Where is the final image in a 2 lens optics problem?

In summary, Homework Equations state that di1=+24cm and di2=+6cm. The object is -12cm from the first lens and +3cm from the second lens. The image is 27cm away from the first lens and 24cm away from the second lens.
  • #1
iurod
51
0

Homework Statement


We have a convex lens with a focal length of 8cm, which is 12cm from the second converging lens that has a focal length of 4cm. An object is 12cm from the first lens. Where is the final image?


Homework Equations


1/do + 1/di = 1/f



The Attempt at a Solution



So i know I have to apply the lens equation to both lens, starting with the first lens:

1/di = 1/f - 1/do = 1/8cm + 1/12cm = 1/24cm

di1=+24cm

here is where I get confused. Applying the lens equation to the second lens
I did:

1/di2= 1/f - 1/do(do here equals di1)
1/di2= 1/4cm - 1/12cm = 2/12cm

di2= +6cm

di1 + di2 = 30cm

The answer is 27cm, I've checked my math several times and still get the incorrect answer.

any help is greatly appreciated.

Thanks
 
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  • #2
iurod said:
di1=+24cm

here is where I get confused. Applying the lens equation to the second lens
I did:

1/di2= 1/f - 1/do(do here equals di1)

No, do is not equal to di1. Both the object distance and the image distance are measured from the lens. Make a drawing to see how far is the first image from the second lens and at which side. If the object is behind the lens, you have to take the object distance with negative sign.
I think that 27 cm is the distance of the final image from the object.
It is not di1+di2.

ehild
 
  • #3
The image after the first lens is 12cm to the left of the second lens. I am confused as what is in front and behind when it comes to the second lens. Do i take the left as front and right as back (like i did for the first lens), or is this reversed when talking about the second lens?

Thank you again for your help.
 
  • #4
It makes things clear if you draw a figure. Is the first image between the lenses, or outside?

ehild
 
  • #5
the image is outside, all the way to the right.
 
  • #6
So the first image would be 24 cm away from the first lens. How far is it from the second one?

ehild
 
  • #7
it is 12 cm to the right of the second lens. So here the object is -12cm to the lens, so instead of using positive 12 for the second lens equation i should have used -12.

1/di2 = 1/4cm - 1/-12 = 4/12cm = 3cm

di1+di2 = 24cm + 3cm = 27cm

?
 
  • #8
It is all right that di1 + di2 is 27 cm, but where is the image? You should give the reference point. So it is 27 cm away from what?

ehild
 
  • #9
I have written in my notes that it should be 27cm from the first lens.

Question, why wouldn't it be considered 27cm from the object?

Is it when us the lens equation we are always dealing with the image length relative to the lens and not the object?

Thanks for your help!
 
  • #10
No, it is 12 +3 cm far from the first lens, but 27 cm far from the object.

The distances in the lens equation always are meant from the lens.

ehild
 
  • #11
Now I get it, Thank you very much for all your help ehild
 

1. What is the "2 lens problem" in optics?

The "2 lens problem" refers to a common scenario in optics where light passes through two lenses in succession, resulting in a final image. This problem is often used in physics and engineering courses to demonstrate principles of refraction, magnification, and image formation.

2. How do you calculate the magnification in a 2 lens problem?

The magnification in a 2 lens problem can be calculated by multiplying the individual magnifications of each lens. The magnification of a lens is equal to the ratio of the image distance to the object distance, or M = -di/do. For example, if the first lens has a magnification of 2 and the second lens has a magnification of 3, the overall magnification would be 2 x 3 = 6.

3. What is the difference between a converging lens and a diverging lens?

A converging lens, also known as a convex lens, brings parallel rays of light together at a focal point. This type of lens is thicker in the middle and thinner at the edges. On the other hand, a diverging lens, also known as a concave lens, spreads parallel rays of light apart. This type of lens is thinner in the middle and thicker at the edges.

4. How does the placement of lenses affect the final image in a 2 lens problem?

The placement of lenses in a 2 lens problem can affect the size, orientation, and location of the final image. For example, if the lenses are placed closer together, the final image will be larger and closer to the lenses. If the lenses are placed further apart, the final image will be smaller and farther from the lenses.

5. Can a 2 lens problem have a virtual image?

Yes, a 2 lens problem can have a virtual image. A virtual image is formed when the light rays do not actually converge or intersect to form an image, but they appear to originate from a point behind the lens. This can occur in a 2 lens problem when the object is placed within the focal length of the first lens and the final image is formed on the same side of the lenses as the object.

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