The Physical World I

Spring 2005
MWF 11:30

Instructor: Dr. Bruce Law, CW 327, Tel: 532-1618.


R28.5.  The angle of incidence is equal to the angle of reflection.

R28.6. The image is the same distance behind the mirror as the object is in front of the mirror.

R28.24. The critical angle corresponds to that angle of incidence (when going from a more dense medium (eg. water) to a less dense medium (eg. air)) where the refracted (or transmitted beam) travels along the surface. NOTE: for angles of incidence above the critical angle there is NO refracted beam.

R28.27. The light “bends” in an optical fiber because the angle of incidence is always above the critical angle and therefore ALL of the light is reflected (and none of it is refracted or transmitted through the surface of the fiber).



E28.7. In the “nighttime” position the glare from the headlights, which comes from the silvered BACK surface of the mirror, is reflected above the head of the person (see diagram in book on right). The driver can still see the cars behind him because he is looking at the much weaker reflection from the wedged non-silvered FRONT surface of the mirror.

E28.25. (THIS QUESTION WILL NOT BE IN ANY EXAM OR TEST – IT IS FOR YOUR ENTERTAINMENT!) You would aim the spear below the fish. Light from the fish, as it passes through the water/air surface, is refracted AWAY from the normal into your eyes. Your brain thinks that light travels in a straight line – however because of the refraction of light at the water/air surface the fish is in reality below the position that you actually see it. To actually hit the fish with a laser beam you would point the laser beam directly at the fish because (as with light from the fish) the laser beam is refracted in passing through the air-water surface.


PROBLEMS OF Home Work 17

P28.4. You approach the mirror at 2m/s and your image approaches the mirror at 2m/s. Therefore, you and your image approach each other at 4m/s.