# Focal length of a mirror

1. Oct 15, 2009

### tiger1

[SOLVED] Focal length of a mirror

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.

Last edited: Oct 15, 2009
2. Oct 15, 2009

### rl.bhat

If image is virtual, magnification is also negative.
So -m = -s'/s
Now try to solve the problem.

3. Oct 15, 2009

### tiger1

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.

4. Oct 15, 2009

### rl.bhat

-0.5 = - s'/s
Or 0.5 = s'/s

5. Oct 15, 2009

### tiger1

-.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.

6. Oct 15, 2009

### rl.bhat

Image is virtual.
so 1/f = 1/s - 1/s'

7. Oct 15, 2009

### tiger1

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