The angle of a prism is 60^{circ}. When light | vedclass

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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 t

Vedclass · Accepted Answer

$40^{\circ} < i < 50^{\circ}$

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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 $A = 60^{\circ}$,$i = 60^{\circ}$,and $e = 40^{\circ}$.Thus,the deviation $\delta = 60^{\circ} + 40^{\circ} - 60^{\circ} = 40^{\circ}$.According to the principle of reversibility of light,if $i = 60^{\circ}$ gives $e = 40^{\circ}$,then $i = 40^{\circ}$ will give $e = 60^{\circ}$,resulting in the same deviation $\delta = 40^{\circ}$.The graph of deviation $\delta$ versus angle of incidence $i$ is a $U$-shaped curve where the minimum deviation $\delta_m$ occurs at the angle of incidence $i = i_m$.Since the deviation is the same $(40^{\circ})$ at $i = 40^{\circ}$ and $i = 60^{\circ}$,the minimum deviation must occur at an angle of incidence between these two values.Therefore,the angle of incidence for minimum deviation lies in the range $40^{\circ} < i < 60^{\circ}$. Looking at the provided graph,the minimum occurs at $i \approx 48^{\circ}$,which falls in the range $40^{\circ} < i < 50^{\circ}$.

The angle of a prism is $60^{\circ}$. When light is incident at an angle of $60^{\circ}$ on the prism,the angle of emergence is $40^{\circ}$. The angle of incidence $i$ for which the light ray will deviate the least is such that

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