Optics - distance between object and image

In summary: Yes, exactly. It's just algebra at this point.sorry for the late reply but, thanks a lot for your help, I really appreciated it!
  • #1
Smile101
29
0
hello,

Homework Statement


A lens has a focal length of 20 cm and a magnification of 4. How far apart are the object and the image?

Homework Equations


m=di(image)/do(object)
1/di + 1/do = 1/f

The Attempt at a Solution


Well, I don't really know what to do, I feel like I don't have enough information. Your help will be very appreciated!
 
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  • #2


Since you have 2 equations in 2 unknowns ... solve 1 equation for 1 of the unknowns, then substitute that into the 2nd equation.
 
  • #3


Redbelly98 said:
Since you have 2 equations in 2 unknowns ... solve 1 equation for 1 of the unknowns, then substitute that into the 2nd equation.

Yes, I was thinking about that but how do I do it? Seeing as if i did m=di/do 4=di/do --but it leads me to a dead end

or if i used 1/di+1/do=1/f i can only go as far as 1/di+1/do=1/20 then once again i get to a dead end!
 
  • #4


Smile101 said:
4=di/do

Okay, so far so good. Now solve that equation for di:

di = _____ ?​
 
  • #5


Redbelly98 said:
Okay, so far so good. Now solve that equation for di:

di = _____ ?​

umm.. di=4(do) but since we don't have do we can't solve for di
 
  • #6


Smile101 said:
umm.. di=4(do) but since we don't have do we can't solve for di

You can now substitute 4do for di in the other equation ... that eliminates di, leaving do as the only unknown.
 
  • #7


Redbelly98 said:
You can now substitute 4do for di in the other equation ... that eliminates di, leaving do as the only unknown.

so the equation is : 1/di +1/do =1/f

1/4do + 1/do = 1/20

Now I'm guessing we want to make do single... but i have to get rid of the 4 so since I'm dividing...

1/do^2 = 1/20 - 1/4
 
  • #8


Smile101 said:
so the equation is : 1/di +1/do =1/f

1/4do + 1/do = 1/20
Multiply do to both sides and solve for do.
 
  • #9


rl.bhat said:
Multiply do to both sides and solve for do.

Yes, exactly. It's just algebra at this point.
 
  • #10


sorry for the late reply but, thanks a lot for your help, I really appreciated it! :)
 

1. What is the distance between an object and its image?

The distance between an object and its image is called the object-image distance. It is the distance from the object to the point where the image is formed by the optical system. It is typically denoted as d.

2. How is the distance between an object and its image determined?

The distance between an object and its image is determined by the type of optical system being used. For example, in a simple convex lens system, the distance can be calculated using the thin lens equation, 1/f = 1/do + 1/di, where f is the focal length, do is the distance from the object to the lens, and di is the distance from the lens to the image.

3. How does the distance between an object and its image affect the size of the image?

The distance between an object and its image has a direct impact on the size of the image. As the distance increases, the image size decreases, and vice versa. This relationship is described by the magnification equation, m = -di/do, where m is the magnification.

4. Can the distance between an object and its image be negative?

Yes, the distance between an object and its image can be negative. This occurs when the image is formed on the same side of the lens as the object, and is known as a virtual image. In this case, the distance is denoted as a negative value.

5. How can the distance between an object and its image be changed?

The distance between an object and its image can be changed by altering the focal length of the optical system. This can be done by changing the position of the lens or by using a different lens with a different focal length. Additionally, the distance can also be changed by adjusting the position of the object relative to the lens.

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