When two lenses are used in combination, the first one forms an image that then

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 1.40 cm -tall object is 58.0 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

Explanation / Answer

focal length of the first lense, f1=40cm

focal length of the second lense, f2=60cm

separation between the lense, d=300cm

object distance. u1=58cm (to the left of the lens)

a)

use,

1/u1+1/v1=1/f1

1/58+1/v1=1/40

==> v1=128.9 cm

image distance from the first lens, v1=128.9 cm

now,

height of the object, h1=1.4cm

magnefication m1=-h1'/h1 = v1/u1

h1'/1.4=128.9/58

====> h1'=-3.11 cm

height of the image, h1'=3.11 cm

b)

image due to first lens become the object for the second lens,

object distance for the second lens,

u2=d-v1

u2=300-128.9

u2=171.1 cm

now,

1/u2+1/v2=1/f2

1/171.1 + 1/v2 =1/60

===> v2=92.4cm

final image forms at v2=92.4 (from the second lens)

here,

total magnification M=m1*m2=h'/h

magnificatin of first lense is m1=-v1/u1=-128.9/58=2.22

and

magnificatin of second lense is m2=-v2/u2=-(92.4)/171.1)=-0.54

now,

M=m1*m2

=(-2.22)*(-0.54)

=1.19

but M=h'/h

===>

1.19=h'/1.4

===> h'=1.67 cm

height of the final image is, h'=1.67cm

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