- #1
DDRchick
- 27
- 0
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
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