Power of Two Lenses: Find d & l

[Aelo]

Homework Statement

Design a thin-lens telephoto lens with a back focal distance of 50 mm and an effective focal length of 200 mm, and a maximum system length (distance between first lens and image) of 100 mm. Find the power of each of the two lenses. (Lenses are in air.)

Homework Equations

Power, K = 1/(focal length, f)
Max system length = d + l', where d is the distance between the two lenses and l' is the distance from the second lens to the image. (This is not an equation I "know" - I came up with it myself, so it could be faulty.)

The Attempt at a Solution

I know of only two equations with the effective focal length - one comes from constructing the system vertex matrix (AKA ABCD), and the other comes from doing a paraxial ray tracing (AKA YNU). However, I don't believe I have enough information to use either method, although - making a guess - I would say that I would use the vertex matrix method. There are more equations for that, which I've attached in an image.

Using the given effective focal length and the BFD, I could find the A_v element of the matrix. But with both d and l' unknown, I don't know how I could use that to solve for either K.

 

[jbsteele]

So, I'm stuck. Can anyone help me out here? Thanks! A:I think the way to solve this is by using the lens maker equation. $$\frac{1}{f}=\frac{n-1}{R_1} + \frac{n-1}{R_2}$$where $f$ is the focal length, $n$ is the refractive index of the material, and $R_1$ and $R_2$ are the radii of curvature of the surfaces of the lenses. We have that $$f=200\mathrm{mm}, \quad n=1.$$Therefore, we can rewrite the equation as:$$\frac{1}{200}=\frac{1}{R_1} + \frac{1}{R_2}$$Now, we can rearrange the equation to get $$R_1R_2 = 200(R_1+R_2)$$From the given data, we know the sum of $R_1$ and $R_2$ which is$$R_1+R_2 = \frac{50}{2}+\frac{100}{2} = 75\mathrm{mm}$$Substituting this in the above equation, we get $$R_1R_2 = 15000\mathrm{mm}^2$$Now, we can solve for the individual radii of curvature. $$R_1 = \sqrt{15000} = 122.5\mathrm{mm}, \quad R_2=-75\mathrm{mm}$$Now, we can calculate the power of each lens using the lens maker equation$$K_1 = \frac{1}{R_1} = \frac{1}{122.5} \approx 0.00815\mathrm{D}, \quad K_2 = \frac{1}{R_2} = -\frac{1}{75} = -0.0133\mathrm{D}$$