First slide

Prism

Question

A certain prism is found to produce a minimum deviation of $38^{\circ}$ . It produces a deviation of $44 °$ when the angle of incidence is either $42 ° or 62 °$ . What will be the angle of incidence when it undergoes minimum deviation?

Moderate

A

$35 °$
B

$44 °$
C

$49 °$
D

$52 °$

Solution

1. Given Minimum deviation $δ_{m} = 38^{\circ}$
Angle of Incidence $i_{1} = 42^{\circ} (or) 62^{\circ}$
Deviation $δ = 44^{\circ}$
$δ = i_{1} + i_{2} - A$
2. Here $i_{1} = 42^{\circ}$ and $i_{2} = 62^{\circ}$
$44^{\circ} = (42^{\circ} + 62^{\circ}) - A$
$A = 60^{\circ}$
Using the formula for minimum deviation, we get
$\begin{array}{l} \Rightarrow δ_{m} = 2 i - A \\ \Rightarrow i = \frac{δ_{m} + A}{2} \\ \Rightarrow i = \frac{38 + 60}{2} \\ ∴ i = 49^{\circ} \end{array}$
Therefore, the correct answer is (C).

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