Two converging lenses (f_1 = 9.30 cm and f_2 = 6.10 cm) are separated by 18.4 cm

Question

Two converging lenses (f_1 = 9.30 cm and f_2 = 6.10 cm) are separated by 18.4 cm. The lens on the left has the longer focal length. An object stands 12.3 cm to the left of the left-hand lens in the combination. Locate the final image relative to the lens on the right. Obtain the overall magnification. Is the final image real or virtual? With respect to the original object, is the final image upright or inverted and is it larger or smaller? (a) Locate the final image relative to the lens on the right. 4 69 cm to the right of the right-hand lens (b) Obtain the overall magnification. 0 7901.

Explanation / Answer

a)
for the first lens,

f1 = 9.3 cm

u1 = 12.3 cm (object distance)

let v1 is the image distance

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

1/v1 = 1/f1 - 1/u1

1/v1 = 1/9.3 - 1/12.3

v1 = 38.13 cm

magnifcation due to the first lens, m1 = -v1/u1

= -38.13/9.3

= -4.1

now object distance for the second lens,

u2 = -38.13 + 18.4

= -19.73 cm

f2 = 6.1 cm

let v2 is the final image distance

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

1/v2 = 1/f2 - 1/u2

1/v2 = 1/6.1 - 1/(-19.73)

v2 = 4.66 cm to the right of righ-hand lens <<<<<<<<<<-----------------------------Answer

magnifcation due to the second lens, m2 = -v2/u2

= -4.66/(-19.73)

= 0.236

b)

overall magnification, M = m1*m2

= -4.1*0.236

= -0.968 <<<<<<<<<<-----------------------------Answer

final image is real and inverted.

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