# Homework Help: Optics(Converging lens and Diverging lens)

1. Dec 20, 2011

### BSElectrician

1. The problem statement, all variables and given/known data
http://www.imageurlhost.com/images/r0its2lyef7v3s0ppms2.png

2. Relevant equations
Mirror Equation
(1/q) + (1/p) = (1/f) = (2/r)
Magnification
M = (q/p)

for two lenses
M1 & M2
Total magnification = M1*M2

3. The attempt at a solution
we have p1 = 40cm and f1 = 30cm
then the image distance of lens 1 will be q1 = 120

then we have diverging lens 2
with focal length f2 = 20cm

then im stuck i dont know what to do with the object distance of the diverging lens
i also dont know what to do with the fact that the object is 2cm high

Last edited: Dec 20, 2011
2. Dec 20, 2011

### grzz

The image of the converging lens acts as a VIRTUAL object for the diverging lens.

3. Dec 20, 2011

### BSElectrician

is it between the diverging and converging lens?

this is how i do to find the image distance of the divergng lens
between C lens and D lens (110cm)
but there is an image distance q1 from the C lens (120cm)
the object distance from the D lens p2 might be (110cm - 120cm)

(1/q2) + (1/(110-120)) = (1/20)
q2 = 6.67
then i can find the magnification of the Divirging lens
M2 = q2/p2
total
M = M1M2

whats the use of that 2cm thing above the object?

4. Dec 20, 2011

### grzz

If the formula (1/u) + (1/v) = 1/f is to be used then the focal length of the diverging lens is taken as -20.