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The refractive index of the material of an equilateral prism is 1.6. The angle of minimum deviation due to the prism would be
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The correct answer is:
between $45^{\circ}$ and $60^{\circ}$
Here, $\mu=1.6, A=60^{\circ}$
$\begin{aligned} & \text { As, } \mu=\frac{\sin \left(\frac{A+\delta_m}{2}\right)}{\sin \left(\frac{A}{2}\right)} \\ & \therefore \quad \sin \left(\frac{60^{\circ}+\delta_m}{2}\right)=1.6 \sin \left(\frac{60^{\circ}}{2}\right) \\ & \Rightarrow \sin \left(\frac{60^{\circ}+\delta_m}{2}\right)=0.8 \Rightarrow \delta_m=46.3^{\circ}\end{aligned}$
$\begin{aligned} & \text { As, } \mu=\frac{\sin \left(\frac{A+\delta_m}{2}\right)}{\sin \left(\frac{A}{2}\right)} \\ & \therefore \quad \sin \left(\frac{60^{\circ}+\delta_m}{2}\right)=1.6 \sin \left(\frac{60^{\circ}}{2}\right) \\ & \Rightarrow \sin \left(\frac{60^{\circ}+\delta_m}{2}\right)=0.8 \Rightarrow \delta_m=46.3^{\circ}\end{aligned}$
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