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.
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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 -.
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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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