JAMB - Physics (2017 - No. 40)
Calculate the angle of minimum deviation of a 60º prism of a refractive index 1.5[ \(sin^{-1}\) (0.75) = 49º]
38o
19.47o
16.25o
49o
Explanation
For a refractive index () = \(\frac{\sin\frac{1}{2} (A + D)}{\sin\frac{1}{2}A}\)
D = angle of minimum deviation
A = refractive angle of the prism
1.5 = \(\frac{\sin\frac{1}{2} (60 + D)}{\sin\frac{1}{2} \times 60}\)
1.5 = \(\frac{\sin\frac{1}{2} (60 + D)}{\sin 30}\)
Sin \(\frac{1}{2}\) (60 + D) = 1.5 * Sin 30
Sin \(\frac{1}{2}\) (60 + D) = 0.75
\(\frac{1}{2}\) (60 + D) = Sin-1 (0.75)
but Sin-1 (0.75) = 49o
\(\frac{1}{2}\) (60 + D) = 49
60 + D = 2 * 49 = 98o
D = 98o - 60o
D = 38o
D = angle of minimum deviation
A = refractive angle of the prism
1.5 = \(\frac{\sin\frac{1}{2} (60 + D)}{\sin\frac{1}{2} \times 60}\)
1.5 = \(\frac{\sin\frac{1}{2} (60 + D)}{\sin 30}\)
Sin \(\frac{1}{2}\) (60 + D) = 1.5 * Sin 30
Sin \(\frac{1}{2}\) (60 + D) = 0.75
\(\frac{1}{2}\) (60 + D) = Sin-1 (0.75)
but Sin-1 (0.75) = 49o
\(\frac{1}{2}\) (60 + D) = 49
60 + D = 2 * 49 = 98o
D = 98o - 60o
D = 38o
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