II. Long problems (32 points each) . A converging lens with a focal length of 8.
ID: 2305047 • Letter: I
Question
II. Long problems (32 points each) . A converging lens with a focal length of 8.0 cm is placed 12.0 cm to the right of a diverging lens -- of focal length-6.0 cm. An object of height 2.0 cm is 4.0 cm to the left of the diverging lens. T a) Where is the image from the diverging lens alone? For this part of the problem, ignore the sence of the converging lens Where is the final image, that is the image produced by both lenses together? Is it or virtual, and is it upright or inverted? What is the size of the final image?Explanation / Answer
(A) for diverging lens,
f1 = - 6 cm
object distance, p1 = 4cm
Applying lens equation,
1/f1 = 1/p1 + 1/q1
1/(-6) = 1/4 + 1/q1
q1 = - 2.4 cm
Ans: 2.4 cm left to the diverging lens
(B) for converging lens,
f2 = 8 cm
object distance, p2 = 2.4 + 12 = 14.4 cm
1/8 = 1/14.4 + 1/q2
q2 = 18 cm
18 cm right to the converginglens.
image is real.
m = (q1/p1)(q2/p2) = - 0.75
m is negative so image is inverted.
(C) hi = m ho = 0.75 x 2 = 1.50 cm