Calculating Radius of Curvature for Upright Image Mirror

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The discussion centers on calculating the radius of curvature for a mirror that produces an upright image with a magnification of 7.9, positioned 10.0 mm from an object. The correct approach involves using the mirror formula, 1/f = 1/do + 1/di, and the magnification formula, m = -di/do. The final calculation for the radius of curvature (r) is determined to be 17.76 mm when the correct image distance (di) is used as -79 mm. The importance of correctly interpreting the problem statement and applying the negative sign in the magnification formula is emphasized.

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  • Knowledge of magnification concepts: m = -di/do
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DDRchick
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A production line inspector wants a mirror that produces an upright image within magnification of 7.9 when it is located 10.0 mm from a machine part.
What is its radius of curvature?




I used:
r/2 = f
1/f=(1/do) + or - (1/di)
m=di/do




first i did 7.9=10/do
then i found that do=1.27mm
So then i did 1/f=1/1.27+1/10
I got f=1.3
Then i multiplied that by 2
(r/2)=f
So then i got 2.6
I tried -2.6 for the heck of it
and then i did it all again except i did:
1/f=1/1.27-1/10
and calculated it all out and it was still wrong.
Even when it was negative.
Argh. Help? =( Because i have 2 more like this. x.x
 
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Hi DDRchick,

DDRchick said:
A production line inspector wants a mirror that produces an upright image within magnification of 7.9 when it is located 10.0 mm from a machine part.


The question here is using the word "it" which can often cause trouble (and I think might be leading to a mistake in your work). Is it referring to the mirror's distance from the object, or the image's distance (because I've seen problems with the meaning both ways). From the wording though, I think it means that the mirror is 10mm from the object.

What is its radius of curvature?




I used:
r/2 = f
1/f=(1/do) + or - (1/di)


I don't think there would ever be a negative sign in this formula (though sometimes do or di can be negative values).

m=di/do

This formula need a minus sign:

<br /> m=-\ \frac{d_i}{d_o}<br />

first i did 7.9=10/do


d_i is the image distance (distance from image to mirror); I don't think that's what the 10mm refers to in the problem statement. I think when you change these three things you'll get the right answer (because your basic approach looks right to me).
 
Ugh It's still wrong. :/
Crap bucket.
di/10=7.9
di=79
1/f=1/79+1/10
f=8.88
r/2=8.88
8.88(2) = 17.76
I tried it with a negative sign too just for kicks, that was marked wrong also.
:(
 
DDRchick said:
Ugh It's still wrong. :/
Crap bucket.
di/10=7.9

Remember that I mentioned above that this formula needs a minus sign. So di will be -79mm.

di=79
1/f=1/79+1/10
f=8.88
r/2=8.88
8.88(2) = 17.76
I tried it with a negative sign too just for kicks, that was marked wrong also.
:(

Do you mean you just put in -17.76, or that you used the minus sign in the magnification formula? If you use di= -79mm, I believe the rest of your procedure is fine.
 
I mean i just put in -17.76...
Garr.
Well thanks for your help, it was due a day ago unfortunately, but i survived aha.
:)
 
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