What is the magnification problem with two lenses?

In summary, the problem involves two lenses, one diverging with a focal length of -7 cm and one converging with a focal length of 16 cm, placed 38 cm apart. The task is to determine the distance between the first lens and the final image, as well as the magnification of the image. After correcting some values, the final answer for the distance is 63.3365 cm and the magnification is 0.1244.
  • #1
FlipStyle1308
267
0
Here is the problem I am working on:

Two lenses that are L = 38 cm apart are used to form an image, as shown in Figure 26-52. Lens 1 is diverging and has a focal length f1 = -7 cm; lens 2 is converging and has a focal length f2 = 16 cm.
26-53alt.gif


[a] Determine the distance from lens 1 to the final image. (Include the sign of each answer.)
What is the magnification of this image?

I got [a] correct with an answer of 62.3003 cm, using di1 = (1/f1 - 1/do1)^-1. For part I did the following:

m = (m1)(m2) - (-do/do1)(-di2/do2) = (di1)(di2)/(do1)(0.38m-di1) = -0.188, which is wrong. Anyone able to help me out? Thanks.
 
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  • #2
Not clear what you did wrong. What did you get for di1, do2, di2? And thus m1 and m2?
 
  • #3
Doc Al said:
Not clear what you did wrong. What did you get for di1, do2, di2? And thus m1 and m2?

di1 = -8.8421 cm
do2 = (38cm - di1)
di2 = 24.3003 cm
m1 = -do/do1
m2 = -di2/do2

I used my values for di1 and di2 for part [a] and got the answer right, so I'm sure at least those 2 are correct. I am pretty sure the rest are as well, but let me know if it is an error in any of these values that is giving me a wrong answer.
 
  • #4
FlipStyle1308 said:
di1 = -8.8421 cm
do2 = (38cm - di1)
di2 = 24.3003 cm
I get a different value for di1 (and thus di2 and part [a]), so one of us is making an error.
 
  • #5
FlipStyle1308 said:
Here is the problem I am working on:

Two lenses
I'm curious about what happened to this thread a couple of days ago. I was in the middle of responding and found that my preview was showing up in the middle of some totally unrelated thread. I restarted my browser to see if that would clear the problem and the whole thing was gone. I notified the webmaster that something had gone wrong, but I'd like to know if you deleted your first post of this problem, or if the system just lost it somehow. If you deleted it, I want to pass that infomation along to the webmaster.

And of course Doc Al is right. Your di1 is not correct; that error propegates through your calculation. Also, although your final expression for the magnification is correct there are some (probably just typographical) errors in the intermediate steps. As a check on your yourself it would be good to keep in mind that the virtual image of a diverging lens is between the lens and the focal point. Your di1 is not.
 
  • #6
OlderDan said:
I'm curious about what happened to this thread a couple of days ago.
I was wondering the same thing myself. My guess is that when we moved to the new server, some recent data was vaporized. Several threads that I responded to just disappeared; in some other threads, my posts were gone but the thread remained. I hope it's stable now.
 
  • #7
Yeah, I think when the servers were moving, the thread was deleted, so I made a new one, because I couldn't find it anymore.

If my di1 is not correct, how did I get a correct answer for part [a]? What did you get for di1 and di2? And are my other values that I am plugging in correct?
 
  • #8
Don't ask me how you got a correct answer for part [a]--I don't think you did! :smile: (It's not off by much.)

Rather than debate it, just recalculate di1. It'll take you less than a minute.
 
  • #9
Using di1 = (1/f1 - 1/do1)^-1 = (1/-14cm - 1/24cm)^-1, I still get the same answer of -8.8421 cm.
 
  • #10
FlipStyle1308 said:
Using di1 = (1/f1 - 1/do1)^-1 = (1/-14cm - 1/24cm)^-1, I still get the same answer of -8.8421 cm.
Where did that 14 cm come from?
 
  • #11
Double the 7cm, still keeping the negative sign. Is that incorrect? Am I supposed to just use -7cm instead of -14cm?
 
  • #12
Why in the world would you double it? The focal length is given as 7 cm, so use -7 cm (negative since it's a diverging lens).
 
  • #13
FlipStyle1308 said:
Yeah, I think when the servers were moving, the thread was deleted, so I made a new one, because I couldn't find it anymore.

If my di1 is not correct, how did I get a correct answer for part [a]? What did you get for di1 and di2? And are my other values that I am plugging in correct?
Stick with Doc Al on this, but FYI I don't think your first answer was correct. It was only off by about a centimeter, but it was off.
 
  • #14
Oh, my mistake lol. So the real answer for [a] is 63.3365 cm?, and di1 = -8.4194 cm and di2 = 25.3365 cm?

So my answer for should be -0.2028?
 
  • #15
Please correct that value of di1, once and for all. (Yes, part [a] looks OK.)
 
  • #16
LOL okay. And is my part correct as well? Whoops, I meant, 0.1244 as my answer for .
 
Last edited:
  • #17
FlipStyle1308 said:
So my answer for should be -0.2028?

How did you arrive at that answer?

Whoops, I meant, 0.1244 as my answer for .

Or that one?
 
  • #18
-0.1244, final answer?
 
  • #19
FlipStyle1308 said:
0.1244, final answer?
How did you get that answer? List all your values, and your final calculation.
 
  • #20
m = (m1)(m2) - (-do/do1)(-di2/do2) = (di1)(di2)/(do1)(0.38m-di1) = -0.1312 (sorry, this is my final answer lol)

di1 = -0.054 m
di2 = 0.253 m
do1 = 0.24
do2 = 0.38 m + 0.054 m
 
  • #21
Much better. (Whew! :tongue: )
 
  • #22
Thanks for all your help, Doc Al!
 

1. What is the lens magnification problem?

The lens magnification problem refers to the discrepancy between the magnification stated on a lens and the actual magnification achieved in real-world use. This can be caused by various factors, such as lens quality, camera settings, and shooting conditions.

2. How does lens magnification affect image quality?

Lens magnification can have a significant impact on image quality. A higher magnification can result in a more detailed and enlarged image, but it can also lead to distortions and loss of sharpness. On the other hand, a lower magnification may produce a smaller image with less detail, but it can also result in a more accurate representation of the subject.

3. How can I determine the actual magnification of a lens?

To determine the actual magnification of a lens, you can use a magnification chart or perform a test shot of a standardized subject at a known distance. You can then compare the resulting image size to the size of the subject in real life to calculate the actual magnification.

4. Does lens magnification vary between different camera brands?

Yes, lens magnification can vary between different camera brands due to differences in lens design, sensor size, and other factors. It is important to consider the specific camera and lens combination when determining the actual magnification.

5. How can I avoid the lens magnification problem?

To avoid the lens magnification problem, it is important to research and choose high-quality lenses that have been tested and reviewed for their accuracy in achieving the stated magnification. Additionally, using proper shooting techniques and understanding the relationship between camera settings and lens magnification can help minimize the issue.

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