Finding Object Distance for Upright Image with Magnification +1.51 for Concave Mirror

[longcatislong]

Homework Statement

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.

Homework Equations

1/Do + 1/Di= 1/f

m=-di/do=hi/ho

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!

 

[collinsmark]

Homework Statement

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.

Homework Equations

1/Do + 1/Di= 1/f

m=-di/do=hi/ho

The Attempt at a Solution

First i convered all cm into m.

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

Okay, everything seems right so far.

1/.025=40

I'm not sure why you would want to invert that at this point in the process, but okay.

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

Substitution is the right idea.

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

Now you've lost me.

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 )

 

[longcatislong]

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

 

[Redbelly98]

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.