}{d_o}$
• plug in known values:
• $m=-\frac{-f}{0.5\,f}$
• double negative makes a positive and $\frac{1}{0.5}\!=\!2$, so
• $m=+2$
• Because the magnification is positive we know the image is enlarged and upright.
• Also we know the image is upright based on its orientation to the principal axis. If line drawing has lines that converge at the image location above the principal axis, the image is upright, if lines converge below the principal axis, image is inverted.

### Problem 4.5

The object of height $h_o$ is at a distance $\;d_o\!=\!\frac{3}{2}f\;$ from the concave mirror of focal distance $f$. Find the image distance, size and other properties.

Steps:

1. Find the image distance $d_i$ from the thin-lens equation:
• $\frac{1}{f}=\frac{1}{1.5\,f}+\frac{1}{d_i}$
• $d_i=\frac{1}{\frac{1}{f}-\frac{1}{1.5\,f}}$ $=\frac{f}{1-\frac{2}{3}}$ $=3\,f$
• Positive $d_i$ means the image is real (in front of the mirror)
2. Find the magnification
• $m=-\frac{d_i}{d_o}$ $=-\frac{3f}{\frac{3}{2} f}= -2$

### Problem 4.6

A fireman moving at a constant speed $v=1$ north of the river, starting at point A (1 minute shortest distance from the river), needs to get water at point M and bring it to the house at point B (2 minutes shortest distance from the river if moving with speed $v=1$, horizontal distance between A and B is 3 minutes). His speed south of the river is only 75% of his speed north of the river.

1. Tabulate the fireman's travel time for the following choices of x (the horizontal displacement between A and M):
1. 1.0 min
2. 1.1 min
3. 1.2 min
4. 1.3 min
5. 1.4 min
6. 1.5 min
7. 1.6 min
2. Plot the fireman's travel time as a function of x

Steps:

1. Noting that the horizontal distance between M and B is $3-x$, the total travel time is (in minutes):
• $\sqrt{1+x^2}+\frac{4}{3}\sqrt{2^2+(3-x)^2}$. Using this formula,
x (min) Total travel time (min)
1.0 5.19
1.1 5.16
1.2 5.15
1.3 5.14
1.4 5.14
1.5 5.14
1.6 5.14

### Problem 4.7

The object of height $h_o$ is at a distance $\;d_o\!=\!2\,f\;$ from the convex lens of focal distance $f$. Find the image distance, size and other properties.

Steps:

1. Solve Thin lens for $d_i$
• $\frac{1}{f}=\frac{1}{d_o} + \frac{1}{d_i}$
• $\frac{1}{d_i}=\frac{1}{f}-\frac{1}{2f}$ $=\frac{1}{f}-\frac{1}{2}\!\cdot\!\frac{1}{f}$
• $\frac{1}{d_i}=\frac{0.5}{f}$
• $d_i=2f$
2. Solve for magnification:
• $m=\frac{-d_i}{d_o}=$ $\frac{-2f}{2f}=-1$, so the image is of the same size, but inverted

### Problem 4.8

The object of height $h_o$ is at a distance $\;d_o\!=\!1.1\,f\;$ from the convex lens of focal distance $f$. Find the image distance, size and other properties.

Steps:

1. Solve for $d_i$
• Using the thin-lens equation, substitute 1.1f for $d_o$. 1/f = 1/(1.1f) + 1/$d_i$. Solve the algebra for $d_i$. $d_i$ = 11f
2. Solve for $h_i$
• Using the magnification equation, substitute 11f for $d_i$. $h_i$/$h_o$ = -11f/1.1f. $h_i$ = -10$h_o$

### Problem 4.9

The object of height $h_o$ is at a distance $\;d_o\!=\!0.9\,f\;$ from the convex lens of focal distance $f$. Find the image distance, size and other properties.

Steps:

1. Solve for di: Like before using the thin lens equation 1/f = 1/di +1/do,
2. plug in the value for do in the problem for the object distance.
3. 1/.9f +1/di = 1/f, now solving for just di using algebra we find (1-(1/.9))/f = 1/di
4. solve and take the inverse which will give us 9f = di this value should be negative because the image is projected back towards the front side.
5. Solve for magnification:
6. M = -di/do = hi/ho
7. We find a magnification of 10 which indicates the object is not only virtual and upright but also magnified by 10.

### Problem 4.10

The object of height $h_o$ is at a distance $\;d_o\!=\!\big|\,f\big|\;$ from the concave lens of focal distance $\;f\!<\!0$. Find the image distance, size and other properties.

Steps:

## Homework Questions Ch.26

### Problem 26.5

Sunlight enters a room at an angle of 32$^\circ$ above the horizontal and reflects from a small mirror lying flat on the floor. The reflected light forms a spot on a wall that is 2.0 m behind the mirror. If you now place a pencil under the edge of the mirror nearer the wall, tilting it upward by 5.0$^\circ$, how much higher on the wall $(\Delta y)$ is the spot?

I was working on problem 26.5 and I set it up the following way:

• height 1:
• $\tan(32^\circ)$ $=\frac{y_1}{2\,{\text m}}$
• $\Rightarrow\;$ $2\,{\text m}\cdot\tan(32^\circ)=y_1$
• height 2:
• $\tan(32^\circ\!+\!5^\circ)$ $=\frac{y_2}{2\,{\text m}}$
• $\Rightarrow\;$ $2\,{\text m}\cdot\tan(32^\circ\!+\!5^\circ)=y_2$
• after doing that I would find the change in height by subtracting height 1 from height 2.
• However, according to the site height 2 is found by doing:
• $y_2=2\,{\text m}\cdot\tan(32^\circ\!+\!10^\circ)$.
• Why is the angle in the second instance increased by 2$\times$ the change change in the tilt of the mirror (10$^\circ$) as opposed to just the angle that the mirror is tilted (5$^\circ$)? This doesn't make intuitive sense to me.

So, basically, your question is: if the mirror is rotated by a small angle $\alpha$, and the incident ray is the same relative to the room, how much is the reflected ray rotated by. Try to think through the following easier questions (in terms of the angle $\alpha$):

• After the mirror is rotated, what is the new incidence angle relative to the mirror?
• What is the new reflected angle relative to the mirror's normal?
• What is the new reflected angle relative to the room's vertical direction?

#### Another Question Pr. 26.5

I have tried to work through why the angle of the reflected ray is double the angle that the mirror is rotated by but still can't figure it out. If the mirror is rotated by 5°, I would think that would add 5° to the angle of incidence which always equals the angle of reflection. Is it because the new reflected angle is +5° relative to the mirror and +5° relative to the room’s horizontal? Any hints?

You are right, that would add 5° to the angle of accidence relative to the mirror's normal.

• The angle of reflection will also increase by 5° relative to the normal. But then, the outgoing beam relative to the original beam is deflected by 10° more, since the normal has turned 5° and relative to that new normal the outgoing beam gained an extra 5°.
• Another way of looking at it, the angle between the outgoing beam and the incoming beam is the angle of incidence plus the angle of reflection, that is, it is twice the angle of incidence. If the latter is increased by 5°, the outgoing beam should deflect by 10°.

Prof. Nicholas Kuzma 2015/05/24 15:17

### Problem 26.13

The rear window in a car is approximately a rectangle, 1.6 m wide and 0.31 m high. The inside rearview mirror is 0.51 m from the driver's eyes, and 1.51 m from the rear window. What is the minimum length for the rearview mirror if the driver is to be able to see the entire width and height of the rear window in the mirror without moving her head?

• Q: I tried 0.51/ (1.6 + 0.51) * (0.31 + 1.67) I'm not sure where to go after this
• A: Not sure why you tried this… See problem 4.3 in this wiki, it explains the idea.
• Hint: the small triangle formed by the mirror and the driver's head is similar to the larger triangle formed by the back window and the reflection of the driver's head in the mirror. — Prof. Nicholas Kuzma 2015/05/10 22:00

### Problem 26.18

A section of a sphere has a radius of curvature of 0.90 m. If this section is painted with a reflective coating on both sides, what is the focal length of the convex side and the concave side?

• Q: I used $f = -\frac{R}{2}$ and my radius of curvature is 0.9 m they want to know the focal length of the convex side. I got $-0.45$, but it says that it is not that right answer.
• A: Check the units. You answer on the website is expected in the units that are shown next to the input box. These might be cm, not m. — Prof. Nicholas Kuzma 2015/05/10 22:13

### Problem 26.33

A magician wishes to create the illusion of a 2.74 m tall elephant. He plans to do this by forming a virtual image of a 50.0 cm tall model elephant with the help of a spherical mirror. The model is placed 3.00 m from the mirror.

• Q: All of the calculations make sense in this problem. My question is how would a virtual image be projected through a mirror? Wouldn’t a mirror block any image behind it, which means the audience wouldn’t be able to see it?
• A: A flat mirror always creates a virtual image, and while it is true that sometimes the mirror image is obscured by the object (but not by the mirror!), it is often possible to see mirror images if the object is a little bit off axis. In this case, the virtual image is actually much larger than the object, so projecting a little bit off-axis shouldn't be a big problem for the audience to see it.

Prof. Nicholas Kuzma 2015/05/24 22:02

### Problem 26.69

A laser beam enters one of the sloping faces of the equilateral glass prism ($n=1.42$) and refracts through the prism. Within the prism the light travels horizontally.

I am unsure where to begin with this problem. I know $n_1\sin\theta_1 = n_2\sin\theta_2$, but I'm not sure where to go from there. Just a little nudge in the right direction would be helpful Thank you! — Thomas Ewing

I also had trouble with this problem. I don't understand the answer key's approach at all. My problems:

1. How do we know that $\beta = 30^\circ$? That is not obvious to me - is there a trig 101 thing I'm forgetting?
2. Where do we measure the angles of incidence and refraction from? From the surface of the prism? From the vertical? Is there some standard that I'm missing?

• The key word here is equilateral, that means all the sides of the cross-section triangle are equal, and all its angles are $60^\circ$.
• See the diagram to the right.
• The definitions of $\theta_1$ and $\theta_2$ include the words relative to the normal: that is, relative to that thin black line perpendicular to the refracting surface
• The angle $\theta_2$ denoted by ? is “obviously” $30^\circ$, because, added to the adjacent $60^\circ$, it forms the right angle ($90^\circ$).

### Problem 26.83

• What is the magnification of this image?

#### Steps I took so far

• distance from lens 1 to the final image is 34 cm.
• object to lens 1 is 24 cm
• lens 1 to f1 is 14 cm
• lens 1 to lens 2 is 35 cm
• f2 to lens 2 is 7 cm
• What is the magnification?

#### Question

• I did $(m-f)/(f-d_o)$ and got 1.4
• I did $-d_i/d_o=$ $-34/24$ and got $-1.4$.
• they were both wrong
• So, I still can't get the magnification part of this answer. I keep getting the same answers.

#### Suggestion

• This is a two-step problem, because it has two lenses.
• The magnification is the product of individual magnifications:
• $m=m_1\cdot m_2$
• For the first lens, use $m_1=-\frac{d_i^{(1)}}{d_o}=$ $-\frac{33.6}{24}=$ $-1.4$
• For the second lens, use $m_2=-\frac{d_i}{d_o^{(2)}}=$ $-\frac{-1.17\,{\text{cm}}}{1.4\,{\text{cm}}}$ $=0.833$
• The total magnification is then $m=m_1\cdot m_2$

### Problem 26.92

A horizontal incident beam consisting of white light passes through an equilateral prism…

I'm having similar problems with 26.92. Can someone please clarify how to identify angles in these equilateral triangles and how to approach Snell's Law in dispersion questions? I've read and done the example problem on dispersion on pg 934 of the book and I still don't see what's happening.

If someone can clarify these issues, I'd be very grateful!

• Again, equilateral means all angles of the triangle are $60^\circ$ each.
• That means, similar logic to 26.69, that $\theta_1=30^\circ$ (measured from the normal)
• From the Snell's law, find $\theta_2$
• Finding angle 3 in the diagram, needed to figure out the Snell's law for the second refraction, is a bit tricky:
• Use the following facts:
• Angles $\alpha$, $\beta$, and $60^\circ$ form a triangle and therefore all add up to $180^\circ$
• Angles $\alpha$ and $\theta_2$ add up to $90^\circ$
• Angles $\beta$ and 3 add up to $90^\circ$ as well

I hope it helps! — Prof. Nicholas Kuzma 2015/05/12 23:17