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Image with an Plano-Convex lens

  • #1
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Homework Statement



This isn't really a homework, but I am trying to figure out how this would work.

I have an plano-convex lens with an focal length of 500 mm. I have an laser source that I like to focus on, to the infinity. I.e. I like to see the laser lobe.

Case: What happens to image of the lobe if the object is placed closer to the lens than the focal length?
For example 100 mm from the lens?
The curved side is towards the object.

Homework Equations



The lens formula gives,
1/f = 1/o + 1/i

The Attempt at a Solution



f = 500 mm
o = 100 mm
i = 1/(1/500-1/100) = -125, which makes no sense.

I tested this in the lab and I have seen the lobe at infinity but when I got back and tested again I couldn´t re-produce the same result. The lobe at infinity wasn´t at the focal length of the lens. I saw the laser lobe but not at the focal point. So I am trying to figure out what I am doing wrong.

For example, if the object is 3 meters in front of the lens, then the image of that object is 600 mm behind the lens.

f = 500 mm
o = 3000 mm
i = 1/(1/500-1/3000) = 600

The lobe at infinity, for the object placed 3 meters in front of the lens, would then be at the focal length of the lens (500 mm behind the lens).

What happens if the object is placed closer than the focal length of the lens? Can i see the image with the lens or does the object distance have to be larger than the focal length of lens?

Its weird, because I know I have seen the lobe when I first did the measurement. - 125 mm should indicate the image is formed in the opposite side of the lens.. but that doesn't make sense to what I have observed.
 
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Answers and Replies

  • #2
Merlin3189
Homework Helper
Gold Member
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I don't understand what you mean by "laser lobe"? But your calculation is fine.
I have an plano-convex lens with an focal length of 500 mm. I have an laser source that I like to focus on, to the infinity. I.e. I like to see the laser lobe.

Case: What happens to image of the lobe if the object is placed closer to the lens than the focal length?
For example 100 mm from the lens?
The curved side is towards the object.

Homework Equations

[/B]The lens formula gives, 1/f = 1/o + 1/i

The Attempt at a Solution



f = 500 mm
o = 100 mm
i = 1/(1/500-1/100) = -125, which makes no sense.
.
Why no sense? A virtual image.
 
  • #3
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I don't understand what you mean by "laser lobe"? But your calculation is fine.

Why no sense? A virtual image.
Well the laser source is not a point light source. Instead it has an rectangular profile, which I want to image onto a camera.

Would I see the virtual image behind the lens? With image distance of -125 mm, i picture the image to be seen 125 mm in front of the lens (and not behind the lens like in the normal case). For example with the object placed 3 meters in front of the lens, the image is seen at 600 mm behind the lens.

Where would I see the virtual image when the object is placed 100 mm in front of the lens (with focal length f = 500 mm)?
Likewise what would I see if I then place my camera at the focal length?

/ Thanks
 
  • #4
Merlin3189
Homework Helper
Gold Member
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You can't go looking at the virtual image - it is much too risky. You must project it onto a screen.
If your focal length is 100 mm, then obviously you need to have the lens more than 100 mm from the laser.
200 mm would give an equal sized image, but I guess you need a larger image, so that you can see it better.
Between 100 mm and 200 mm is what you need.
You can do the sums (as you've shown.) Otherwise, set the screen up 1 m away and adjust the spacing to focus the image.
 
  • #5
berkeman
Mentor
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Well the laser source is not a point light source. Instead it has an rectangular profile, which I want to image onto a camera.
How do you make a rectangular profile laser light source? What is the source? Laser diode, gas laser, etc.?
You can't go looking at the virtual image - it is much too risky. You must project it onto a screen.
Agreed. @nordmoon -- what laser safety measures do you have in place for these experiments?

A laser beam would more normally have a more Gaussian distribution of energy, no?

https://commons.wikimedia.org/wiki/File:Laser_gaussian_profile.svg

647px-Laser_gaussian_profile.svg.png
 

Attachments

  • #6
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Well obviously I’m not watching the image directly, but rather on a screen or with a camera and some filters. And I am using safety googles at all times.

The shape of my light source isn’t important but for simplicity we can say it’s a waveguide. I just want to image it onto my screen or camera.

I guess what I’m asking is what an virtual image is? Does it mean I can’t find it with my camera?

Am I correct that the virtual image will appear at the same side of the lens as the object, for the example case with the object placed 100 mm from lens, with focal length 500 mm.

Is there no way for me to see this image projected on the other side of the lens, which I would see if the object is at distances > focal length?
According to above the object distance have to be larger than the focal length.
 
  • #7
Merlin3189
Homework Helper
Gold Member
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634
...
The shape of my light source isn’t important but for simplicity we can say it’s a waveguide. I just want to image it onto my screen or camera.

I guess what I’m asking is what an virtual image is? Does it mean I can’t find it with my camera?
Yes you could photograph it with your camera by placing the camera pointing at the lens and laser and focused at the range of the virtual image.
But I think you would be better to get a real image on a screen and photo that.

Am I correct that the virtual image will appear at the same side of the lens as the object, for the example case with the object placed 100 mm from lens, with focal length 500 mm.

Is there no way for me to see this image projected on the other side of the lens, which I would see if the object is at distances > focal length?
According to above the object distance have to be larger than the focal length.
Yes. No. Yes !

Yes - closer than the focal length you won't get a real image.

So No - there is a way to project it onto a screen: exactly as you say, put the lens more than the focal length from the laser. (NB. I just realised in my first post I mistook the focal length as 100 mm, so my distances were wrong.)

Yes - the object distance needs to be greater than the focal length - and the image distance needs to be even greater, in order to have some magnification.

With your 500 mm lens you'll need to have the laser over 500 mm away and project the image onto a screen as far away as you can manage, maybe 5m or more if possible. That would give you nearly 10x magnification.

Maybe you could find another lens with a much shorter focal length. maybe borrow an eyepiece from a telescope or microscope? That would allow you to get a greater magnification at a shorter distance.
 
  • #8
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I see, Thank you. I will just change the lens then.
 

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