Calculate the angle of minimum deviation of a 60o prism of a refractive index [sin-10.75 = 49o]

Assessment: JAMB UTME - Physics - 2017 Subject: Physics

Question 1 Report

38.00o

19.47o

16.25o

49o

Answer Details

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

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