Lens: object distance vs. image distance?

[Seadragon]

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!

 

[berkeman]

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!

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

Can you show us your graph?

 

[Seadragon]

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?

 

[berkeman]

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 into 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

.

 

[rwisz]

The thin lens equation, 1/s+1/s'=1/f, is indeed applicable to both convex and concave lenses. The difference lies in the sign convention used for the distances. For a convex lens, the focal length is positive and the distances are measured from the center of the lens. For a concave lens, the focal length is negative and the distances are measured from the vertex of the lens.

Now, looking at the graph of s' versus s, we can see that as s increases, s' decreases. This is because the image distance is inversely proportional to the object distance. As the object moves farther away, the image distance decreases. This is true for both convex and concave lenses, as long as the object is placed beyond the focal point.

Physically, this can be explained by the way a lens bends light rays. As the object moves away from the lens, the light rays become more parallel, causing the image to form at a closer distance from the lens. This can be seen in the thin lens equation, where as s increases, 1/s decreases, causing 1/s' to also decrease.

In conclusion, the graph of s' versus s varies in a linear fashion due to the inverse relationship between the object and image distances, as described by the thin lens equation.