S10P1CLec23A

# S10P1CLec23A - Physics 1C Lecture 23A "If Dracula cant see...

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Physics 1C Lecture 23A "If Dracula can’t see his reflection in the mirror, how come his hair is always so neatly combed?" --Steven Wright

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Mirror Equation You can mathematically relate the object distance, p, and the image distance, q, by using similar triangles. This gives us the following relationship: 1 p + 1 q = 2 R But since f=R/2, we get: 1 p + 1 q = 1 f where f is the focal length of the mirror.
Mirror Equation The lateral magnifcation, M, oF the image height compared to the object height can also be Found geometrically. This gives us the Following relationship: where the negative sign comes From the Fact that h’ is inverted with respect to h. M = ʹ h h = q p

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Ray Diagrams The most important thing to remember when dealing with the mirror equations is to remember the sign conventions. For a concave mirror f is +, for a convex mirror f is -.
Spherical Mirrors Example A convex spherical mirror of radius of curvature R that is 20.0cm produces an upright image precisely one-quarter the size of an object, a candle. What is the separation distance between the object and its image? Answer The coordinate system is already deFned by the sign convention. The center of the mirror is the origin.

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Spherical Mirrors Answer First, let’s fnd the ±ocal length o± the mirror: where the negative sign comes ±rom the ±act
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## This note was uploaded on 06/03/2010 for the course PHYS PHYS 1C taught by Professor F during the Spring '00 term at UCSD.

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S10P1CLec23A - Physics 1C Lecture 23A "If Dracula cant see...

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