Lens: object distance vs. image distance?

  • Thread starter Seadragon
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  • #1
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Homework Statement



An object is at distance s from a thin glass lens of focal length f. The lens is surrounded
by air. The image is at distance s' from the lens. Draw a graph of s' versus s. Explain
physically why the graph varies as it does.


Homework Equations



I think I am supposed to use the thin lens equation of 1/s+1/s'=1/f.


The Attempt at a Solution



For a fixed focal length, an increase in s causes the linearly proportional decrease in s'? As an object is moved farther away from the lens, the image also moves farther away.

I don't understand how this equation is independent from the case of a convex or concave lens.

Thanks for any help!

Homework Statement





Homework Equations





The Attempt at a Solution

 

Answers and Replies

  • #2
berkeman
Mentor
59,039
9,141

Homework Statement



An object is at distance s from a thin glass lens of focal length f. The lens is surrounded
by air. The image is at distance s' from the lens. Draw a graph of s' versus s. Explain
physically why the graph varies as it does.


Homework Equations



I think I am supposed to use the thin lens equation of 1/s+1/s'=1/f.


The Attempt at a Solution



For a fixed focal length, an increase in s causes the linearly proportional decrease in s'? As an object is moved farther away from the lens, the image also moves farther away.

I don't understand how this equation is independent from the case of a convex or concave lens.

Thanks for any help!

Homework Statement





Homework Equations





The Attempt at a Solution

What is the difference in f for a convex versus concave lens?

Can you show us your graph?
 
  • #3
8
0
A diverging lens has a negative focal length and a virtual image. So f and s' are negative.

A converging lens has a positive focal length and a real image. So f and s' are positive.

1/s + 1/s' = 1/f.

Let's say s is the independent variable on my graph:

For a converging lens:

1/s + 1/s' = 1/f

1/s = 1/f - 1/s'

s = (1/f - 1/s')^-1

For a diverging lens:

1/s - 1/s' = -1/f

1/s = 1/s' - 1/f

s = (1/s'-1/f)^-1.


I get a different answer for either converging or diverging =(. I think the question suggests that it shouldn't make a difference whether converging or diverging.

I have no idea how I would graph:

y = (1/x - c)^-1


Well, I guess if I say that 1/f=0 then the graph is y=x. But 1/f doesn't equal zero because if it did, there would be no lens.

Am I on the right track?
 
  • #4
berkeman
Mentor
59,039
9,141
A diverging lens has a negative focal length and a virtual image. So f and s' are negative.

A converging lens has a positive focal length and a real image. So f and s' are positive.

1/s + 1/s' = 1/f.

Let's say s is the independent variable on my graph:

For a converging lens:

1/s + 1/s' = 1/f

1/s = 1/f - 1/s'

s = (1/f - 1/s')^-1

For a diverging lens:

1/s - 1/s' = -1/f

1/s = 1/s' - 1/f

s = (1/s'-1/f)^-1.


I get a different answer for either converging or diverging =(. I think the question suggests that it shouldn't make a difference whether converging or diverging.

I have no idea how I would graph:

y = (1/x - c)^-1


Well, I guess if I say that 1/f=0 then the graph is y=x. But 1/f doesn't equal zero because if it did, there would be no lens.

Am I on the right track?
The easiest way to start with the graph is to just graph a few points for each lens system. Start with the converging lens system, and start with the Object at infinity to the left. Where is the image? Now move the Object in to 2*f distance to the left. Where is the Image now? And what happens when you move the Object inside 1*f to the left of the lens?

And do the same thing for the diverging lens. Do the graphs match the equations you've written?

http://en.wikipedia.org/wiki/Lens_equation#Lensmaker.27s_equation

.
 

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