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JAMB Physics 2017 Paper

Question : 40

Total: 49

\sin^{-1}

(0.75) = 49º]

Solution:

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\left(\frac{1}{2}\times 60\right)}

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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