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- Feb 10, 2011
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Lens tend to be a stickly point for me but I have learned some facts.
I don't find the draw the ray method very helpful.
I tend to use the 1/f = 1/i + 1/o equation with the following:
1) f is negative for diverging lense, +f for converging lens
2) In a lens the object should be on the left of the lens. If it's on the right, -o. If it's on the left, it's +o. Same with mirror.
3) I have memorized that mirror have virtual images on the right of the mirror, so it's -i for mirror with image on the right. it's +i for a mirror with image on the left.
In lens, its +i for image on the right(real side) and -i for images on the left(virtual side).
I just plug that those signs in and even if a problem doesn't give a number for focal length, the signs and the nature of fractions (ie. 1/o goes to zero as object distance increases etc) help me figure out the locations of where the image is formed.
Any tips, comments on this method ? Improvement, suggestions question are welcome.
I don't find the draw the ray method very helpful.
I tend to use the 1/f = 1/i + 1/o equation with the following:
1) f is negative for diverging lense, +f for converging lens
2) In a lens the object should be on the left of the lens. If it's on the right, -o. If it's on the left, it's +o. Same with mirror.
3) I have memorized that mirror have virtual images on the right of the mirror, so it's -i for mirror with image on the right. it's +i for a mirror with image on the left.
In lens, its +i for image on the right(real side) and -i for images on the left(virtual side).
I just plug that those signs in and even if a problem doesn't give a number for focal length, the signs and the nature of fractions (ie. 1/o goes to zero as object distance increases etc) help me figure out the locations of where the image is formed.
Any tips, comments on this method ? Improvement, suggestions question are welcome.