For a transparent prism,if the angle of minim | vedclass

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(B) The formula for the refractive index $n$ of a prism is given by $n = \frac{\sin((A + \delta_{\min})/2)}{\sin(A/2)}$.Given that the angle of minimu

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$1 < n < 2$

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(B) The formula for the refractive index $n$ of a prism is given by $n = \frac{\sin((A + \delta_{\min})/2)}{\sin(A/2)}$.Given that the angle of minimum deviation $\delta_{\min}$ is equal to the refracting angle $A$,we substitute $\delta_{\min} = A$ into the formula.$n = \frac{\sin((A + A)/2)}{\sin(A/2)} = \frac{\sin(A)}{\sin(A/2)}$.Using the trigonometric identity $\sin(A) = 2 \sin(A/2) \cos(A/2)$,we get $n = \frac{2 \sin(A/2) \cos(A/2)}{\sin(A/2)} = 2 \cos(A/2)$.Since the refracting angle $A$ of a prism must satisfy $0 Therefore,$0 Multiplying by $2$,we get $0 However,for a physical prism,$A$ is typically small or less than $180^{\circ}$,and $n > 1$. Thus,$1 < n < 2$.

For a transparent prism,if the angle of minimum deviation is equal to its refracting angle,the refractive index $n$ of the prism satisfies.

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