Homework Help: Concave Mirror

1. May 3, 2012

longcatislong

1. The problem statement, all variables and given/known data

Suppose the radius of curvature of a concave mirror is 5.0 cm
a) Find the object distance that gives an upright image with a magnification of +1.51.

2. Relevant equations

1/Do + 1/Di= 1/f

m=-di/do=hi/ho

3. The attempt at a solution

First i convered all cm into m.

f=R/2, so .05/2 = .025.
1/.025=40

then I used m=-Di/Do. Since don't know Do or Di, I substituted for the unknowns using 1/Do + 1/Di= 1/f

I came up with...
m=-[40-(1/Do)]/[40-1/(40-1/Do)] and solved for Do.

I got Do=.02506m.

Online HW says it's wrong.

Any help is GREATLY appreciated. What I did makes sense to me but obviously there's an error or just some major concept I don't understand. Thank you so much for taking the time!

2. May 3, 2012

collinsmark

Okay, everything seems right so far.
I'm not sure why you would want to invert that at this point in the process, but okay.
Substitution is the right idea.
Now you've lost me. :uhh:

So far, you've already figured out
$$\frac{1}{D_o} + \frac{1}{D_i} = \frac{1}{f}$$
and
$$m = -\frac{D_i}{D_o}.$$
Rather than going the other way around, try substituting $D_i = -mD_0$ into the first equation. Also substitute $f = R/2$ into the first equation. It might make it a little easier to solve for $D_o$ that way. (Hint: then find a common denominator for the left side of the equation )

3. May 3, 2012

longcatislong

Oh god that's SO much simpler than what I was trying to do. Thanks a million!

4. May 4, 2012

Redbelly98

Staff Emeritus
Glad it worked out. By the way, I find it easier to work in cm rather than meters for this problem. For example, "1/2.5" is easier (for me) to work with than "1/0.025". In the end you get the same answer, of course.