Magnification and focal length

In summary: However, in this case there are too many unknowns, so it is not possible to solve for the distance of the image.
  • #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 - can't do with only 1 known variable.

I don't think my approach is right at all. Any help will be appreciated.
 
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  • #2
there are two known variables right?

actually no. of equations>greater than no. of unknowns if I'm reading the question correctly
 
  • #3
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
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
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
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
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
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
Celer said:
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.
 
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1. What is magnification?

Magnification is the process of making an image appear larger than its actual size. It is a measure of how much larger an object appears when viewed through a lens or optical instrument compared to its actual size.

2. How is magnification calculated?

Magnification is calculated by dividing the size of the image by the size of the object. It can also be calculated by dividing the focal length of the lens by the focal length of the eyepiece.

3. What is the relationship between magnification and focal length?

The relationship between magnification and focal length is inverse. This means that as the focal length increases, the magnification decreases, and vice versa.

4. How does changing the focal length affect the image?

Changing the focal length of a lens affects the image by changing the field of view and the magnification. A longer focal length will result in a narrower field of view and a higher magnification, while a shorter focal length will result in a wider field of view and a lower magnification.

5. What is the difference between magnification and resolution?

Magnification refers to the apparent size of an object, while resolution refers to the amount of detail that can be seen in an image. Increasing magnification does not necessarily improve resolution, as it may only make the image appear larger without increasing the level of detail.

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