Optics

laczlacylaci

2+ Year Member The fastest way to solve this problem is to know that a convex mirror and concave lens will have virtual, upright and reduced.

I just wanted to confirm this with calculations.
1/f=1/o+1/i
(1/-30)=(1/12)+(1/i) (focal length is -, because of convex mirror)
i=-8 ish
m=-i/o=8/12=0.66 <---- m=+ indicates upright=virtual, Since m is <1 it is reduced.

Is this the correct way of calculating this or do I put a negative in front of the do?

theonlytycrane

5+ Year Member
I never bothered memorizing the facts about which are virtual / upright / etc. I just do the calculation and confirm based on what I get like you did.

The object distance "o" will always be positive. The mirror is convex (diverging) so the f is negative. The calculated "i" and magnification from -i / o will give you the correct answer. You did it right- no negative sign in front of the "o".

• laczlacylaci

laczlacylaci

2+ Year Member
I never bothered memorizing the facts about which are virtual / upright / etc. I just do the calculation and confirm based on what I get like you did.

The object distance "o" will always be positive. The mirror is convex (diverging) so the f is negative. The calculated "i" and magnification from -i / o will give you the correct answer. You did it right- no negative sign in front of the "o".
For some reason, I'm not getting notifications telling me that people have commented on my post! Thank you for the help on all these questions !