Approximating the Edge of a Lens as a Prism

[Mr_Allod]

Homework Statement

A convex planer lens has Diameter D = 3cm, focal length f = 10cm and refractive index n = 1.5
a. Find the minimum thickness of the lens which maintains the diameter D = 3cm
b. Approximate the top/bottom edge of the lens as a prism and find the angle $\theta$ shown.

Relevant Equations

Lensmaker's Formula: $\frac 1 f = \frac {n_2-n_1} {n_1} \left( \frac 1 R_1 - \frac 1 R_2 \right)$

Hello there, for part a. of this problem I thought I should try to find the radius of curvature R of the lens using the Lensmaker's Formula. Then it would be quite easy to find the minimum thickness T by just finding the thickness of the circle segment using Pythagoras' Theorem. But part of deriving the Lensmaker's Formula is making the assumption that the thickness of the lens is negligible, so $T \to 0$. So I'm not sure if I can actually use it here?

For part b., assuming that I know the thickness T, my first thought was to find the tangent to the circle at the point where the curved and planer surfaces meet. Then I would have the relationship $$90^{\circ} = \theta +\phi$$
And I would be able to find $\theta$ with trigonometry. But I don't know if this is the correct way to approximate the angle between the curved and planer surfaces, its just a guess that made sense to me at the time. So if there is a more accurate way to approximate it I would appreciate it if you could let me know. Thank you in advance!

 

[hutchphd]

Sounds like a plan. Your expression for the angle is exact for the edge as drawn. Is there a part (c) to the question? (I know what I would want for extra credit!)

 

[Mr_Allod]

Sounds like a plan. Your expression for the angle is exact for the edge as drawn. Is there a part (c) to the question? (I know what I would want for extra credit!)

Ah thank you, I'm glad to hear my hunch made sense.

Yes there is in fact a part (c) and also a part (d) actually, for part (c) we are asked to find the Angular Deviation of this approximated prism.

Then in part (d) the objective is to recalculate the focal length of the lens by finding the point of intersection of rays passing through the top and bottom edges. I'm guessing the new focal length will be slightly different from the original 10cm because of the prism approximation.

 

[hutchphd]

Yes that's a nice question and instructive. And you get to do the trigonometry, and are introduced to spherical aberration.