The image distance and magnification for a lens are q=-8.57 cmand M=.423. A conv
ID: 1748605 • Letter: T
Question
The image distance and magnification for a lens are q=-8.57 cmand M=.423. A convergent lens with focal length of 10.0 cm is placed 15.0cm from (and on the side opposite the object) the lens mentionedabove. Complete the follwing statement: "The image formed by thefirst lens is treated as the _________ for the second lens." Usingthis statement and the given information, determine the objectdistance for the second lens. Then locate the image formed by thesecond lens and calculate the overall magnification. State whetherthe final image is real or virtual, upright or inverted, andenlarged or reduced. Any and all help is greatly appreciated!!!! The image distance and magnification for a lens are q=-8.57 cmand M=.423. A convergent lens with focal length of 10.0 cm is placed 15.0cm from (and on the side opposite the object) the lens mentionedabove. Complete the follwing statement: "The image formed by thefirst lens is treated as the _________ for the second lens." Usingthis statement and the given information, determine the objectdistance for the second lens. Then locate the image formed by thesecond lens and calculate the overall magnification. State whetherthe final image is real or virtual, upright or inverted, andenlarged or reduced. Any and all help is greatly appreciated!!!!Explanation / Answer
for thefirst lens
u=8.57cm
m=0.423
m=v/u=0.423 hence v=0.423*8.57=3.6251cm
now forthe second lens
u=15-3.6251=11.3748cm
f=10cm
v=?
1/v=1/f-1/u=1/10-1/11.3748=>v=82.73cm
final image will be formed at a distance of 82.73cm from the secondlens
magnification m=m1*m2
where m1 and m2 are the magnifications produced by theindividual lenses.
so m=0.423*82.73/11.3748=3.0768
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