wo converging lenses (f1 = 9.00 cm and f2 = 6.00 cm) are separated by 18.0 cm. T

Question

wo converging lenses (f1 = 9.00 cm and f2 = 6.00 cm) are separated by 18.0 cm. The lens on the left has the longer focal length. An object stands 13.5 cm to the left of the left-hand lens in the combination. (a) Locate the final image relative to the lens on the right. (b) Obtain the overall magnification. (c) Is the final image real or virtual? With respect to the original object, (d) is the final image upright or inverted and (e) is it larger or smaller? Could you please show all of the formulas used and please tell me why the answer is what it is for C, D, and E? Thanks!

Explanation / Answer

a) for the first lens

f1 = 9 cm

u1 = 13.5 cm

let v is the image distance.

apply, 1/u1 + 1/v1 = 1/f1

1/v1 = 1/f1 - 1/u1

1/v1 = 1/9 - 1/13.5

v1 = 27 cm

magnification, m1 = -v1/u1 = -27/13.5 = -2

for the seocnd lense,

f2 = 6 cm

u2 = -27 + 18

= -9 cm

let v2 is the final image

Apply, 1/u2 + 1/v2 = 1/f2

1/v2 = 1/f2 - 1/u2

1/v2 = 1/6 - 1/(-9)

v2 = 3.6 cm

magnification, m2 = -v2/u2 = -3.6/(-9) = 0.4

a) final image distance = 3.6 cm from right lense

b) overall magnification, m = m1*m2

= -2*0.4

= -0.8

c) the image is real. because out going rays passes through the image.

d) inverted.

e) smaller (because, |m| < 1)

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