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Magnification and focal length

  • Thread starter Celer
  • Start date
  • #1
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Homework Statement


A lens has a focal length of 120 cm and a magnification of 4.0. How far apart are the object and the image?

Homework Equations


I used the thin lens equation 1/f = (1/di + 1/do)^-1 and M = -d image/d object


The Attempt at a Solution


I have been trying this by rearranging the thin lens equation...didnt work. Tried finding di and do given only magnification - cant do with only 1 known variable.

I dont think my approach is right at all. Any help will be appreciated.
 

Answers and Replies

  • #2
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there are two known variables right?

actually no. of equations>greater than no. of unknowns if i'm reading the question correctly
 
  • #3
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yes 2 known variables...but i have no idea which equation to exactly use. I have been trying going through all the equations in my optics unit...

So how do I do this?
 
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  • #4
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assume it to be convex....so f is -120 cm i guess and M= -di/d0 is given

you just have to use the above equations..get di in terms of d0 from M equation and put it into f equation
you should get d0 to be negative..take its absolute value and add to di to get your answer..
 
  • #5
238
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Let f= focal length and let [tex]d_{0}[/tex] distance from lens to the object
Let [tex]d_{1}[/tex] distance from object to image and let m=magnification.

Then, [tex]M= \frac{f}{f-d_{0}} = - \frac{d_{1}/dist{0}[/tex]

Then m = f/(f-d_0) = -d_1/d_0
Now, you need to solve for d_1

Wait up. Sorry, something's wrong with my latex code.
 
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  • #6
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Sorry, I have never seen these equations...my teacher rushed through the Optics units pretty fast.

Do you mind if you explain what formula's you are using...I have only learned the Thin Lens and the Magnification one.
 
  • #7
238
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Sure. M is the magnification. Since you've already know that

[tex]M=- \frac{d_{i}}{d_{o}}[/tex]
where [tex]d_{i}[/tex]= distance to image, [tex]d_{o}[/tex]= distance to object

M can also be written as [tex]M= \frac{f}{f-d_{o}}[/tex](you should memorize this equation), where f is the focal length.
If you've set those two equal to each other:
[tex]M= \frac{f}{f-d_{o}} = - \frac{d_{i}}{d_{o}}[/tex]
then you will find it easier to solve for the distance of the image, [tex]d_{i}[/tex] than with the thin lens equation.
 
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  • #8
13
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Thanks for your help. Just one more minor question...I've read in my textbook that (hi/ho)= (di/do).

If another question pops up with only height of image/object given could I substitute do with ho in the equation you showed?
[tex]
M= \frac{f}{f-d_{o}}
[/tex]

Also, would [tex]
M= \frac{f}{f-d_{o}}
[/tex] work with di replacing do?
 
  • #9
238
0
Thanks for your help. Just one more minor question...I've read in my textbook that (hi/ho)= (di/do).

If another question pops up with only height of image/object given could I substitute do with ho in the equation you showed?
[tex]
M= \frac{f}{f-d_{o}}
[/tex]

Also, would [tex]
M= \frac{f}{f-d_{o}}
[/tex] work with di replacing do?
Yeah, that's another way to write as magnification. Btw, it's -d_i/d_o ;a).

Well, if you're only give the height of the image or height of object, then there's too many unknowns so it's not possible to solve for the distance of the image (since you do not know the distance of the object nor distance of the image (and the other height)

Usually, physics problems have just the right number of known variables that allows you to manipulate equations to solve for unknown variables.
 
Last edited:

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