1025_Physics ProblemsTechnical Physics

1025_Physics ProblemsTechnical Physics - 366 P36.45 Image...

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366 Image Formation P36.45 Ray h 1 is undeviated at the plane surface and strikes the second surface at angle of incidence given by θ 1 1 1 1 0500 143 = F H G I K J = F H G I K J −− sin sin . . h R cm 20.0 cm . Then, 100 160 20 0 21 .s i n i n . . . θθ == F H G I K J a f cm FIG. P36.45 so 2 229 .. The angle this emerging ray makes with the horizontal is 0860 −= ° It crosses the axis at a point farther out by f 1 where f h 1 1 33 3 = = ° = tan . tan . . bgaf cm cm . The point of exit for this ray is distant axially from the lens vertex by 200 000625 22 . . cm cm cm cm −−= af a f so ray h 1 crosses the axis at this distance from the vertex: x 1 33 3 0 006 25 33 3 =− = . cm cm cm . Now we repeat this calculation for ray h 2 : = F H G I K J sin . . 1 12 0 36 9 cm 20.0 cm 12 00 20 0 i n i n . . . F H G I K J a f 2 73 7 . f h 2 2 12 12 0 36 8 16 0 = = ° = tan
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This note was uploaded on 12/14/2011 for the course PHY 203 taught by Professor Staff during the Fall '11 term at Indiana State University .

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