4500HWM03

# 4500HWM03 - Minimum Deviation by a Prism A ray in air...

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Unformatted text preview: Minimum Deviation by a Prism A ray in air passes through a prism as shown in the ﬁgure. The angle of incidence on the prism is 01. The prism has refractive index n and an apex angle A. It can be shown that for all possible angles of incidence, the deviation 6 is min- imum when the ray inside the prism is normal to the bisector of the apex angle. Since the the minimum deviation can be easily measured, the resulting symmetric situation can be used to determine the value of the refractive index n. For the case of minimum deviation, derive, showing all work, the value of the re— fractive index n as a function of the minimum devation angle 6mm and the prism apex angle A only (eliminate all other variables). Simplify your result as much as possible taking maximum advantage of the symmetry that occurs at minimum deviation. Put your ﬁnal answer in the space provided. 3 || Minimum Deviation by a Prism Using the notation deﬁned in the problem statement, the four equations devel— oped in the class notes become (S = 61 + 92 — A (1) 817162 = nsinﬂg (2) 6"2 = A — 9'1 (3) a; = sin—1(3m61) (4) n Combining (1) and (3) 5Z01+92—0/1—6/2 (5) From symmetry, 61 = 02 and 0’1 = 6’2 and these correspond to 6 2 6mm which may be expressed as 6min = 201 — 20/1 (8) Also I / A 141—201 or 61—; (9) From (8) and (9) ‘ min A 91 : 6—:— (10) 2 and so (2) becomes sin(6mm + A) = nsin(:4—) (11) 2 2 or 31n(émJE+—A) n — g (12) ...
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