# 2 converging lenses

1. Nov 4, 2008

### chickenoodle

1. The problem statement, all variables and given/known data
There are 2 lenses separated by 5 cm. Very far to the left of the first lens is an object (the distance of the object from lens one is NOT given). The focal length of lens 1 is 20 cm and the focal length of lens 2 is 10 cm. Where is the image due to the first lens? State the image's distance from lens 1 and whether this image is on the right of left of lens 1.

2. Relevant equations

the mirror equation:

1/s + 1/s' = 1/f

s is the distance of the object
s' is the distance of the image
f is the focal length

3. The attempt at a solution

for the first lens:

1/s' = 1/f - 1/s = (1/20cm) - 1/s

s is not given and I don't know how to find it so I used a crude method of setting up an equation with 2 unknown variables.

if d is the total distance; d = x + 5cm

then x is the distance between the object and lens 1; x = d - 5cm

so I set x = s = d - 5cm and then solved for s'.

s' = [(1/20cm) - {1/(x-5cm)}]^-1

This is the whole solution I came up with but you get the idea, the equation just gets longer and uglier.

this doesn't seem right. Is there some way to solve for s and get a numerical value?

This question is just part "a" from one very long question and I'm hoping that if someone can help me figure this first part out I'll be able to do the rest on my own.

Thanks

2. Nov 5, 2008

### Redbelly98

Staff Emeritus
Welcome to PF.

Though s is not given, we are told that it is very far away. So s is very large in comparison with f.

Given that, what can you say about 1/s when compared to 1/f?

3. Nov 5, 2008

### chickenoodle

I just found this in the book:

as s --> infinity s' --> 0.5r = f

so I can just neglect s then and s' becomes

1/s + 1/s' = 1/f

0 + 1/s' = 1/f

s' = f = 20 cm

Last edited: Nov 5, 2008
4. Nov 5, 2008

### physics girl phd

For the first lens in the compound lens system, your above math statement shows that the image will be formed at the focus because the object is is far away... and doesn't that make sense based on what you know about simple optics? In fact, an easy way to find the focal length of a converging lens is to try to see where it focuses in image of something really far away (like the sun... focusing it into a tiny dot-like image!)

I think this is what you were trying to find in your original statement of the question. Now I suspect the problem moves on to the second lens:

So, what does your second lens do? It takes the image from the first as its object and makes an image of that. So, where is the image from the first lens relative to the second lens?

5. Nov 5, 2008

### Redbelly98

Staff Emeritus
1/s --> 0 as s --> infinity, so yes.

EDIT: responding to Post #3.

6. Nov 5, 2008

### chickenoodle

So the "intermediate object" is the image due to the first lens and it is located -15 cm (from 5cm - 20cm) on the incident-light side of the second lens.

the focal length of the second lens is 10cm and s = -15cm so s' is

1/s' = 1/f - 1/s = 1/10cm - (1/-15cm) = 0.167 cm

s' = 6 cm to the right of the second lens

7. Nov 5, 2008

### physics girl phd

Without checking actual numbers, that seems right for the location of the image after the second lens:

why? This makes some real sense, especially with far away (therefore nearly parallel) approaching rays. Adding the first, "weaker," longer focus lens starts to focus the light just a bit before it hits the second "stronger" shorter focal length lens which will focus it more "quickly"... so with this first lens in place, the light focuses closer to the second lens than it would without the weaker lens there!

"Quickly" is perhaps a funny term to use, but since the light does focus closer with the stronger lens, it actually does focus more quickly also! :rofl: