lens equation problem

[JimmyRay]

Hi my teacher assigned a few challenging optics problems, I got most of them but im 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?

[Doc Al]

Another way to phrase the question: When is the linear magnification equal to 1 (or -1)?

[soccerjayl]

find the answer here:

http://www.physicsclassroom.com/Class/refrn/U14L5db.html

[JimmyRay]

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

[dextercioby]

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.

[JimmyRay]

lol I know Doc knows his stuff im 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... thats not a position on the principal axis, I dont see how they (in the website soccer gave me) they got 2F for the point where |hi| = ho

[Doc Al]

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.)

[JimmyRay]

Ohh I see, okay thanks...