A prism of refractive index sqrt{2} has a ref | vedclass

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(A) For a prism,the condition for minimum deviation is given by the formula: $\mu = \frac{\sin(\frac{A + \delta_m}{2})}{\sin(\frac{A}{2})}$.At minimum

Vedclass · Accepted Answer

$45$

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(A) For a prism,the condition for minimum deviation is given by the formula: $\mu = \frac{\sin(\frac{A + \delta_m}{2})}{\sin(\frac{A}{2})}$.At minimum deviation,the angle of incidence $i$ is related to the refractive index $\mu$ and the prism angle $A$ by the relation: $\mu = \frac{\sin i}{\sin(A/2)}$.Given: $\mu = \sqrt{2}$ and $A = 60^{\circ}$.Substituting the values: $\sqrt{2} = \frac{\sin i}{\sin(60^{\circ}/2)}$.$\sqrt{2} = \frac{\sin i}{\sin(30^{\circ})}$.Since $\sin(30^{\circ}) = 0.5$,we have: $\sin i = \sqrt{2} 	imes 0.5 = \frac{\sqrt{2}}{2} = \frac{1}{\sqrt{2}}$.Therefore,$i = \arcsin(\frac{1}{\sqrt{2}}) = 45^{\circ}$.

$A$ prism of refractive index $\sqrt{2}$ has a refracting angle of $60^{\circ}$. At what angle must a ray be incident on it so that it suffers minimum deviation (in $^{\circ}$)?

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