Convergent Lenses: 2-Part Problem and Solution | Focal Point and Image Location

[DrMcDreamy]

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?

 

[DrMcDreamy]

*bump*

 

[CINA]

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.

 

[DrMcDreamy]

Thank you!

My work:

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

 

[DrMcDreamy]

I have a similar problem here's 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?