estradiol9

5+ Year Member
Feb 3, 2010
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Pre-Medical
.All upright images are virtual. All inverted images are real..
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.As the object moves closer to R (2f) from an infinite point, the image increases in size. As the object then moves toward f, the image continues to become larger. These images are all inverted and real. Once you move past the focal point toward the lens/mirror, then the image becomes very large and is virtual and upright. As you move closer to the lens/mirror, the image becomes smaller:

So for a converging lens or mirror.. such as a concave mirror OR a convex lens..

- If the object is closer than the focal point, the image created from the ray trace backs is a virtual image and the image will be upright and very magnified.

-If the object is AT the focal point, there is no real or virtual image because the rays continue to be parallel and do not converge at any point.

-If the object is between the focal point and twice the focal point, the image created from the converging rays is a real image that is inverted and magnified.

-If the object is AT the point twice the focal length, the image created from the converging rays will be a real image will be inverted and will be the same size as the object.

-If the object is beyond twice the focal point, the image created from the converging rays is a real image that is inverted and will be shrunken.

-If the object distance is infinitely far, the image created from the converging rays will be a real image at or very near the focal point and very shrunken..
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.1/P + 1/Q = 1/F.
.M= -Q/P.
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where f is the focal length and is (+) for converging and (-) for diverging;
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q is the image distance and is (+) for real images behind the lens/mirror which are formed from converging rays of outgoing light and (-) for virtual images in front of the lens/mirror which are formed from trace-backs on the opposite side of outgoing light;
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p is the object distance for the lens and is (+) when there is only 1 lens/mirror
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.m is the magnification and the magnitude is >1 if the image is magnified or <1 is the image is shrunken, if m is <0 then the image is inverted and if m >0 then the image is upright..
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.Converging refers to f>0 of a convex lens or a concave mirror..
Diverging refers to f<0 of a concave lens or a convex mirror.


I tried this out with an example of a converging lens that had a +15cm f and +10cm p. Using the formula, q is determined to be -30cm. Since it is negative, the image is virtual and it is front of the lens/mirror. And since the image is virtual, it is upright. The magnification can then be determined from the above formula to be +3, so this states it is magnified 3x and upright. And since the q is calculated to be 30cm and he f was 15cm, this also means that the image is at the R (2f) point.


.Sorry if that was long.. but I was just want to make sure I am doing this right..
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