# Homework Help: Spherical aberration

1. Oct 29, 2006

### jnimagine

How does spherical aberration cause an image to appear when the object is placed at the focus???

In theory, when the object is placed at the focus, no image should appear but when i did an experiment with a concave mirror and a candle, image did appear. But the image was very very far away from the focus but still on the image side.

Does this have anything to do with spherical aberration? or the thickness of the candle? because in theory, ur only talking about a single line and the candle posesses a thickness... maybe that has something to do with the image appearing???

2. Oct 29, 2006

### andrevdh

Spherical abberation can be thought of as different regions of the optical component (ring shaped regions of the mirror symmetrically located about the optical axis) having slightly different focal planes for an object located at a particular distance from the mirror. This means that for some parts of the mirror the object is at the focal plane, but for some parts it is not. This means that some part of the mirror could be responsible for the appearance for the image, although my guess is that the object was not exactly at the focal point yet and if youmoved it a little bit closer it would have moved further away and eventually disappear to become a virtual image on the other side of the mirror.

3. Oct 29, 2006

### andrevdh

Spherical abberation can be thought of as different regions of the optical component (ring shaped regions of the mirror symmetrically located about the optical axis) having slightly different focal planes for an object located at a particular distance from the mirror. This means that for some parts of the mirror the object is at the focal plane, but for some parts it is not. This means that some part of the mirror could be responsible for the appearance for the image, although my guess is that the object was not exactly at the focal point yet and if you moved it a little bit closer the image would have moved further away and eventually disappeared to become a virtual image on the other side of the mirror.

4. Oct 29, 2006

### OlderDan

Yes, spherical aberration has something to do with it. I went away and wrote something about this, and did a calculation to show how spherical aberration can contribute to this effect. I will post it as I wrote it without trying to connect it to andrevdh's reply. I hope the connection will be obvious when you read them both.

I fully expect that the image you saw was blurry, but the variation in the focal length of a spherical mirror for off-axis rays could explain why an image forms at all. I dare say you would have had a very difficult time deciding where the image that you observed was located. If you were seeing a sharp image, your candle was not at the focal point.

If you look carefully at the pictures of the light beams at the link I posted

http://www.physics.montana.edu/demonstrations/video/6_optics/demos/sphericalaberrationinamirror.html

you will observe another interesting effect. The focal point for one of the individual narrow off-axis beams is itself off axis. If you blocked off the inner part of the mirror, so that only the outer beams were visible, and if the surface were rotated about the axis to make a three dimensional image, the sharpest image would be a ring around the axis. It would not even be a point on the axis.

This is a complicated effect to analyze in detail. Your candle could not possibly have been at the focal point even if there was a precise focal point. The candle is a three dimensional object. The focal point is a point (ideally; not actually). Some of the candle was no doubt in front (closer to mirror) of the focal point and some of it was behind the focal point. An image of such an object would actually be three dimensional, even if the focal point were ideal. There is no way you can get a very precise image of a three dimensional object on a screen. What you were seeing could be in part due to the part of the candle that was outside the focal point coming to focus far from the mirror.

It is a rather complicated calculation to figure out where light that leaves the nominal focal point of a spherical mirror comes to focus on the mirror axis. I have done that calculation, and I could post the equations that shows the location of the “image” in terms of the angle between the mirror axis and a radius to a point on the mirror. For small angles, the distance from the mirror is enormous, approaching infinity as the angle approaches zero. But as the angle increases, the image gets closer and closer to the center of the circle. If you went out as far as 60 degrees, the image would be only one radius from the center of the sphere. Of course your mirror does not go to 60 degrees.

If you stop there, you would conclude that the image is just smeared along the entire length of the axis, and that is sort of true. But there is more to it than that. The fact is that each ring of say one-degree width at larger distances from the axis has greater area, and that means that more light energy is reflected from off-axis rings. Not only that, the light from an off-axis ring of is more tightly focused along the length of the axis than the weaker light from a ring closer to the axis. The figure I am posting here is a good approximation to the light energy per unit length along the axis from rings of one-degree width from zero to 15 degrees. At 15 degrees the “image” is only about 15 radii, or 30 focal lengths from the mirror, The light from the inner rings is so much less intense and so smeared out that it has almost no effect on image formation. There is much more energy from the last few degrees of the mirror, and that energy is relatively well focused.

There is enough focusing from the aberration effect to create a blurry image of the candle far from the mirror. The three dimensional nature of the candle flame would also make a contribution. I would be interested to know how many focal lengths it was from the mirror to the image you were seeing, and how well you think you could measure that distance. My guess is it was pretty far away, and not very well defined.

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5. Oct 29, 2006

### jnimagine

My focus was at 12.7cm and the image distance i got when i placed the candle at the focus is 170cm..!
So, what you're saying is that the reasons that account for image appearing when the object is placed at the focus are that the candle is three-dimensional and that the spherical aberration causes a varying focal points and could have possibly caused an image to appear.. right?

6. Oct 29, 2006

### OlderDan

Yes, both of those things could contribute. Based on the numbers you quoted, it seems more likely that your candle was not accurately placed at the focus. If the focal length is about 12.5 cm, the radius is about 25cm, so the image is only about 6 radii from the center of the circle. You would need part of the mirror to be off axis by about 23 degrees for light from the focal point to reach the axis that close to the mirror. I doubt your mirror was that big. I should have asked you before. Do you know the diameter of the mirror?

7. Oct 29, 2006

### jnimagine

Well, i'm guessing this but it was a very small mirror... the diameter might only be like 6cm..? or even 5cm?

Last edited: Oct 29, 2006
8. Oct 29, 2006

### OlderDan

In that case, the edges of the mirror are only about 6 or 7 degrees off axis. Spherical aberration should not be producing an image. You must have had your candle a bit too far from the mirror.

Do you know how to calculate the object distance from a known image distance and focal length? You would only need to be about 1cm from the focal point to produce that image at 170 cm.

9. Oct 30, 2006

### jnimagine

but the three dimesionality still accounts for this right?
and one more question about concave mirrors...
when the candle was placed between the vertex and the focus of the mirror (f=12.7cm) a virtual image appeared behind the mirror.
We measured the image distance approximately by pointing a finger towards where the image might be and got 6.0cm (-6.0 when used in mirror equation) and the object was at 6.35cm. However, when I use the mirror equation to calculate the focus, i get a negative answer that is not even close to the actual focus 12.7cm. Do u think this is just due to human errors of not measuring the image distance accurately or is there some other factors that can account for this? OR is the image distance supposed to be almost the same as the object distance in this case?

10. Oct 30, 2006

### OlderDan

Maybe a combination of slight misplacement, a 3-D effect, and a bit of help from aberration at the outer edges of your mirror would be enough to do it. I doubt your candle flame had a radius of 1cm. Then again, I don't know how precisely that image was located.

The image distance should have been one focal length when the object was placed at one half the focal length from the mirror. I can't think of any reason other than human error why you would be this far off. Locating a virtual image can be tricky. You have to be sure to move your head from side to side until there is no relative motion between the virtual image you see and the real object co-located with the image.

11. Oct 31, 2006

### jnimagine

The image distance should have been one focal length when the object was placed at one half the focal length from the mirror. I can't think of any reason other than human error why you would be this far off. Locating a virtual image can be tricky. You have to be sure to move your head from side to side until there is no relative motion between the virtual image you see and the real object co-located with the image.[/QUOTE]

But then in one of the discussion questions, it asks you why is the value of image position negative 0.5f? it's not supposed to be 0.5f.....

12. Oct 31, 2006

### jnimagine

can someone plzzzz help me.....??? T.T

13. Oct 31, 2006

### OlderDan

Have you used the mirror equation to relate focal length, object distance, and image distance?

http://www.glenbrook.k12.il.us/gbssci/phys/class/refln/u13l3f.html [Broken]

1/f = 1/o + 1/i

As a check to verify this is correct, note that if the object distance is f, then the image distance must be infinite. If the object distance is 2f, the image distance is 2f. If the object distance is less than f, the image distance must be negative, which indicates a virtual image.

If the object distance is f/2, the image distance must be -f, which is what I said earlier; a virtual image forms one focal length behind the mirror. To get an image distance of -f/2 you would need an object distance of f/3

Your discussion question is either an error, or it is related to an object distance of f/3, not f/2. The point of the question is probably to make sure you understand the signifacance of the negative distance (virtual image)

Last edited by a moderator: May 2, 2017