# Feynman Exercise 27-1: Shape of glass surface to converge a light beam inside the glass

[/B]

## Homework Equations

1/s+n/s′=1/f
where
s is distance from source to diffracting surface,
s' is distance from diffracting surface to focus,
f is the focal length,
n is the refractive index.

## The Attempt at a Solution

Since we have parallel beams, we have s = infinity so the equation reduces to
n/s' = 1/F' <=> s' = n*F'
If our coordinate system is placed so that we have origo at F', we get:
s' = sqrt(x^2 + y^2)
inserting and solving for y we get:
sqrt(x^2 + y^2) = n*F'
x^2 + y^2 = (n*F')^2
y = +/- sqrt( (n*F')^2 - x^2 )
But the solution in the book says:

What am i doing wrong?

Also, i notice that my solution is a spherical surface with a radius n*F', but the books solution is ellipsical. From optics we know that it should be a spherical surface, so is the book simply wrong?

#### Attachments

17.2 KB · Views: 413
6.8 KB · Views: 163
Last edited by a moderator:

## Answers and Replies

mjc123
Homework Helper
I don't know the answer, but from the diagram it seems that when y = 0, x = F', and that is not a solution to your equation or the book's.

I don't know the answer, but from the diagram it seems that when y = 0, x = F', and that is not a solution to your equation or the book's.
I don't think that's right. In the problem the origin is not specified. In the books' solution we have a point at x=0, y=0, so the origin must be on the surface of the lens itself.

haruspex
Homework Helper
Gold Member
1/s+n/s′=1/f
Isn't that an approximation?
How about you start from first principles and say that the optical path length is the same for all rays?

aa_o and Charles Link
Homework Helper
Gold Member
The known result in Optics is that the surface is an ellipse and not spherical for the optical path length to be precisely the same for all rays. And I agree with the book's answer. ##\\ ## There is a simple trick to computing the result for this if you begin with the result that the solution must be an ellipse: The ray that comes in at the very top of the ellipse must also pass through the far focal point of the ellipse, which is located at (c,0) in an x-y coordinate system with (0,0) at the center of the ellipse. By Snell's law ## n \sin{\theta}=1 \sin{90^{\circ}}=1 ##. Using ## \tan{\theta}=\frac{c}{b} ##, where the ellipse is ## \frac{(x-a)^2}{a^2}+\frac{y^2}{b^2}=1 ##, with ## a^2-b^2=c^2 ##, I was able to verify the book's answer within about 5 minutes. ## \\ ## Setting the optical path length, (where the length in the material gets multiplied by ## n ##), equal for all paths to the point ## (F',0) ## would take a little bit of work, but that is what the problem is asking for.

Last edited:
I figured that deriving from first principles would give me the best understanding of the concept.
I did this and came up with the same answer as the book. Thanks for the help guys!