How Do You Calculate Image Distance with Magnification in a Converging Lens?

[ben488]

If I have a converging lens, of focal length f, and magnification of the image m, where q is the distance of the image from the lens show that q = (1+m) f

m = - q/p
1/f = 1/p + 1/q


I just need to show that they are equal, but I am getting confused with the maths..

q= (1 - (q/p)) (1/(1/p + 1/q))
q= (1 - (q/p))(qp/(p+q)
q= (qp/p + q) - (q/p)(qp/(p+ q)

 

[Chi Meson]

If by "they" you mean p=q, you could look at the scene from the ray diagram point of view. The only situation where p=q for a converging lens, is when both are at a point at 2x the focal length of the lens. At this point, the gemoetry is perfectly symmetrical, and image is the same size as the object, therefore, magnification is 1 (w

 

[Doc Al]

I just need to show that they are equal, but I am getting confused with the maths..

q= (1 - (q/p)) (1/(1/p + 1/q))
q= (1 - (q/p))(qp/(p+q)

Good. Rearranging, this becomes:
q= [(p-q)/p][qp/(p+q)] = [(p-q)/(p+q)]q

Well, looks like it's not true!

Perhaps they meant to write: q = (1-m)f

 

[Chi Meson]

Actually I get unity each time I do it (actually 1= -1). So then I noticed that the first equation:
q= (1+m)f
is derived directly from the thin-lens and magnification equations, if you use m=q/p (no negative).

So the equation q=(1+m)f is a general equation (true for all real situations) not a specific equation, as long as you are using the non-negative convention for the magnification equation.

There is a problem with the question, as stated.

 

[ben488]

thanks so much for all your help, its great.

Ben