The thin lens has some basic properties that have some redundancy but they are the ones I remember and deploy when needed:

- a ray through the lens center is not bent
- all rays through a point that then pass through the lens converge at some other point
- a ray through the focal point (a point on the optical axis at distance from lens
, the focal length of the lens) will be parallel to the optical axis after being bent by the lens*f* - all rays from a point a distance
from the lens will converge at a distance**A**on the other side of the lens and obey the thin lens law:*B**1/***A**+ 1/**B**= 1/**f**

So if I have a (purple) ray with origin

*that hits the lens at point*

**a***, how does it bend? It bends towards point*

**m***no matter what the ray direction is. So the new ray is:*

**b**

**p**(t)**= m +**t**(b-m)**So what is point

**b?**It is in the direction of point

*but extended by some distance.*

**c,***We know the center of the lens*

*so*

**c,***we can*

**use the ratios of the segments to extend to that:**

**b = a + (c-a)**(B+A) / A?*So what is*

*(B+A) /A?*

*We know 1/A + 1/B = 1/f, so B = 1/(1/f-1/A). So the point*

*is:*

**b**

**b = a + (c-a) (**1/(1/f-1/A) + A)/A =

**b = a + (c-a)**(1/(A/f - 1) + 1) =

**b = a + (c-a)***(A/(A - f))**OK lets try a couple of trivial cases. What if A = 2f?*

**b = a + (c-a)**(2f/(2f-f)) =

**b = a +**2**(c-a)**That looks right (symmetric case-- A = B there)

*So final answer, given a ray with origin*

*that hits a lens with center*

**a***and focal length*

**c***f*at point

*, the refracted ray is:*

**m**p(t) = m + t(

**a + (c-a)***(A/(A - f)) -***m**)*There is a catch. What if*

*This happens when*

*B < 0?**Address that case when it comes up :)*

*A < f.*

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