Write the equation for the refractive index of a prism in terms of the angle of minimum deviation $(\delta_m)$ and the angle of the prism $(A)$.
(N/A) For a prism with refractive index $n$, angle of prism $A$, and angle of minimum deviation $\delta_m$, the relationship is given by Snell's Law applied at the minimum deviation condition.
At minimum deviation, the angle of incidence $i$ is equal to the angle of emergence $e$, and the angle of refraction $r_1$ is equal to $r_2 = r = A/2$.
The angle of incidence is given by $i = (A + \delta_m) / 2$.
Using Snell's Law, $n = \frac{\sin(i)}{\sin(r)}$.
Substituting the values, we get the refractive index equation: $n = \frac{\sin((A + \delta_m) / 2)}{\sin(A / 2)}$.
At minimum deviation, the angle of incidence $i$ is equal to the angle of emergence $e$, and the angle of refraction $r_1$ is equal to $r_2 = r = A/2$.
The angle of incidence is given by $i = (A + \delta_m) / 2$.
Using Snell's Law, $n = \frac{\sin(i)}{\sin(r)}$.
Substituting the values, we get the refractive index equation: $n = \frac{\sin((A + \delta_m) / 2)}{\sin(A / 2)}$.
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[$A$] At minimum deviation,the incident angle $i_1$ and the refracting angle $r_1$ at the first refracting surface are related by $r_1=\left(i_1 / 2\right)$.
[$B$] For this prism the refractive index $\mu$ and the angle of prism $A$ are related as $A=\frac{1}{2} \cos ^{-1}(\mu / 2)$.
[$C$] For this prism,the emergent ray at the second surface will be tangential to the surface when the angle of incidence at the first surface is $i_1=\sin ^{-1}\left[\sin A \sqrt{4 \cos ^2 \frac{A}{2}-1}-\cos A\right]$.
[$D$] For the angle of incidence $i_1=A$,the ray inside the prism is parallel to the base of the prism.
[$A$] At minimum deviation,the incident angle $i_1$ and the refracting angle $r_1$ at the first refracting surface are related by $r_1=\left(i_1 / 2\right)$.
[$B$] For this prism the refractive index $\mu$ and the angle of prism $A$ are related as $A=\frac{1}{2} \cos ^{-1}(\mu / 2)$.
[$C$] For this prism,the emergent ray at the second surface will be tangential to the surface when the angle of incidence at the first surface is $i_1=\sin ^{-1}\left[\sin A \sqrt{4 \cos ^2 \frac{A}{2}-1}-\cos A\right]$.
[$D$] For the angle of incidence $i_1=A$,the ray inside the prism is parallel to the base of the prism.
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