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

  • #1
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4

Homework Statement


upload_2018-10-15_17-5-19.png
[/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:
upload_2018-10-15_17-22-31.png

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?
 

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Answers and Replies

  • #2
mjc123
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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.
 
  • #3
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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.
 
  • #4
haruspex
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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?
 
  • #5
Charles Link
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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.
 
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  • #6
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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!
 

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