# 2 convergent lenses

## Homework Statement

This is a 2 part problem but I figure out the first part. Heres the 1st problem and the solution:
7- Given a convergent lens which has a focal point f. An object is placed at distance p = $$\frac{4}{3}$$f to the left of the lens. See the sketch.
Solution: q1 = 4f, and is a real image

8- Now place a convergent lens with a same focal length f at a distance d = f behind the first lens. Determine q2, i.e., the image location measured with respect to the second lens # 2.
I will upload the actual image later.

Given:
p1 = $$\frac{4}{3}$$f
q1 = 4f
f1 = f

p2 = ?
q2 = ?
f2 = f

## Homework Equations

$$\frac{1}{f}$$ = $$\frac{1}{p}$$+$$\frac{1}{q}$$

## The Attempt at a Solution

I would show my attempt at the solution but I don't know what is p2. How do I get p2?

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*bump*

The image formed by the first lens is now the object for the second lens. Find out where the image forms with respect to the second lens. Without drawing a picture it looks like p2 is -3f. That is:

|--|------q1/p2

where "|" is a lens and "--" is one distance f. As you can hopefully see, q1/p2 is 4f infront of lens one and 3f behind lens two. So do what you did in part one but with p2=-3f.

Thank you!

My work:

3$$\frac{1}{f}$$-$$\frac{1}{-3f}$$ = $$\frac{3-1}{3f}$$ = $$\frac{4}{3}$$f (inverse) = $$\frac{3}{4}$$f

I have a similar problem heres my work but its wrong cause its not one of the answer choices:

Problem:

Consider the setup of the two-lens system shown in the figure, where the separation of the two lenses is denoted by d = 1.5 a. Their focal lengths are respectively f1 = a and f2 = 2 a. An object is placed at a distance 2 a to the left of lens # 1. Find the location of the final image q2 with respect to lens #2. Take q2 to be positive if it is to the right of lens #2 and negative otherwise.

Formula:

$$\frac{1}{f}$$ = $$\frac{1}{p}$$+$$\frac{1}{q}$$

My Work:

p1 = 2a
q1 = ?
f1 = a

p2 = ?
q2 = ?
f2 = 2a

$$\frac{1}{q1}$$ = $$\frac{1}{f}$$-$$\frac{1}{p}$$ = 2($$\frac{1}{a}$$) - $$\frac{1}{2a}$$ = $$\frac{2-1}{2a}$$ = $$\frac{1}{2a}$$ (inverse) = 2a = q1

$$\frac{q1}{p2}$$ = q1 = $$\frac{2a}{p2}$$ = 2a $$\rightarrow$$ $$\frac{2a x p2}{2a}$$=$$\frac{2a}{2a}$$ $$\rightarrow$$ p2 = 1

$$\frac{1}{2a}$$-($$\frac{1}{1}$$)2 = $$\frac{1-2a}{2a}$$ = $$\frac{a}{2a}$$ = $$\frac{2a}{a}$$ = 2 = q2

What am I doing wrong?