Focal length of a mirror

by tiger1
Tags: focal, length, mirror
 P: 10 1. The problem statement, all variables and given/known data A 2.0-cm-tall object is placed in front of a mirror. A 1.0-cm-tall upright image is formed behind the mirror, 150 cm from the object. What is the focal length of the mirror? 2. Relevant equations 1/f=1/s+1/s' m=h'/h m=-s'/s 3. The attempt at a solution .5=-s'/s s+s' = 1.5m Combining those 2 equations, I get: s'=3 s=-1.5 1/f=(1/-1.5)+(1/3) f=3m When I calculate it the focal length, f, it comes out to 3m, which is obviously way too big.
 HW Helper P: 4,430 If image is virtual, magnification is also negative. So -m = -s'/s Now try to solve the problem.
P: 10
 Quote by rl.bhat If image is virtual, magnification is also negative. So -m = -s'/s Now try to solve the problem.
Setting m negative gave me the same answer
-.5=s'/s
s=s'/-.5
s+s'=1.5
s'/.5+s'=1.5
-(1/2)s'+s'=1.5
s'/2=1.5
s'=3

According to my book, s' should be negative, though, because we have a virtual image on the opposite side of the object.

Continuing anyways...
s=1.5-s'
=-1.5

1/f=(1/-1.5)+(1/3)=-.3
f=-3m

Still not correct.

HW Helper
P: 4,430

Focal length of a mirror

-0.5 = - s'/s
Or 0.5 = s'/s
P: 10
 Quote by rl.bhat -0.5 = - s'/s Or 0.5 = s'/s
-.5=-s'/s
s=-s'/-.5

s+s'=1.5

-s'/-.5 + s' = 1.5
(1/2)s' + s' = 1.5
(3/2)s'=1.5
s'=1

s+s'=1.5
s=1.5-s'
s=.5

1/f=(1/.5)+(1/1)=3
f=1/3=.33m

Which is still wrong.
 HW Helper P: 4,430 Image is virtual. so 1/f = 1/s - 1/s'
 P: 10 Solved: m=-s'/s=1/2 -s'=s/2 |s'|=s/2 |s|+|s'|=1.5 |s|+|s/2|=1.5 3/2s=1.5 s=1 |s|+|s'|=1.5 |s'|=1.5-|s| s'=1.5-1 s'=.5 but since s' is negative for virtual, upright images s'=-.5 1/f=(1/1)+(1/-.5)=-1 f=-1m

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