The angle of deviation through a prism is min | vedclass

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Question

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

Vedclass · Accepted Answer

Statements $(A), (B)$ and $(C)$ are true

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$(A)$ For a prism, the angle of 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.At the angle of minimum deviation:$1$. The angle of incidence is equal to the angle of emergence, i.e., $i = e$. This implies that the incident ray and emergent ray are symmetric with respect to the prism, making statement $(A)$ and $(C)$ true.$2$. The refracted ray inside the prism becomes parallel to its base, which implies $r_1 = r_2$. This makes statement $(B)$ true.Since statements $(A)$, $(B)$, and $(C)$ are all correct conditions for minimum deviation, the correct option is $(A)$.

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