Question

A 1.4-cm-tall object is located 3.0cm to the left of a converging lens with a focal length of 4.0cm . A diverging lens, of focal length -7.4cm, is 14cm to the right of the first lens.

Find the position of the final image.

s2` = -5.7 cm

Find the size of the final image.

h2` = ?

Find the orientation of the final image.

a) real, upright

b) real, inverted

c) virtual, upright

d) virtual, inverted

Answer 1

image due to first lense(converging lens)

object distance u=3 cm

f1=4cm

let image distance be v

1/3+1/v=1/f1

1/3+1/v=1/4

===>

v=-12cm is the image distance due to first lense

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

now,

object distance for the second lense(diverging lens) is,

u'=14-12 cm

u'=2 cm

let image distnace be v'

and

f2=-7.4 cm

now

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

1/2+1/v'=1/-7.4

===>

v'=1.57 cm

image distance due to second lense is v'=1.57 cm

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

now,

final image distance from the actual object is

=3+14+1.57=18.57 cm .......is answer

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

here

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

here,

object height h=1.4cm

final image height is h'

and

magnificatin of first lense is m1=-v/u=-(-12/3)=

magnificatin of second lense is m2=-v'/u'=-(1.57)/2=-0.79

now,

M=m1*m2

=4*-0.79

=-3.15

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

but M=h'/h

===>

h'=h*(M)

=1.4*(-3.15)

=-4.41

h'=-4.41 cm    .......is answer

and

final image is real,inverted