# Object Space vs Image Space

1. Oct 24, 2016

### Staff: Mentor

1. The problem statement, all variables and given/known data
For an object distance z = 200 mm, an object space index n = 1.5, and a front focal length ff = −80 mm, what is the effective focal length f and the magnification m? Is the image upright or inverted?

2. Relevant equations
Lots and lots of gaussian and newtonian imaging equations!
But I'm guessing one of the equations is: $\frac{z}{f_f} = 1-\frac{1}{m}$

3. The attempt at a solution

My confusion here has to do with the optical spaces. Since z is positive, the object is to the right of the lens. But is the object space to the right of the lens as well? I've read in my book that optical spaces actually extend from negative infinity to positive infinity, but I don't quite understand it yet. Also, which index of refraction, n or n', is 1.5?

Nevermind, I'm stupid. They tell me it's $n$ right in the question.

Last edited: Oct 24, 2016
2. Oct 24, 2016

### Staff: Mentor

Using the equation above, I get:

$\frac{z}{f_f} = 1-\frac{1}{m}$

$\frac{200}{-80} = 1-\frac{1}{m}$

$\frac{5}{2} = \frac{m-1}{m}$

$5m=2m-2$

$3m=-2$

$m=-\frac{2}{3}$

3. Oct 24, 2016

### Staff: Mentor

I'm stupid, they tell me $n$ right in the question. Using the following equation:

$\frac{z}{n}=\frac{1-m}{m}f_e$

Plugging in 1.5 for $n$, 200 for $z$, and $-\frac{2}{3}$ for m gives me 53.33 mm.