Homework Help: Location of object that forms an image in 2 lens system

1. Mar 11, 2015

Phynos

1. The problem statement, all variables and given/known data

Two converging lenses having focal lengths of f1 = 12.5 cm and f2 = 19.5 cm are placed a distance d = 49.0 cm apart as shown in the figure below. The image due to light passing through both lenses is to be located between the lenses at the position x = 33.0 cm indicated. (a) At what value of p should the object be positioned to the left of the first lens? (b) What is the magnification of the final image? (c) Is the final image upright or inverted? (d) Is the final image real or virtual?

2. Relevant equations

1/f = 1/p + 1/q

SIGN CONVENTION:
p - positive because object is in front of mirror 1
q - positive because image is behind mirror 1
P - positive because object is in front of mirror 2
Q - negative because image is in front of mirror 2

3. The attempt at a solution

I will be using capital letters for the quantities specific to the second lens.

For lens one:

1/f = 1/p + 1/q
Since the image of the first lens is the object of the second (q = P)
1/f = 1/p + 1/P
1/P = 1/f - 1/p

For lens two:

1/F = 1/P + 1/Q
The image of the second lens (final image) is given in the question (Q = d-x)
Also, this quantity is negative since it appears in front of the lens (-Q = x-d)
1/F = 1/P + 1/(x-d)
1/P = 1/F - 1/(x-d)

Putting together the two equations I just solved:

1/f -1/p = 1/F - 1/(x-d)

Solving for p (object for first lens):

1/p = 1/(x-d) +1/f - 1/F
p = [ 1/(x-d) + 1/f - 1/F ]^(-1)

Subbing in values:

p = -29.6cm

The magnitude seems reasonable, my problem is the sign. I followed the sign convention so why isn't this value positive? Object distance (p) is positive when the object is in front of the lens.

2. Mar 11, 2015

TSny

Hello, Phynos.
Yes, the image of the 1st lens is the object of the 2nd lens. But, that doesn't imply that q = P. Think about the relation between q and P.

3. Mar 12, 2015

Phynos

Oh of course... I overlooked that.

Equation 1 modification:

1/f = 1/p + 1/q

[q = P - d]

1/f - 1/p = 1/(P-d)
P (1/f - 1/p) - d (1/f - 1/p) = 1
P = (1 + d (1/f - 1/p) )/(1/f - 1/p)
P = (1/f - 1/p)^(-1) + d
1/P = [ (1/f - 1/p)^(-1) +d ]^(-1)

Set equal to the second equation...

[ (1/f - 1/p)^(-1) + d ]^(-1) = 1/F - 1/(x-d)
(1/f - 1/p)^(-1) + d = [ 1/F + 1/(d-x) ]^(-1)
1/f - 1/p = [ [ 1/F + 1/(d-x) ]^(-1) -d ]^(-1)
1/p = 1/f - [ [ 1/F + 1/(d-x) ]^(-1) -d ]^(-1)

p = [ 1/f - [ [ 1/F + 1/(d-x) ]^(-1) -d ]^(-1) ] ]^(-1) <-- This has to be one of the messiest things I've ever written.

Subbing in values gives me:

p = 9.54cm

Which seems reasonable, and it's the right sign! Thanks for the catch.

EDIT:

Oh no... q = d - P

so I had to do some sign changes. (Did it on my whiteboard here, not going to write it all out again, I'm fairly certain it's right this time...)

p = [ [ [ 1/(x-d) + 1/F ]^(-1) - d ]^(-1) - 1/f ]^(-1)

p = 22.7cm

I think I'm going to spend some time this weekend figuring out how to make equations neater on forums like this.

Last edited: Mar 12, 2015
4. Mar 12, 2015

TSny

Almost. Check the signs in the equation q = P - d.

I find it easier to work out data for the second lens (plugging in numbers as I go) and then figure out the data for to the first lens.

5. Mar 12, 2015

Phynos

I didn't think you'd be so quick to reply again, I just edited the last post. Noticed as soon as I hit submit. Thanks!

Yeah, I suppose I didn't do this the most efficient way... I usually like to rearrange the variables without putting in the quantities until I've solved for the thing I need. Sometimes the result is really nice. Other times not so much.

6. Mar 12, 2015

Phynos

This time it's much simpler, but I get a different answer. Perhaps I made a mistake. (This is driving me crazy, I'm not fond of optics!)

Equation 1:
p = [ 1/f - 1/(d-P) ]^(-1)

Equation 2:
P = [ 1/F + 1/(x-d) ]^(-1)

Inserting 2 into 1:
p = [ 1/f - 1/ (d - [ 1/F + 1/(x-d) ]^(-1) ) ]^(-1)

p = 13.7cm

Working backwards from both of my solutions yields results that don't match up with the information the question gives. Clearly my algebra is failing me.

7. Mar 12, 2015

Phynos

OK, so if anyone looking at this could just completely ignore the other posts... That would be great. I'm abandoning that horrid solution for a slightly simpler one, but I'm getting nonsense. It's slightly easier to read this time.

As before, here are the original equations and subsequent manipulations:
EQ 1:
1/f = 1/p + 1/q
1/p = 1/f - 1/q
p = [ 1/f - 1/q ]^(-1)

EQ 2:
1/F = 1/P + 1/Q

We have (Q = d-x), Q must be negative, since it's behind the mirror, thus (-Q = x-d)
Also (P = d-q)

1/F = 1/(d-q) + 1/(x-d)

Solving for q.

1/(d-q) = 1/F - 1/(x-d)
d-q = [ 1/F - 1/(x-d) ]^(-1)
q = d - [ 1/F - 1/(x-d) ]^(-1)

I need inverse q to put into the other equation, so...

1/q = [d - [ 1/F - 1/(x-d) ]^(-1) ]^(-1)

Inserting this into EQ 1:

p = [ 1/f - [d - [ 1/F - 1/(x-d) ]^(-1) ]^(-1) ]^(-1)
p = 18.1cm

Now if I work backwards with that value...

q = [1/f - 1/p]^(-1)
q = 40.4

P = d-q = 49.0 - 40.4 = 8.6

Q = [1/F - 1/P]^(-1)
Q = -16.0

49 - 16 = 33, the x position given in the question. So this is the correct answer. I figured I would post it in case someone else gets this problem.

8. Mar 12, 2015

Phynos

b) Total Magnification. (I need assistance with this one)

The total magnification is the product of each individual magnification.

m1 = m and m2 = M

m = -q/p
where q = [1/f - 1/p]^(-1)

m = -[ p(1/f - 1/p) ]^(-1)
m = -[ p/f - 1 ]^(-1)
m = -2.23

Shouldn't this be smaller than one?

M = -Q/P
Q in this case is negative, so:
M = Q/P

1/F = 1/P + 1/(-Q)
1/P = 1/F + 1/Q

M = Q [ 1/F + 1/Q ]
M = Q/F + 1
M = 1.82

Shouldn't this also be less than one?

If I'm not mistaken, the image size decreases each time. Greater than one image magnification on either lens seems inconsistent with this.

Total Magnification = 4.06

Yeah, there's something wrong with that...

9. Mar 12, 2015

TSny

Q must be negative, as you say. But that means you should write it as Q = x - d since that will make Q negative.

-Q would be a positive number.

However, it turns out that you actually did let Q = x - d when you plugged into the formula and got:
So, everything looks good here.

10. Mar 12, 2015

TSny

You can show that if the object distance of a converging lens lies between f and 2f, then the magnification will be greater than 1. I agree with your value of m here.

Q is negative. But you should still write M = -Q/P. Then M will be positive, as it should be in this case. Remember Q = -16 cm, so you should plug in -16 cm for Q.