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?