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Lenses question

by physicsdawg
Tags: lenses
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physicsdawg
#1
Aug3-08, 03:10 PM
P: 25
1. The problem statement, all variables and given/known data

An object is 18cm in front of a diverging lens that has a focal length of -12cm. How far in front of the lens should the object be placed so that the size of its image is reduced by a factor of 2.0?

2. Relevant equations

1/do + 1/di = 1/f

m= -di/do



3. The attempt at a solution

i keep solving and getting 36 using those equations, but the answer says its 48 ...i donno how to get 48

1/18 + 1/di = 1/-12
di=7.2

m= -7.2/18
m=-.4

1/2m=-0.2
-.2= -7.2/do
do=36
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G01
#2
Aug3-08, 03:16 PM
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Can you show your calculations, please? I can't find your mistake if I can't see your work.
physicsdawg
#3
Aug3-08, 03:19 PM
P: 25
k i edited it

G01
#4
Aug3-08, 03:30 PM
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Lenses question

Hmmm. I am also getting 36 as my answer. I know this is a stupid question, but are you sure your numbers are correct? It doesn't hurt to check. Also where are you getting the answer from?
physicsdawg
#5
Aug3-08, 03:38 PM
P: 25
physics textbook
tiny-tim
#6
Aug4-08, 06:46 AM
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Thanks
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Quote Quote by physicsdawg View Post
An object is 18cm in front of a diverging lens that has a focal length of -12cm. How far in front of the lens should the object be placed so that the size of its image is reduced by a factor of 2.0?

i keep solving and getting 36 using those equations, but the answer says its 48 ...i donno how to get 48
Hi physicsdawg!

Try using 8cm instead of 18cm.
alphysicist
#7
Aug4-08, 11:06 PM
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P: 2,249
Hi physicsdawg,

Quote Quote by physicsdawg View Post
1. The problem statement, all variables and given/known data

An object is 18cm in front of a diverging lens that has a focal length of -12cm. How far in front of the lens should the object be placed so that the size of its image is reduced by a factor of 2.0?

2. Relevant equations

1/do + 1/di = 1/f

m= -di/do



3. The attempt at a solution

i keep solving and getting 36 using those equations, but the answer says its 48 ...i donno how to get 48

1/18 + 1/di = 1/-12
di=7.2
I believe this answer is incorrect. The numerical value is right, but it needs to be a negative number.


m= -7.2/18
m=-.4

1/2m=-0.2
With di=-7.2, all of these magnifications will be positive numbers (which checks out, since a single diverging lens creates upright images).

-.2= -7.2/do
The second image will not be in the same place as the first image, and so the di for the second case is an unknown quantity. This equation should therefore be:

[tex]
0.2 = - di/do
[/tex]
and you need to find do. Do you see how to find it?
G01
#8
Aug4-08, 11:21 PM
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Quote Quote by alphysicist View Post
Hi physicsdawg,


I believe this answer is incorrect. The numerical value is right, but it needs to be a negative number.


With di=-7.2, all of these magnifications will be positive numbers (which checks out, since a single diverging lens creates upright images).


The second image will not be in the same place as the first image, and so the di for the second case is an unknown quantity. This equation should therefore be:

[tex]
0.2 = - di/do
[/tex]
and you need to find do. Do you see how to find it?

Ahh, of course. The second image is not in the same place as the first. I made that mistake as well. ( For some reason, I thought that was a condition in the problem. Guess I read into it too much.) Nice catch alphysicist.


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