Image Formed by a System of Lenses

[Renaldo]

Homework Statement

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?

Homework Equations

1/d_o + 1/d_i = 1/f

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?

 

[SammyS]

Homework Statement

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?

Homework Equations

1/d_o + 1/d_i = 1/f

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?

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 ?

 

[Renaldo]

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

 

[SammyS]

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]

Added in Edit:

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 ?

 

[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.

 

[SammyS]

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.

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 ?​

 

[Renaldo]

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

 

[SammyS]

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

Correct.

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

 

[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.

 

[SammyS]

Show your numbers.

It works out for me.

 

[Renaldo]

1/f = 0.4
1/d_o = 1.923

1/(0.4 - 1.923) = -0.66 cm

(f = 2.5)
(do = 0.52)

 

[SammyS]

1/f = 0.4
1/d_o = 1.923

1/(0.4 - 1.923) = -0.66 cm

(f = 2.5)
(do = 0.52)

d_O is negative.

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

This gives approximately

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

 

[Renaldo]

Thanks. That works.