Question:
How does one determine critical angles for light entering a prism.
More specifically, the apex angle of this prism is 70 degrees, and index
of refraction n=1.58. How can I determine the minimum incident angle
for a ray if it is to pass through the prism and out the opposite side?
I was trying: Angle = ARCSIN(1/1.58), but this did not work. Would
I use the formula: n1(sin(THETA1)) = n2(sin(THETA2)) ?
Thanks.
Answer:
Draw a picture of the prism with a ray entering one side.
Draw a normal at the point of entry and continue the ray (it will refract toward
the normal upon entry)
to the other side. Draw a normal at this point of exit, and
continue the ray out of the second side, this time refracting away from the
normal. Let W denote the angle between the entry-pt normal and the entering
ray; let X denote the angle between the entry-pt normal and the ray in the
prism; let Y denote the angle between the exit-pt normal and the ray in the
prism; and let Z denote the angle between the exit-pt normal and the exit-
ing ray. Let A denote the apex angle, n1 the index of refraction outside
the prism, and n2 the index of refraction of the prism (with n1 < n2).
Consider the triangle formed from the apex angle and the ray inside the
prism; note that its "base" angles are (90-X) and (90-Y). Then A + (90-X)
+ (90-Y) = 180, so X + Y = A. Thus, X = A - Y and so sin(X) = sin(A - Y)
= sin(A)*cos(Y) - cos(A)*sin(Y). Call this equation Eq. 1.
From Snell's law, n1*sin(W) = n2*sin(X) and n2*sin(Y) = n1*sin(Z). These
give sin(X)=(n1/n2)*sin(W) and sin(Y)=(n1/n2)*sin(Z). Inserting these into
Eq. 1 we get, after a bit of algebra, sin(W)=sin(A)*sqrt(N^2-[sin(Z)]^2) -
cos(A)*sin(Z), where N=n2/n1 and where cos(Y)=+sqrt(1-[sin(Y)]^2) since
-90 < Y < 90. The critical W corresponds to Z=90. Taking Z=90, n1=1,
n2=1.58, and A=70 gives N = 1.58 and sin(W) = .8075, so W = 53.85 degrees.
rcwinther
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