27_Ch 23 College Physics ProblemCH23 Mirrors and Lenses

27_Ch 23 College Physics ProblemCH23 Mirrors and Lenses -...

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Mirrors and Lenses 293 23.50 Since the object is midway between the lens and mirror, the object distance for the mirror is . The mirror equation gives the image position as 1 12.5 cm p =+ 11 121 20 qRp 2 1 54 1 .0 cm 12.5 cm 50.0 cm 50.0 cm =− = = = 2 25.0 cm p , or This image serves as the object for the lens, so 1 50.0 cm q 1 25.0 cm q = −= . Note that since , this is a virtual object. The thin lens equation gives the image position for the lens as 2 0 < p ( )( ) () 22 25.0 cm 16.7 cm 25.0 cm 16.7 cm pf −− == 2 q 2 0 q < 50.3 cm = Since , this is a virtual image that is located 50.3 cm in front of the lens or 25.3 hind the cm be mirror . The overall magnification is ( ) 12 50.3 cm 50.0 cm 12.5 cm 25.0 cm MMM  == − =   +8.05 = qq pp    Since 0 M > , the final image is upright 23.51 As light passes left-to-right through the lens, the image position is given by
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