(8c34p109) A diverging lens with a focal length of -18 cm and a converging lens
ID: 1649290 • Letter: #
Question
(8c34p109) A diverging lens with a focal length of -18 cm and a converging lens with a focal length of 15 cm have a common central axis. Their separation is 15 cm. An object of height 4.0 cm is 10 cm in front of the diverging lens, on the common central axis. Where does the lens combination produce the final image of the object (the one produced by the second converging lens)? Where is the image located as measured from the converging lens? (cm) Submit Answer Incorrect. Tries 1/7 Previous Tries What is the height (absolute value) of that image?Explanation / Answer
From the given question,
object distance(u1)= -10 cm
focal length of diverging lens(f1)=-18 cm
1/v-1/u=1/f
1/v-[1/(-10)]=1/(-18)
v=-45/7 cm=-6.42 cm
This is object distance for second lens
u2=-6.42-15=-21.42 cm
f2=15 cm
1/v-1/u=1/f
1/v-[1/(-21.42)]=1/(15)
v=-50.05 cm
h2/h1=(v2/u2)(v1/v1)
=(50.05/21.42)(6.42/10)
=1.5
h2=4*1.5
=6.0 cm
Image is located at 50.05 cm in left from converging lens.
height of image is 6.0 cm
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