Optics problem involving a converging lens

[thezac11]

Homework Statement

An Object and a screen are located a fixed distance D apart.

(A) Show that a converging lens of focal length F placed between the object and screen will form a real image on the screen for two positions that are separated by a distance d=(D(D-4F))^1/2

(B) Show that the ratio of image sizes for these two lens positions is ((D-d)/(D+d))^2

Homework Equations

The Attempt at a Solution

I really don't see how you can prove this equation without having values for F or D.

I just don't think I'm thinking on the right track. If anyone could give me some useful pointers or examples that would be great. Thanks a lot

 

[anathema]

!

Dear fellow scientist,

Thank you for your post. I understand your confusion about the equations provided without any specific values for F or D. However, these equations are general equations that apply to any situation where an object and a screen are located a fixed distance apart and a converging lens is placed in between.

To help you better understand the equations, let's break them down step by step.

(A) To show that a converging lens of focal length F placed between the object and screen will form a real image on the screen for two positions separated by a distance d=(D(D-4F))^1/2, we can use the thin lens equation: 1/f = 1/do + 1/di, where f is the focal length of the lens, do is the distance from the object to the lens, and di is the distance from the lens to the image.

In this case, we have a fixed distance D between the object and screen, so do = D. We also know that the image formed by the lens will be on the screen, so di = D. Plugging these values into the equation, we get: 1/f = 1/D + 1/D, which simplifies to 1/f = 2/D. Solving for f, we get f = D/2.

Now, using the lens formula: 1/f = 1/do + 1/di, we can find the image distance di for the two positions separated by a distance d. For the first position, do = D and di = D - d. For the second position, do = D and di = D + d. Plugging these values into the lens formula and solving for f, we get: f = D/(2-D/d) and f = D/(2+D/d) respectively.

Since we know from our previous calculation that f = D/2, we can equate the two equations and solve for d. This gives us d = (D(D-4F))^1/2, which is the desired result.

(B) To show that the ratio of image sizes for these two lens positions is ((D-d)/(D+d))^2, we can use the magnification formula: m = -di/do, where m is the magnification, di is the image distance, and do is the object distance.

For the first position, do = D and di = D - d. For the second position,