Algebraic Explanation of Image Formation in Convex Lenses

In summary, when an object is moved from the focal length towards a convex lens, the image increases in size and appears to be on the same side of the lens as the object. This is confirmed by the thin lens equation, which shows that as the object moves closer, the image also appears to get closer. The ray diagram may show a different result, but the lens equation is the more accurate and intuitive method for predicting the behavior of the image.
  • #1
Matt M
1
0

Homework Statement



An object is in the focal length of a convex lens. As the object is moved from the focus towards the lens, what happens to the object?

Homework Equations


1/f=1/d0+1/di

The Attempt at a Solution


I know how to solve this problem using ray diagrams, and the solution is:

The image increases in size and moves farther from the lens

I am trying to conclude this algebraically instead of just with a ray diagram. Just choosing numbers though, I let f=5 and do=4. Plugging this into the thin lens equation, I find that d0=-20, showing a virtual image on the same side of the lens as the object. However, then when I move the image closer and let f=5 and d0=3, the thin lens equation gives that d0=-7.5. Hence it seems to me that as the object is moving closer, it is still past the focal point, but moving closer to the lens as opposed to farther. Why is the numerical example leading to different results than the ray diagram? Thanks in advance!
 
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  • #2
Matt M said:

Homework Statement



An object is in the focal length of a convex lens. As the object is moved from the focus towards the lens, what happens to the object?

Homework Equations


1/f=1/d0+1/di

The Attempt at a Solution


I know how to solve this problem using ray diagrams, and the solution is:

The image increases in size and moves farther from the lens

I am trying to conclude this algebraically instead of just with a ray diagram. Just choosing numbers though, I let f=5 and do=4. Plugging this into the thin lens equation, I find that d0=-20, showing a virtual image on the same side of the lens as the object. However, then when I move the image closer and let f=5 and d0=3, the thin lens equation gives that d0=-7.5. Hence it seems to me that as the object is moving closer, it is still past the focal point, but moving closer to the lens as opposed to farther. Why is the numerical example leading to different results than the ray diagram? Thanks in advance!

Because you drew the wrong ray diagram?

The result given by the lens equation is intuitively right. If you have a magnifying glass, and you're looking at something small (say, an ant) through the lens, then
  1. The image is larger than the object.
  2. The image appears to be on the same side of the lens as the object (so it's a virtual image).
  3. As you move the object closer, the image appears to get closer, as well.
That's exactly what the lens equation tells you. So what ray diagram is telling you something different?
 

1. What is the algebraic formula for calculating the focal length of a concave lens?

The algebraic formula for calculating the focal length of a concave lens is f = -R/2, where f is the focal length and R is the radius of curvature of the lens.

2. How do you determine the power of a concave lens using algebra?

The power of a concave lens can be determined using the formula P = -1/f, where P is the power of the lens and f is the focal length in meters.

3. Can you use algebra to calculate the image distance of an object placed in front of a concave lens?

Yes, the image distance (i) can be calculated using the formula 1/i + 1/o = 1/f, where o is the object distance and f is the focal length. Solving for i will give you the image distance.

4. How does the algebraic formula for a concave lens differ from that of a convex lens?

The algebraic formula for a convex lens is f = R/2, while for a concave lens it is f = -R/2. This is because the focal length of a concave lens is negative, indicating that the light rays are diverging rather than converging.

5. Can algebra be used to determine the magnification of an image produced by a concave lens?

Yes, the magnification (M) can be calculated using the formula M = -i/o, where i is the image distance and o is the object distance. A negative value for M indicates that the image is inverted compared to the object.

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