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I am trying to figure out where is the focal point and where is the image. I read some information online about the point where you see the image is the focal point, but however, my supervisor mentioned the point where you see the image is on "point 2" on the drawing. But as I remembered from school, the "point one" is where the retina is in order to have a good vision in our eyes.

The main purpose I am asking this because I want to get as much light as possible to a small area, and me and him were trying to figure out which point has the maximum light. I managed to obtain a focused image on a sheet of paper when doing the experiment (I believe that's point 1). Please help, Thanks a lot!!

Basically, could you tell me which point has the maximum intensity of light and if possible, point one or two is the image is. Thanks Convex lens, determine where is maximum light

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Mentioning "retina" makes this question confusing. Your eye has a lens in it. So sometimes you need to think about two lenses: One lens made of glass plus one lens which is your eye. The two-lens problem is different from the one-lens problem. I don't think that's what you meant. I think you're asking about the one-lens problem: Just a single glass lens and no human eyes. – Steve B Sep 17 '11 at 2:52

The focal point is generally not the image point. That is only true if the incoming rays are parallel, as in your drawing.

If the general case, use the thin lens formula, which is a ray-optics approximation that assumes an infinitely thin lens:

$$ \frac{1}{s_0} + \frac{1}{s_1} = \frac{1}{f} $$

where $s_0$ is the distance from the lens to the point where the incoming rays converge (the object point), and $s_1$ is the distance from the lens to the point where the outgoing rays converge (the image point), and $f$ is the focal distance of the lens.

You can see that if the incoming rays are parallel, then $s_0 \to \infty$, so $s_1 = f$.

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Er...I seem to recall that the focal point is where the image forms (that is what it means), but it is not generally at the focal length. – dmckee Sep 15 '11 at 20:04
"focal point" isn't really a well defined term, but most people would take it to be the point on the optical axis, one focal length behind the lens. The place where the image forms is called the "image plane" – Colin K Sep 15 '11 at 21:00
alright thanks, in details, I have two fresnel lens each have the same focal length, 220mm, and the placed like this: one curved surface facing the source and the other curved surface facing the other way, so they are back-to-back to each other. Say if I set the S0 to be 30cm, i got the focused image on a piece of paper at 14cm. Also If I set the S0 to be 22cm, the focused image on a piece of paper is at 16cm. Is this due to the lens abbreviation? Thanks – Aaron Kwong Sep 16 '11 at 7:55
I'm having a hard time understanding what it is you're doing. Placing two f=22 cm Fresnel lenses (essentially, thin plano-convex lenses) back-to-back doesn't give you one biconvex f=22 lens. – ptomato Sep 16 '11 at 9:54

First thing to note, is that there is a distinction between real and virtual images here, and that the treatment of the question depends on which you are dealing with and how you wish to examine it.

  • Virtual images Must be re-focused by a second optical system, but your eye is exactly that. For this purpose you want to be off the focal point and in a region where the rays are diverging. (Your diagram does not cover this case!)
  • Real images (as in your diagram) can be treated two ways:
    1. You can project them onto a viewing surface (like a theater screen). For that you want to place the screen at the focal point (point one in you diagram) so that things will be in focus.
    2. You can view them directly by re-focusing them with your eye. This case is just like virtual images: you want to be off the focal point and in a region where the beams are diverging (this is point 2 in your diagram) because your eye is a converging system and you want the image to have a finite size.
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Yep, sorry for my misleading question, I am using a piece of paper to find the real image point, where I will put the optical fiber at the same spot to let the maximum light into the fiber, is this theory correct? – Aaron Kwong Sep 16 '11 at 7:57

as I remembered from school, the "point one" is where the retina is in order to have a good vision in our eyes.

There's probably confusion between you and the supervisor w.r.t. the terms you're using, and your explanation above is a bit unclear anyway (I don't mean any disrespect, I'm just stating a fact the way I see it).

If your drawing represents the structure of the human eye, then yes, point 1 is where the retina is. The lens creates a so-called real image that gets projected on the retina, just like a digital camera projecting the image on a sensor.

However, if your drawing represents a lens on your bench, with the source very far away to the left, and you put your eye in point 1, then you won't see much. You need to put a piece of paper in point 1, then the image will form on it - again, just like at the movies. This is because the image is real, not virtual - so therefore you can't put your eye in its middle and expect to see it. That would be like going at the movies and standing with your back against the silver screen - do you think you'll see the movie? Of course not.

I am trying to figure out where is the focal point and where is the image.

For a convergent lens, talking about real images, those that you can see if you insert a piece of paper:

If the source is very very far away (I'm talking a few hundred meters or more), the image is formed exactly at the focal point on the other side of the lens. The location of the image and the location of the focal point coincide.

If the source starts moving closer to the lens, the image starts to form further away - the image is moving away from the lens as the source is moving closer in.

If the distance between source and lens is 2x bigger than the focal length, then the image is formed at the same distance (2x focal length), on the other side of the lens.

If the source is at exactly 1x the focal length (the source is in the focal point), then the image is formed very very far away on the other side.

If the source moves closer to the lens than the focal length, then no real image is formed.

Look again at the diagram I posted on the page of your previous question a few days ago.

The main purpose I am asking this because I want to get as much light as possible to a small area, and me and him were trying to figure out which point has the maximum light.

In theory, that point is where the real image of the source is formed. See above.

In practice, because the source is not a point of light, and because lenses are not perfect, there might be smaller variations. But don't expect to find the max light point too far away from where you can clearly see an image of the source if you insert a piece of paper.

So it's really easy. Insert a piece of paper, move it back and forth until you get a clear image, and start from there.

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so in one simple sentence: when I insert a piece of paper into the system, and the point where I can see a clear Image, thats the max light point? so I can put my optical fiber at the same place? Thanks!! – Aaron Kwong Sep 16 '11 at 7:50
Short answer: yes. Long answer: it depends, but in most situations it's either exactly the same place, or very close. I'm trying to figure out a counterexample, but I'm drawing blanks. So, there you have it. Put a piece of paper in front of it, and start your investigations at that point. BTW, you could jury-rig a photodiode with a measuring device, put it at the other end of the fiber, and move the lens-end of the fiber until the photodiode shows maximum light. This method is absolutely exact, leaves nothing to guessing or trial. The scheme should be simple enough. – Florin Andrei Sep 16 '11 at 20:16
Ask a friend who knows electronics to put the circuit together. It's a photodiode and a resistor, connected to a 9 V battery - it's like $5 worth of components. Just measure the current through the diode with a multimeter - when the current is maximum, the light is maximum. – Florin Andrei Sep 16 '11 at 20:20

It might be possible that the bulb you are placing is actually not at infinity(i mean at a distance comparable to the focal length). In which case your image formation, or point of maximal intensity would be a bit farther from focal point, as your supervisor mentioned.

But otherwise, the image has to be on focal point.

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It's usually at "point one." The image appears at the focal distance, but only if the object is far away.

There is no "focal point." That's a misconception promoted by k-6 grade texts.

Beware of widespread grade-school misconceptions:

  • A convex lens has a "focal point." (no, it doesn't)
  • illuminated objects send out parallel rays (nope.)
  • The image enters the lens as parallel rays. (no, only light from a single distant point does that)
  • Behind the lens is the place where the image turns upside-down. (wrong.)
  • The object must be smaller than the lens diameter. (um. what?!)
  • A lens in sunlight can form an intense burning point (no, it forms a small inverted image of the sun.)


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Ok thanks, what if the object is close? is there one clear point or its in a range due to the reflection of the lens? – Aaron Kwong Sep 16 '11 at 7:56

The point 1 , where incoming parallel rays all meet, is the focal point. At this point all the incoming rays from a far object are just a single point , not an image (this is the distance you would hold a magnifying glass away from something you wanted to burn with rays from the sun) . The image of a far away object will focus on a screen at some point 2 beyond the focal point and it will be rotated 180 degrees with respect to the object .

BTW in the eye the image on the retina is rotated 180 degrees(its upside-down, and rightside-left ) , the brain rotates it back. There are people , either at birth or from brain damage who do not have this capacity to rotate retinal images . Also see ( copy and paste )

YEP, we focus on an upside-down world !


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