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Calculate the angle of minimum deviation of a 60º prism of a refractive index 1.5[ \(sin^{-1}\)...

Physics
JAMB 2017

Calculate the angle of minimum deviation of a 60º prism of a refractive index 1.5[ \(sin^{-1}\) (0.75) = 49º]

 

  • A. 38o
  • B. 19.47o
  • C. 16.25o
  • D. 49o
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Correct Answer: Option A
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

There is an explanation video available below.


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Explanation Video

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Post-UTME Past Questions - Original materials are available here - Download PDF for your school of choice + 1 year SMS alerts
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