Solving for Equal Image and Object Height on the Principle Axis | Optics Problem

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Homework Help Overview

The discussion revolves around an optics problem concerning the conditions under which the height of an image equals the height of an object, specifically focusing on the principle axis and magnification concepts.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • Participants explore the relationship between image height and object height through the concept of linear magnification, questioning when this magnification equals 1 or -1. There are discussions about the implications of negative magnification and the significance of absolute values in the context of the problem.

Discussion Status

The discussion includes various attempts to understand the conditions for equal heights, with some participants providing insights into the lens equation and the relationship between object and image distances. There is an acknowledgment of differing interpretations regarding the conditions for achieving equal heights, and some guidance has been offered regarding the use of specific distances in relation to the lens.

Contextual Notes

Participants mention a specific point of interest at 2F, which is derived from the lens equation, but there is uncertainty about how this conclusion is reached. The original poster expresses confusion about the mathematical steps taken and the implications of their findings.

JimmyRay
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Hi my teacher assigned a few challenging optics problems, I got most of them but I am stuck on this one..

"Find a position on the principle axis that will ensure |hi| = ho ".

When is the height of the image equal to the height of the object? How would I go about solving this?
 
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Another way to phrase the question: When is the linear magnification equal to 1 (or -1)?
 
magnification equal to negative 1? how? ...

how can you have negative magnification? I understand the positive 1, and doesn't the absolute value bars around "hi" mean we're only concidering positive answers? ...

Socceryjayl thanks for the website ... but lol I still don't know how they found 2F to be that point where hi = ho
 
Doc knows what he's talking about.The transversal linear magnification is defined by a ratio between two real numbers.As far as i know,such ratio should yield a positive number,or a negative one or zero.(In this special case,+-infinty is an accepted solution).In your case,for your relation to hold it could be as well "-1" as "+1".

Daniel.

Show us your work,to see what u're doing wrong.
 
lol I know Doc knows his stuff I am not doubting him.

ok well I used m = |hi|/ho to get m = 1.

and m = -di/do, so 1 = -di/do ... I cross multiplied to get do = -di ...

I subbed do = -di into the equation 1/do + 1/di = 1/f

When I solved for f by doing -1/di + 1/di = 1/f

I got 0 = 1/f ...

But uhh... that's not a position on the principal axis, I don't see how they (in the website soccer gave me) they got 2F for the point where |hi| = ho
 
The key is to realize that the heights will be equal when the distances are equal. Take a convex lens (f is positive) as an example. What condition will allow di = do? Using the lens equation: 1/di + 1/do = 1/f, so 1/do + 1/do = 1/f, thus do = 2f. This means that if we put the object at distance of 2f in front of the lens, the image will be a distance 2f behind the lens: and the heights will be equal. (In this case, m = - di/do = -1. The image is upside down.)
 
Ohh I see, okay thanks...
 

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