t1Asolspr2010 - Physics 1112 Spring 2010 University of...

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Unformatted text preview: Physics 1112 Spring 2010 University of Georgia Instructor: HBSchuttler Solution to PHYS 1112 In-Class Exam #1A Thu. Feb. 4, 2010, 9:30am-10:45pm Conceptual Problems Problem 1: An object O is placed before a mirror, and a viewer S observes the objects image produced by the mirror, as shown here: mirror 2 4 1 3 S Where does S perceive the mirror image of O to be located ? (A) Position 1. (B) Position 2. (C) Position 3. (D) Position 4. (E) The image of O cannot be seen by S in the configuration shown above. Answer: (A) By Archimedes law, applied to a planar mirror, the straight line segment connecting any object point (like O ) to its image point must (i) be perpendicular to the plane of the mirror (ii) be bisected (cut in half) by the plane of the mirror. For object point O , Position 1 is the only point satisfying this condition. So, any rays emerging from O which strike the mirror are reflected off the mirror (acc. to Archimedes law) in such a way that, after reflection, they appear to be emerging from Position 1. This is true (a) for all rays emerging from O regardless of where the rays from O strike the mirror; and (b) regardless of how far up or down the mirror extends, i.e. , even if the mirror surface does not cover the wall area immediately opposite of S . It is very important that you draw your own big, clean picture of this set-up, tracing a few (at least two) rays from O to the mirror and then, after reflection with = , tracing the reflected rays backwards to see where they intersect behind the mirror. If you draw this 1 Physics 1112 Spring 2010 University of Georgia Instructor: HBSchuttler carefully, youll find them all intersecting at Position 1. You will also find that you can draw a ray from O to S , such that it (i) stikes the mirror above its bottom end; (ii) gets reflected at the mirror subject to Archimedes law ( = ); and (iii) then reaches S . Therefore the image of O is visible from location S , even though the mirror does no cover the wall immediately opposite to S . This is similar to the in-class quiz on buying the Shortest Possible Mirror on the Wall. You can still see your own feet (on the ground) in this mirror, even though the mirror extends downward to only half of your height above the ground. Problem 2: Sound waves (including ultrasound) have a speed of wave propagation v Air = 346m/s in air and v Water = 1497m/s in water. Also, note that sin(13 . 364 o ) = 346 / 1497 . A narrow ultrasound beam striking the flat water surface of your swimming pool (A) will always have an angle of refraction not exceeding 13 . 364 o if the beam is incident from above the water surface; (B) will have an angle of refraction greater than the angle of incidence if the beam is incident from below the water surface; (C) cannot undergo total internal reflection if incident from above the water surface, regardless of the angle of incidence; (D) will undergo total internal reflection if incident from above...
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This note was uploaded on 04/23/2010 for the course PHYS 1112L taught by Professor Staff during the Spring '10 term at University of Georgia Athens.

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t1Asolspr2010 - Physics 1112 Spring 2010 University of...

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