Calculating Focal Length Using Sign Conventions for Lenses

In summary, the speaker is having trouble with a physics question involving the distance between an object and its upright image. They discuss the potential use of a Double-Concave lens and confusion regarding the sign conventions for object and image distances. The speaker then uses equations to solve for the focal length and expresses gratitude for the assistance.
  • #1
oooride
34
0
Okay for the most part I understand the concepts involved but I'm having some trouble with this question.


The distance between an object and its upright image is 20.0 cm. If the magnification is 0.500, what is the focal length of the lens that is being used to form the image?


Now first I understand since this is an upright image formed (virtual) by the lens, that it is either a Double-Convex (converging) lens with the object between the first focal point (F_1) and the front side of the lens, or it is a Double-Concave (diverging) lens with the object placed anywhere on the front side of the lens.

My guess is that it is a Double Concave lens since the question states the distance between the object and the upright image is +20.0 cm. Since it's positive does this mean that the image formed is to the right (closer to the front of the lens than the object is)?

If so, if it was a -20.0 cm would this mean that it is a Double-Convex (converging) lens with the object between F_1 and the front of the lens, and producing a virtual image to the left of the object?

I'm assuming where I'm confused what type of lens it is in order to use the correct "sign conventions". And the object's distance from the focal point is not given..

What I did was:

M = -q/p
0.500 = -(20.0 cm)/p
p = -(20.0 cm)/0.500
p = -40.0 cm

Also how do I know if 20.0 cm is "p" or "q"? I know that p is the object distance and q is the image distance, but if I have the distance between p and q, how do I know which is which? :yuck:

Then I used 1/p + 1/q = 1/f to find the focal point.

Any help is greatly appreciated.

Thanks in advance.
 
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  • #2
let the equations do the work

Here's how to think this through. First off, assume the standard sign conventions, then let the formulas do the work for you. The standard is that a positive p means the object is in front of the lens and a positive q means that the image is on the other side of the lens.

So the first fact, the distance between object and image = 20 cm, can be written as: p + q = 20. (Note: I just assume q is positive--if it's negative, we'll find out.)

The next fact is magnification = .5 (positive because it's a virtual, upright image). So: m = -q/p = 0.5.

Now combine these two equations and solve for p and q. Then find f.
 
  • #3
I got the right answer and understand it now.

:biggrin: Thanks!
 

What are sign conventions of lenses?

The sign conventions of lenses refer to the rules used to determine the direction and sign of image formation in a lens system.

What is the difference between a positive and negative sign convention in lenses?

In a positive sign convention, distances are measured from the center of the lens towards the object and image. In a negative sign convention, distances are measured from the lens towards the observer.

How do you determine the sign of the focal length in a lens system?

The sign of the focal length is determined by the type of lens being used. For converging lenses, the focal length is positive, while for diverging lenses, the focal length is negative.

What is the significance of the sign of the object and image distances in a lens system?

The sign of the object and image distances determines the type of image formed by the lens. A positive image distance indicates a real image, while a negative image distance indicates a virtual image.

How do you use the sign conventions to determine the magnification of an image in a lens system?

The magnification is determined by the ratio of the image distance to the object distance. A positive magnification indicates an upright image, while a negative magnification indicates an inverted image.

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