Lenses, diminished and magnified

[richnfg]

This is probably a stupidly easy question, but I'm confused so I need to ask about it.

I have this experiment I was doing where I have a light source at one end (which is enclosed in a casing and only let's light get out in the shape of an arrow because of an arrow shaped hole) and a screen at the other. In between is a lens and I move it to get the image of the arrow on the screen in focus. I notice I can get two points of focus, one where the arrow is diminished (smaller) and where it is magnified (larger).

Also, when I move the screen to 40cm from the light source I only seem to be able to get one in focus image, which I cannot decide is diminished or magnified.

I'm very confused as to why this happens, any help would be great.

Thanks!

 

[Andrew Mason]

This is probably a stupidly easy question, but I'm confused so I need to ask about it.

I have this experiment I was doing where I have a light source at one end (which is enclosed in a casing and only let's light get out in the shape of an arrow because of an arrow shaped hole) and a screen at the other. In between is a lens and I move it to get the image of the arrow on the screen in focus. I notice I can get two points of focus, one where the arrow is diminished (smaller) and where it is magnified (larger).

Also, when I move the screen to 40cm from the light source I only seem to be able to get one in focus image, which I cannot decide is diminished or magnified.

I'm very confused as to why this happens, any help would be great.

Are you sure that you are keeping the lens and object fixed while you move only the screen?

For a given object distance and focal length, there is only one point at which the image will be in focus. That is determined by the lens equation:

$$\frac{1}{f} = \frac{1}{o} + \frac{1}{i}$$

Edit and correction:For a fixed screen to object distance, S, there will be two lens positions where the object is in focus provided S>4f. They will be:

$$o = S/2 \pm \frac{\sqrt{S^2 - 4Sf}}{2}$$

If $S = 4f$ there will be only one solution. So I gather that your screen was 80 cm from the object and the lens has a focal length of 20 cm.

AM

 

[richnfg]

Are you sure that you are keeping the lens and object fixed while you move only the screen?

For a given object distance and focal length, there is only one point at which the image will be in focus. That is determined by the lens equation:

$$\frac{1}{f} = \frac{1}{o} + \frac{1}{i}$$

For a given screen to object distance, S = o + i, there will be at most one lens position where the object is in focus. That will be:

$$o = \frac{\sqrt{S^2 - 4Sf}}{2}$$

AM

sorry, probabaly didnt explain very well...

object and screen kept the same then moving the lens. After I have found the two points of focus, move the screen then move the lens again.

 

[jtbell]

Start with

$$\frac{1}{f} = \frac{1}{o} + \frac{1}{i}$$

Suppose you've got the lens in one position that gives you an image on the screen. The lens is a distance $o$ from the object, and a distance $i$ from the image. The other position is the one that has the numbers for $i$ and $o$ switched around.

Exception: if $i$ and $o$ are equal, you don't change anything when you switch them around.

 

[Andrew Mason]

sorry, probabaly didnt explain very well...

object and screen kept the same then moving the lens. After I have found the two points of focus, move the screen then move the lens again.

Ok.
I shouldn't have doubted your experimental results! It seems I was using the midpoint between screen and object as the reference for measuring distances rather than the object location. I have corrected my first answer.

AM