Focal length of a mirror


by tiger1
Tags: focal, length, mirror
tiger1
tiger1 is offline
#1
Oct15-09, 09:54 AM
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.
Phys.Org News Partner Science news on Phys.org
Cougars' diverse diet helped them survive the Pleistocene mass extinction
Cyber risks can cause disruption on scale of 2008 crisis, study says
Mantis shrimp stronger than airplanes
rl.bhat
rl.bhat is offline
#2
Oct15-09, 10:15 AM
HW Helper
P: 4,442
If image is virtual, magnification is also negative.
So -m = -s'/s
Now try to solve the problem.
tiger1
tiger1 is offline
#3
Oct15-09, 10:29 AM
P: 10
Quote Quote by rl.bhat View Post
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.

rl.bhat
rl.bhat is offline
#4
Oct15-09, 10:31 AM
HW Helper
P: 4,442

Focal length of a mirror


-0.5 = - s'/s
Or 0.5 = s'/s
tiger1
tiger1 is offline
#5
Oct15-09, 10:43 AM
P: 10
Quote Quote by rl.bhat View Post
-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.
rl.bhat
rl.bhat is offline
#6
Oct15-09, 10:59 AM
HW Helper
P: 4,442
Image is virtual.
so 1/f = 1/s - 1/s'
tiger1
tiger1 is offline
#7
Oct15-09, 11:08 AM
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


Register to reply

Related Discussions
The focal length of a mirror. Help me maybe? Introductory Physics Homework 5
I have a question about a concave mirror where the focal point is on the outside. Introductory Physics Homework 1
spherical mirror, shortest longest focal length Introductory Physics Homework 7
Focal length of a concave mirror. Introductory Physics Homework 0
Focal length of a concave mirror. Introductory Physics Homework 0