Lenses question

  • #1

Homework Statement



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?

Homework Equations



1/do + 1/di = 1/f

m= -di/do



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
 
Last edited:

Answers and Replies

  • #2
G01
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Can you show your calculations, please? I can't find your mistake if I can't see your work.
 
  • #3
k i edited it
 
  • #4
G01
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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?
 
  • #6
tiny-tim
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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! :smile:

Try using 8cm instead of 18cm. :wink:
 
  • #7
alphysicist
Homework Helper
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Hi physicsdawg,

Homework Statement



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?

Homework Equations



1/do + 1/di = 1/f

m= -di/do



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?
 
  • #8
G01
Homework Helper
Gold Member
2,682
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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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