Finding lens diameter - German physics task

In summary, the focal length of the camera is unknown, but it is assumed that it is far away from the ruler. The rays from the lamps will be parallel and will focus in the focal point.
  • #1
Fabon
14
1
Hey everybody,

I'm right now trying to solve a question I found in a German physics competition 2006 concerning the diameter of a thin camera lens. I'm not able to find a solution, since there is basically nothing given.
Here the task (I underlined phrases which might be important):

A ruler was photographed. There was a chain of lights with small lamps paced far behind the ruler, so that the lamp occur to unclear. You are allowed to assume, that the picture was taken by a single, very thin lens.
Find the diameter of the lens by using the photo.

The only given value is the diameter of the unclear circles caused by the small lamps (by using the ruler on the photo). I know some laws and how to make these "ray diagrams" with lines, but I have no Idea how to solve this task and what is needed, especially since I have never seen a formula (or so) that includes the diameter.

I hope you can help me finding a solution.
 

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  • #2
Fabon said:
I know some laws and how to make these "ray diagrams" with lines
Can you show that?

If the focal length is f, how far away from the ruler is the camera?
 
  • #4
  • #5
Fabon said:
Concerning your question: well, maybe f, since the ruler is focused?

Where would an object at exactly f focus?
 
  • #6
Lok said:
In your diagram how would the far off lights show? Could you draw them and their respective geometrical ray bouncing?

Well, the principle ray and the focal ray would occur to be nearly parallel, since they are so far away, and the rays would meet very close after passing the lens.
 
  • #7
Fabon said:
Well, the principle ray and the focal ray would occur to be nearly parallel, since they are so far away, and the rays would meet very close after passing the lens.
So where would the far off lights focus? And what would happen with the rays afterwards?
 
  • #8
Lok said:
Where would an object at exactly f focus?

Mhh, maybe it is not even possible to project such an object on the other side, since the focal ray would have to point upwards.
 
  • #9
Fabon said:
Mhh, maybe it is not even possible to project such an object on the other side, since the focal ray would have to point upwards.
This just means that the ruler is not at f.
And you can also establish where it is, "between lens and f" or from "f outwards".
 
  • #10
Lok said:
So where would the far off lights focus? And what would happen with the rays afterwards?

Since they are nearly parallel, they would focus (near to or) in the focal point. After meeting, the rays would diverge.
 
  • #11
Lok said:
And you can also establish where it is, "between lens and f" or from "f outwards".

That's already a nice aspect for me to understand the situation - thank you. But how will this knowledge help me to find the diameter?
 
  • #12
Fabon said:
Since they are nearly parallel, they would focus (near to or) in the focal point. After meeting, the rays would diverge.
Now you will get two things on the same screen (there must be one).
One image focused with a graded ruler and one unfocused diverged ray which overlays the ruler on the screen.
You could get the answer by drawing only.
 
  • #13
yes. a nice problem. I haven't seen it before.
 
  • #14
Lok said:
You could get the answer by drawing only.

stedwards said:
What does the image of the point source look like on the image plane?

I hope I got the situation. Sorry for my terrible drawing (as I noted on the picture, the rays coming of the lamp are supposed to be parallel).
 

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  • #15
Fabon said:
I hope I got the situation. Sorry for my terrible drawing (as I noted on the picture, the rays coming of the lamp are supposed to be parallel).
I agree with the adjective. :D
You need a screen, which is also positioned somewhere between lens f and outwards.
I insist on the drawing as the math is really easy to derive form it.
 
  • #16
Lok said:
You need a screen, which is also positioned somewhere between lens f and outwards.

Is it possible to place the screen on the position (behind the lens) where I wrote ruler?
 
  • #17
Fabon said:
I hope I got the situation. Sorry for my terrible drawing (as I noted on the picture, the rays coming of the lamp are supposed to be parallel).
Oh, sorry for not noticing earlier.
But you set the ruler between f and ens, where would it focus and what type of image would it produce?
 
  • #18
Fabon said:
Is it possible to place the screen on the position (behind the lens) where I wrote ruler?
Please check the ruler's (real one) position as stated above.
This would allow you to set a screen behind the lens in the appropriate place for a focused projection to take place.
 
  • #19
Lok said:
what type of image would it produce?

a virtual image?

Lok said:
Please check the ruler's (real one) position as stated above.
This would allow you to set a screen behind the lens in the appropriate place for a focused projection to take place.

Second try! This time I improved my drawing. :D
 

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  • #20
Lok is correct. You need a nice geometric drawing done with a ruler. Then it becomes a matter of comparing triangles. Other than that, I'm only getting in the way, here.
 
  • #21
Fabon said:
a virtual image?
Second try! This time I improved my drawing. :D
Nice. Now the yellow lines are the unfocused rays and they make a speck on the screen of some diameter in accordance with what variables?
The ruler also makes a real image on the screen and in accordance to what variables.
By a bit of playing with the drawing you can get an answer, or you could get 2 equations which will do the same.
 
  • #22
Lok said:
and they make a speck on the screen of some diameter in accordance with what variables?

according to the diameter of the lens?

Lok said:
The ruler also makes a real image on the screen and in accordance to what variables.

Does the size of the image of the ruler depends on the distance to the lens? That irritates me, since we don't know the distance.
 
  • #23
Fabon said:
according to the diameter of the lens?
Aha. So finally you got the diameter of the lens into an equation.
Fabon said:
Does the size of the image of the ruler depends on the distance to the lens? That irritates me, since we don't know the distance.
Competition tip: Whenever you get a problem that you are sure to have a solution ( as in a German phyzy. compet. 2006), and there are only few given values and 100 or so unknown variables, you can bet your *** that they will cancel or become insignificant in the end.
 
  • #24
Lok said:
Aha. So finally you got the diameter of the lens into an equation.

Well, so the ruler measures the diameter D of the light-circles on the picture. So does this mean that the diameter of the lens is D as well?
I think I'm confused...
 
  • #25
Fabon said:
Well, so the ruler measures the diameter D of the light-circles on the picture. So does this mean that the diameter of the lens is D as well?
I think I'm confused...
That would be it.
And at this point one takes his smartphone in one hand and a ruler in the other and with a far reflection or light in the background you could measure your smartphones aperture.
 
  • #26
Is this really so simple? The diameter of the lens would be the diameter of the light-circles? I've read and understanded the thread, but I can't figure out, with which equation you solved the problem.
I've played around with these lens formulas:
1/o + 1/i = 1/f or
O/I = o/i (o=object distance, i=image distance, O=Object size, I = image size)

But I do not get a result. It would be nice if somebody can give me a small hint.

Thanks,
Matt
 
  • #27
Max, look at the drawing in Fabon's post #19. It's a problem best solved using ray diagrams. Note the point sources, projected, out of focus, on the image screen, to the right of the lens.
 
  • #28
stedwards said:
Note the point sources, projected, out of focus, on the image screen, to the right of the lens.
Thanks, I can see that the point sources on the screen are as big as the lens diameter. Is this relevant?
 
  • #30
Matt_ said:
Thanks, I can see that the point sources on the screen are as big as the lens diameter. Is this relevant?

No, it's not relevant. Are you familiar with Geometric Optics? See the drawing in the section "Imaging with a Lens" in http://openspim.org/SPIM_Optics_101/Theoretical_basics. It can be used to solve simple problems geometrically for a thin lens, as is the case in the original question of this thread.

For thin lenses, all you need to know is: 1) A ray of light from a point on an object will pass directly through the center of the lens without change in direction. 2) A ray incident perpendicular to the lens will pass through the focal point of the lens on the other side of the lens. Where these three rays converge is on the image plane of the object.
 
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  • #31
Thank you, due to your explanation I've understand it better.
stedwards said:
Are you familiar with Geometric Optics?
I think yes, I understand the ray diagram (and how a image is transmitted). But somehow I don't get the relation between the lamps and the diameter.

William White said:
Ohh, sorry for my stupidity:
The circle of confusion in the image plane is obtained by multiplying by magnification m: (Wikipedia)
c = diameter of the blur spot * magnification (1)

Is this right?
 
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  • #32
Matt_ said:
I think yes, I understand the ray diagram (and how a image is transmitted). But somehow I don't get the relation between the lamps and the diameter.
Hi Matt_,
The relation you search for is between the spot diameter and the focused image of the ruler. That is to say that for this special case (thin lens and all idealizations), for any lens focal length the result is the same.
Also this would be the reason why a smartphone with it's 2mm (at best) diameter will NEVER take a picture like this one https://en.wikipedia.org/wiki/Bokeh#/media/File:Josefina_with_Bokeh.jpg with the blurred background. It simply cannot do it unless the subject is very small (~10mm) an close to the camera lens (under ~20mm), conditions which cannot be met with the focus of a normal smartphone.
 
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  • #33
Lok said:
Hi Matt_,
The relation you search for is between the spot diameter and the focused image of the ruler. That is to say that for this special case (thin lens and all idealizations), for any lens focal length the result is the same.
Also this would be the reason why a smartphone with it's 2mm (at best) diameter will NEVER take a picture like this one https://en.wikipedia.org/wiki/Bokeh#/media/File:Josefina_with_Bokeh.jpg with the blurred background. It simply cannot do it unless the subject is very small (~10mm) an close to the camera lens (under ~20mm), conditions which cannot be met with the focus of a normal smartphone.
Thanks to all, I understand the problem now :)
 
  • #34
Sorry to question the consensus about a diagram being the solution, (though I always sketch a diagram or two to help me visualise the situation and mark on my symbols.)

I thought of one position for the ruler and screen, where you would not even need a diagram to work out the lens diameter (u = v = 2f, M =1, spot size = D)
Then I thought, there are many other positions where the ruler and screen could be, so I resorted to calculation to show that the special case gives the same result as the general one. But, how can you get this general result from a geometric diagram? Would you not have to draw an infinite number of diagrams? I had to use an inaccurate symbolic diagram and algebra (though pretty simple algebra.)
 
  • #35
The focal points don't matter. There are four rays to consider: two parallel ones from the distant light source, and two parallel ones from two points on the ruler, where the previous two rays go through. We don't know any distances, but we know the two light rays in the pairs ("up", "down") will hit the same spot each (because they have identical trajectories). All other rays from the ruler will hit the same spot as well as the ruler is in the focus.
One diagram catches all possible cases.
 
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<h2>1. How do I find the diameter of a lens in a German physics task?</h2><p>In order to find the diameter of a lens in a German physics task, you will need to use the formula: d = 2f tan(θ), where d is the diameter of the lens, f is the focal length, and θ is the angle of the lens. You will need to know the focal length and angle of the lens in order to calculate the diameter.</p><h2>2. What is the purpose of finding the diameter of a lens in a German physics task?</h2><p>Finding the diameter of a lens in a German physics task is important because it helps determine the size of the lens, which can affect its optical properties and performance. It is also a fundamental calculation in optics and can be used to determine other important parameters of a lens.</p><h2>3. How do I measure the focal length of a lens in a German physics task?</h2><p>The focal length of a lens can be measured by using a lensometer or by using the thin lens equation: 1/f = 1/u + 1/v, where f is the focal length, u is the object distance, and v is the image distance. You will need to know the object and image distances in order to calculate the focal length.</p><h2>4. Can I use any unit of measurement to find the diameter of a lens in a German physics task?</h2><p>Yes, you can use any unit of measurement as long as you are consistent throughout your calculations. However, it is recommended to use units such as millimeters or centimeters for better accuracy and precision.</p><h2>5. Is there a specific method for finding the angle of a lens in a German physics task?</h2><p>There are a few different methods for finding the angle of a lens, depending on the type of lens and the information given in the task. One method is to use a protractor to measure the angle directly. Another method is to use trigonometric functions and the known dimensions of the lens to calculate the angle. It is important to carefully read and understand the given information in order to determine the most appropriate method for finding the angle of the lens.</p>

Related to Finding lens diameter - German physics task

1. How do I find the diameter of a lens in a German physics task?

In order to find the diameter of a lens in a German physics task, you will need to use the formula: d = 2f tan(θ), where d is the diameter of the lens, f is the focal length, and θ is the angle of the lens. You will need to know the focal length and angle of the lens in order to calculate the diameter.

2. What is the purpose of finding the diameter of a lens in a German physics task?

Finding the diameter of a lens in a German physics task is important because it helps determine the size of the lens, which can affect its optical properties and performance. It is also a fundamental calculation in optics and can be used to determine other important parameters of a lens.

3. How do I measure the focal length of a lens in a German physics task?

The focal length of a lens can be measured by using a lensometer or by using the thin lens equation: 1/f = 1/u + 1/v, where f is the focal length, u is the object distance, and v is the image distance. You will need to know the object and image distances in order to calculate the focal length.

4. Can I use any unit of measurement to find the diameter of a lens in a German physics task?

Yes, you can use any unit of measurement as long as you are consistent throughout your calculations. However, it is recommended to use units such as millimeters or centimeters for better accuracy and precision.

5. Is there a specific method for finding the angle of a lens in a German physics task?

There are a few different methods for finding the angle of a lens, depending on the type of lens and the information given in the task. One method is to use a protractor to measure the angle directly. Another method is to use trigonometric functions and the known dimensions of the lens to calculate the angle. It is important to carefully read and understand the given information in order to determine the most appropriate method for finding the angle of the lens.

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