Reading: Chapter 34, GEOMETRIC OPTICS
Study guide: Chapter 34
Webassign #2 is wrong...
At small angles (expressed in Radians), we find that $\sin\theta \approx \theta$:
Also that $\cos\theta \approx 1$, from which you can kinda guess that $\tan\theta=\sin\theta/\cos\theta\approx \theta$:
...eventually we find: $$\frac{1}{s} + \frac{1}{s'}=\frac{2}{R}\equiv \frac{1}{f}$$
This is independent of $\alpha$, so all rays (at small angles) from $P$ converge at $P'$!
If $\alpha$ is not small...
Back to center section of mirror (small angles) the
picture typically looks like:
What if $P$ is far away ($s\to\infty$)?
$$s'=R/2=f$$Parallel rays converge at the focal length, $f$.
Object that extends away from axis... how do we draw
what happens?
$P$- We've done that, rays converge on $P'$.
$Q$?- Draw all the principle rays to see where they meet:
Easiest way to get proportions is to look at the ray that hits the vertex...
Similar triangles $\Rightarrow$: $$\frac{h}{s}=-\frac{h'}{s'}$$ where the negative sign indicates that the image is inverted.
Magnification $m$: $$m\equiv \frac{h'}{h}=-\frac{s'}{s}$$
Try it! -- mirrorexercise.pdf
With the radius of curvature set to $R=2 cm$, use $$\frac{1}{s} + \frac{1}{s'}=\frac{2}{R}\equiv \frac{1}{f}$$ $$m\equiv \frac{h'}{h}=-\frac{s'}{s}$$
Calculate the distance $s'$ to the image, the magnification $m$, and sketch some principal rays to from the candle top to to see where they meet for...
[Example with 5X magnification, 1 m behind candle]
Define focal length as the image distance when the
object is at infinity...
Just like mirrors, the distances to object and image obey $$\frac{1}{s}+\frac{1}{s'}=\text{constant}$$
$$\Rightarrow \frac{1}{s}+\frac{1}{s'}=\frac{1}{f}.$$
The principal rays to use when drawing:
A radius of curvature is
$$\frac{1}{s}+\frac{1}{s'}=\frac{1}{f}$$Still
holds, but now $f$ and $s'$ are negative.
Virtual: When the actual rays which appear to be coming from the image are on the opposite side of the optical element from the image.
Diagram at NIH ...
Characterized by... Nearpoint of order 25 cm (recedes with age from 7 cm at age 10 to > 100 cm after 65
Most common eyesight problem is myopia - (near-sighted
vision)
what kind of corrective lens (positive or negative focal length) is needed?
Color is related to wavelength, one continuous variable. Why do those art folks talk about 3 primary colors? (And how would that be different for cow artists????)
Cones have a variable sensitivity to light depending
on their pigment, characterized by a peak wavelength and a response width:
As
long as your eyeball can focus it, the best strategy to see detail is to move
the object as close as possible to the eye. This makes the image on
your retina as large as possible. The quantity to make as large as possible
is the angle subtended by the object (or image).
But you can't get any closer than your nearpoint (let's
say the usual age-averaged value of ~25 cm), so this angle is...$$\theta=\tan(y/s)\approx\frac{y}{s}.$$
But hold up a magnifying glass. How can you arrange it to get as large an apparent angle $\theta$ as possible?
With the object at the focal point, the virtual image is at $\infty$--but the eye can focus there!
The eye can move much closer to the object, which appears to subtend an angle $$\theta'\approx\frac{y}{f}$$
The net angular magnification is $$M=\frac{\theta'}{\theta}=\frac{y/f}{y/25 cm}=\frac{25 cm}{f}$$ Practically this can be no more than about 3X -5X
The previous magnification we discussed was the lateral magnification $$m=-\frac{s'}{s}$$
An astronomical telescope...
has a net angular magnification of $$M=\frac{\theta'}{\theta}=\frac{f_o}{f_e}$$
