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(B) For a prism,the deviation $\delta$ is given by $\delta = (i + e) - A$,where $i$ is the angle of incidence,$e$ is the angle of emergence,and $A$ is

Vedclass · Accepted Answer

$49$

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(B) For a prism,the deviation $\delta$ is given by $\delta = (i + e) - A$,where $i$ is the angle of incidence,$e$ is the angle of emergence,and $A$ is the prism angle.Given that for $\delta = 44^{\circ}$,the angles of incidence are $i = 42^{\circ}$ and $i = 62^{\circ}$. By the principle of reversibility of light,if $i = 42^{\circ}$ gives $e = 62^{\circ}$,then $i = 62^{\circ}$ gives $e = 42^{\circ}$.Substituting these values into the deviation formula: $44^{\circ} = (42^{\circ} + 62^{\circ}) - A$.$44^{\circ} = 104^{\circ} - A \Rightarrow A = 60^{\circ}$.At minimum deviation $\delta_m$,the angle of incidence $i$ is equal to the angle of emergence $e$,and $i = \frac{\delta_m + A}{2}$.Given $\delta_m = 38^{\circ}$ and $A = 60^{\circ}$,we have $i = \frac{38^{\circ} + 60^{\circ}}{2} = \frac{98^{\circ}}{2} = 49^{\circ}$.

$A$ 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 degrees)

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