Mirrors and Lenses29323.50Since the object is midway between the lens and mirror, the object distance for the mirror is . The mirror equation gives the image position as 112.5 cmp= +1112120qRp21541.0 cm12.5 cm50.0 cm50.0 cm−=−=−==225.0 cmp, or This image serves as the object for the lens, so 150.0 cmq= +125.0 cmq=−= −. Note that since , this is a virtual object. The thin lens equation gives the image position for the lens as 20<p()()()222225.0 cm16.7 cm25.0 cm16.7 cmp fpf−−==−−− −2q20q<50.3 cm= −Since , this is a virtualimage that is located 50.3 cm in front of the lens or 25.3hind thecm bemirror. The overall magnification is ()()12121250.3 cm50.0 cm12.5 cm25.0 cmMM M−==−=−−−+8.05=qqpp−Since 0M>, the final image is upright 23.51As light passes left-to-right through the lens, the image position is given by
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