Question

A converging lens of focal length 19.3 cm is separated by 49.3 cm from a converging lens of the focal length 4.53 cm. Find the position of the final image with respect to the second lens of an object placed 38.6 cm in front the first lens. Answer in units of cm. 021 (part 2 of 3) 10.0 points If the height of the object is 1.7 cm, what is the height of the final image? Answer in units of cm. 022 (part 3 of 3) 10.0 points If the two lens are now placed in contact with each other and the object is 5 cm in front of this combination, where will the image be located? Answer in units of cm.

I dont need the first answer... already got it wrong too many times to submit it... :(

Answer 1

let

f1 = 19.3 cm
d = 49.3 cm
f2 = 4.53 cm

a) u1 = 38.6 cm

let v1 is the image distance from first lens.

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

1/v1 = 1/f1 - 1/u1

1/v1 = 1/19.3 - 1/38.6

v1 = 38.6 cm

magnification due to first lens, m1 = -v1/u1

= -38.6/38.6

= -1

object distance for the second lens, u2 = 49.3 - 38.6 = 10.7 cm

let v2 is the image distance for the second lens.

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

1/v2 = 1/f2 - 1/u2

1/v2 = 1/4.53 - 1/10.7

v2 = 7.86 cm   <<<<<<<<<----------------------Answer

magnification due to second lens, m2 = -v2/u2

= -7.86/10.7

= -0.734

overall magnification, m = m1*m2

= -1*(-0.734)

= 0.734

height of the image, h' = m*h

= 0.734*1.7

= 1.25 cm <<<<<<<<<----------------------Answer

when lenses are in contact.

1/f = 1/f1 + 1/f2

1/f = 1/19.3 + 1/4.53

f = 3.67 cm

u = 19.3 cm (object distance)

v = ? (image distance)

use, 1/u + 1/v = 1/f

1/v = 1/f - 1/u

1/v = 1/3.67 - 1/19.3

v = 4.53 cm <<<<<<<<<---------------------Answer