Page 828, problem 6:
(a) The rays all cross at pretty much the same place, given the accuracy with which we can
draw them.
(b) It could be used to cook food, for instance. All the sunlight is concentrated in a small area.
(c) Put the lightbulb at the point where the rays cross. The outgoing rays will then form a
parallel beam going out to the right.
Page 829, problem 11:
The magnification is the ratio of the image’s size to the object’s size. It has nothing to do
with the person’s location. The angular magnification, however, does depend on the person’s
location, because things farther away subtend smaller angles. The distance to the actual object
is not changed significantly, since it’s zillions of miles away in outer space, but the distance to
the image does change if the observer’s point of view changes. If you can get closer to the image,
the angular magnification is greater.
Page 830, problem 15:
For a flat mirror,dianddoare equal, so the magnification is 1, i.e., the image is the same size
as the object.
Page 830, problem 16:
See the ray diagram below. Decreasingθodecreasesθi, so the equationθf =±θi+±θomust
have opposite signs on the right. Sinceθois bigger thanθi, the only way to get a positiveθfis
if the signs areθf=−θi+θo. This gives 1/f=− 1 /di+ 1/do.
Page 830, problem 19:
(a) The object distance is less than the focal length, so the image is virtual: because the object
is so close, the cone of rays is diverging too strongly for the mirror to bring it back to a focus.