Question

A converging lens with a focal length of 4.8 *cm* is
located 24.8 *cm* to the left of a diverging lens having a
focal length of -15.0 *cm*. If an object is located 9.8
*cm* to the left of the converging lens, locate and describe
completely the final image formed by the diverging lens.

**Where is the image located as measured from the diverging
lens?
What is the magnification?
Also determine, with respect to the original object whether the
image is real or virtual, upright or inverted? Check all statements
that are true.**

The image is virtual

The image is real

The image is upright

The image is inverted

Answer #1

v= image disance , u = object distance m = magnification

first image formation by converging lens

1/v-1/u = 1/f

1/v -1/-9.8= 1/4.8

m1 = v/u = 9.41/-9.8 = -0.96

from here v= 9.41 cm in right of the lens now this image act as object for diverging lens so its distance from diverging lens = 24.8-9.41 =15.39

2nd image formation by diverging lens

1/v-1/u = 1/f

1/v -1/-15.39 = 1/-15

from here **v= -7.596 cm answer**

**in the left of diverging lens**

m2 = v/u =

-7.596/-15.39 = 0.49356

**final magnification = m1*m2 = -0.96*0.49356 =-0.4738176
answer**

**it is virtual and inverted**

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