Question

When two lenses are used in combination, the first one forms an image that then serves as the object for the second lens. The magnification of the combination is the ratio of the height of the final image to the height of the object. A 1.80 cm -tall object is 51.0 cm to the left of a converging lens of focal length 40.0 cm . A second converging lens, this one having a focal length of 60.0 cm , is located 300 cm to the right of the first lens along the same optic axis.

Part A

Find the location and height of the image (call it I1) formed by the lens with a focal length of 40.0 cm .

Enter your answer as two numbers separated with a comma.

Part B

I1 is now the object for the second lens. Find the location and height of the image produced by the second lens. This is the final image produced by the combination of lenses.

Enter your answer as two numbers separated with a comma.

Answer 1

Part (A)

image due to first lense(converging lens)

object distance u=51 cm

f1=40 cm

let image distance be v

1/u+1/v=1/f1

1/51+1/v=1/40

===>

v=200 cm is the image distance due to first lense

now,

let,

object height h=1.8cm

image height due to first lens is h1'

use

magnificatin of first lense is m1=-v/u

m1=-200/51

m1=-3.91

but, m1=h1'/h

h1'=-3.92*h

h1'=-3.92*1.8

height of the image h1'=-7.06 cm

image distance due to first lense v=200 cm

part(B)

object distance for the second lense is,

u'=300-200 cm

u'=100 cm

let image distnace be v'

and

f2=60 cm

now

1/u'+1/v'=1/f2

1/100+1/v'=1/60

===>

v'=150 cm

image distance due to second lense is v'=150 cm

==============================

now,

final image distance from the actual object is

=51+300+150=501 cm

magnificatin of second lense is m2=-v'/u'=-(150)/100=-1.5

and

total magnification M=m1*m2

M=m1*m2

=(-3.92)*(-1.5)

=5.88

but,

M=final image height/object height

M=h'/h

===>

h'=h*(M)

h'=1.8*(5.88)

h'=10.584 is the height of the final image