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(A) For a prism,the deviation $D$ is given by $D = (i_1 + i_2) - A$,where $i_1$ and $i_2$ are the angles of incidence and emergence,and $A$ is the pri

Vedclass · Accepted Answer

$49$

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(A) For a prism,the deviation $D$ is given by $D = (i_1 + i_2) - A$,where $i_1$ and $i_2$ are the angles of incidence and emergence,and $A$ is the prism angle.Given $D = 44^{\circ}$ for $i_1 = 42^{\circ}$ and $i_2 = 62^{\circ}$ (or vice versa),we have:$44^{\circ} = (42^{\circ} + 62^{\circ}) - A$$44^{\circ} = 104^{\circ} - A$$A = 104^{\circ} - 44^{\circ} = 60^{\circ}$.At minimum deviation $(D_m = 38^{\circ})$,the angle of incidence $i$ is given by $i = \frac{A + D_m}{2}$.$i = \frac{60^{\circ} + 38^{\circ}}{2} = \frac{98^{\circ}}{2} = 49^{\circ}$.

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

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