Optics: Point of Refraction
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Given a hemisphere of radius r defined by:
x = r*cos(a)
y = r*sin(a) where -pi/2 <= a <= pi/2
If the hemisphere describes a boundary between two interfaces having indices of refraction n (outside the hemisphere) and m (inside the hemisphere) and assuming the light strikes the hemisphere at point {x,y} with angle a (meaning the light's path is perfectly vertical outside the hemisphere), then it can be shown from Snell's law [n*sin(a) = m*sin(b)] and geometrical reasoning that the equation describing the path the light takes inside the hemisphere (i.e. all {p,q} such that x <= p <= (x - s*y) and 0 <= q <= y) is:
p-x = s*(q-y)
s = tan(a-b)
b = arcsin(n*sin(a)/m)
The problem:
Given any {p,q} inside the hemisphere calculate the point or points {x,y} that cause the refracted light to pass through {p,q}.
Stated another way:
Given any {p,q} inside the hemisphere calculate the angle or angles a that cause the refracted light to pass through {p,q}.
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Solution Summary
This solution solves the expressions isolating x and y. The optics for point of refraction are determined.
Solution Preview
Because,
point (x,y) lies on the hemisphere, therefore,
x^2 + y^2 = r^2 .....(1)
but,
p-x = s*(q-y)
=> x = p - s*(q-y) = (p - sq) + sy ....(2)
Hence, ...
Education
- BEng, Allahabad University, India
- MSc , Pune University, India
- PhD (IP), Pune University, India
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