# Image Formed by a System of Lenses

1. Apr 29, 2013

### Renaldo

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

Three converging lenses of focal length 2.5 cm are arranged with a spacing of 2.2 cm between them, and are used to image an insect 2.3 cm away. Where is the image?

2. Relevant equations

1/do + 1/di = 1/f

3. The attempt at a solution

I used the above equation to find the location formed by the first lens. Then I used that information to calculate the position of the image formed by the second lens. Finally, I used that information to calculate the position of the image formed by the third lens.

My answer is 0.66 cm to the right of the third lens.

I assumed the insect was initially to the right of the system of lenses.

Correct answer is 0.43 cm. What did I do wrong?

2. Apr 29, 2013

### SammyS

Staff Emeritus
It's very hard to say what you did wrong unless you give us more detail.

What position did you get for the image from each lens ?

3. Apr 29, 2013

### Renaldo

I attached my work for this problem. db is the distance from the first first lens to the bug. The arrow represents the bug.

#### Attached Files:

• ###### Bug Optics.jpg
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4. Apr 29, 2013

### SammyS

Staff Emeritus
The image formed by the second lens is about 2.72 cm to the left of the second lens, thus about 5.2 cm to the right of the third lens.

[STRIKE]What sign should you use for 5.2 cm when used as the object distance with lens 3 ?[/STRIKE]

That should be
... thus about 0.52 cm to the left of the third lens.​

What sign should you use for 0.52 cm when used as the object distance with lens 3 ?

Last edited: Apr 29, 2013
5. Apr 29, 2013

### Renaldo

I just want to make sure we are using the same convention.

My L1 is on the far right. L3 is on the far left. L2 is in between. If the image formed by the second lens is 2.72 cm to the left of the second lens, then it would not be 5.2 cm to the right of the third lens. Rather, it would be 0.52 cm to the left of the third lens.

6. Apr 29, 2013

### SammyS

Staff Emeritus
Right. Those were errors -- a typo and a dyslexia error.

That should have said:
The image formed by the second lens is about 2.72 cm to the left of the second lens, thus about 0.52 cm to the left of the third lens.​

The important question is:
What sign should you use for the object distance with lens 3 if the object is to the left of the lens ?​

7. Apr 29, 2013

### Renaldo

negative, because the image is formed on the same side as the object.

8. Apr 29, 2013

### SammyS

Staff Emeritus
Correct.

That should give you the desired result for the location of the final image.

9. Apr 29, 2013

### Renaldo

At first I said the image was to the right of the third lens. I now realize that was incorrect. I still don't understand how that affects the numerical value of my final answer, which is still wrong.

10. Apr 29, 2013

### SammyS

Staff Emeritus

It works out for me.

11. Apr 29, 2013

### Renaldo

1/f = 0.4
1/do = 1.923

1/(0.4 - 1.923) = -0.66 cm

(f = 2.5)
(do = 0.52)

Last edited: Apr 29, 2013
12. Apr 29, 2013

### SammyS

Staff Emeritus
dO is negative.

$\displaystyle \frac{1}{d_i}+\frac{1}{\displaystyle-\frac{2957}{5690}}=\frac{1}{2.5}$

This gives approximately
$\displaystyle d_i=\frac{1}{0.4+1.9242}$​

13. Apr 29, 2013

### Renaldo

Thanks. That works.