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Ch.26: Geometrical optics (Apr 29 - May 1)

Concepts Ch.26

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

Lecture notes 4/29,5/1

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


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

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:


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

Problem 4.6

Steps:


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:


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:


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:


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 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?

Question about Pr. 26.5

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.

Answer

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?

chapter_26.txt · Last modified: 2014/05/17 02:48 by wikimanager