Object 18cm in Front of Lens to Reduce Image by 2.0

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I'll leave the fun for physicsdawg. :rofl:In summary, the object should be placed 36cm in front of the lens to achieve a 2.0 reduction in image size. However, the second image will not be in the same place as the first, and the equation 0.2 = -di/do should be used to find do, instead of di=7.2.
  • #1
physicsdawg
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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
 
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  • #2
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
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?
 
  • #5
physics textbook
 
  • #6
physicsdawg said:
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
Hi physicsdawg,

physicsdawg said:

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
alphysicist said:
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.
 

What is "Object 18cm in Front of Lens to Reduce Image by 2.0"?

"Object 18cm in Front of Lens to Reduce Image by 2.0" refers to a specific distance between an object and a lens that will result in an image being reduced by a factor of 2.0.

Why is this concept important?

This concept is important in understanding the relationship between an object and its resulting image when viewed through a lens. It can also be used in practical applications, such as in photography or microscopy, to manipulate the size of an image.

How is the distance of 18cm determined?

The distance of 18cm is determined by the focal length of the lens and the desired magnification of the image. This distance can be calculated using the formula: Object distance = (magnification + 1) x focal length.

What factors can affect this distance?

The distance of 18cm can be affected by various factors such as the type and quality of the lens, the distance of the object from the lens, and any additional optical elements in the system. These factors can alter the magnification and may require adjusting the object distance to achieve the desired reduction in image size.

Are there any limitations to this concept?

Yes, there are limitations to this concept as it assumes perfect optics and does not account for factors such as aberrations or diffraction. Additionally, the distance of 18cm may not always result in a precise reduction of 2.0 and may vary slightly depending on the specific setup and conditions.

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