(adsbygoogle = window.adsbygoogle || []).push({}); 1. The problem statement, all variables and given/known data

A collimated light beam is incident on the plane side of a of index 1.5, diameter 50mm, and radius 40mm. Find the .

2. Relevant equations

Refraction in a plane surface:

[tex]s'=\frac{-n_2}{n_1}s[/tex]

Refraction on a spherical surface:

[tex]\frac{n_1}{s}+\frac{n_2}{s'}=\frac{n_2-n_1}{R}[/tex]

Spherical Wave Aberrations:

[tex]a(Q)= \frac{h^4}{8} \left[ \frac{n_1}{s} \left( \frac{1}{s}+\frac{1}{R}\right)^2 + \frac{n_1}{s'} \left( \frac{1}{s'}-\frac{1}{R}\right)^2 \right][/tex]

R=Radius of curvature

s=Object distance

s'=Image distance

n=index of refraction

3. The attempt at a solution

So the light is traveling in a straight beam incident on the plane surface. By equation 1, the image distance is virtual and at infinity. so [itex]s'=-\infty mm[/itex]. I then use this value for the calculation of the image of a spherical surface using equation 2. Plugging everything in, I get a distance of 80mm. So s'=80mm.

I then used equation 3 to get the spherical wave aberration and used a height of 25mm since the lens is 50mm in diameter. (Also, R<0 since it is a concave surface from left to right.) I got an aberration of -1.29mm. The book says the correct answer is -0.858mm.

What am I doing wrong? :(?

EDIT: Figured it out! I used 1.5 for the index in equation 3 instead of 1.

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# Homework Help: Spherical Aberrations

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