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

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(A) For a prism,the refractive index $\mu$ is given by the formula: $\mu = \frac{\sin i}{\sin r}$.At the condition of minimum deviation,the angle of i

Vedclass · Accepted Answer

$45$

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(A) For a prism,the refractive index $\mu$ is given by the formula: $\mu = \frac{\sin i}{\sin r}$.At the condition of minimum deviation,the angle of incidence $i$ is equal to the angle of emergence $e$,and the angle of refraction $r$ is equal to $A/2$,where $A$ is the refracting angle of the prism.Given: $\mu = \sqrt{2}$ and $A = 60^{\circ}$.Therefore,$r = \frac{A}{2} = \frac{60^{\circ}}{2} = 30^{\circ}$.Using Snell's law at the first surface: $\mu = \frac{\sin i}{\sin r}$.$\sqrt{2} = \frac{\sin i}{\sin 30^{\circ}}$.$\sin i = \sqrt{2} 	imes \sin 30^{\circ} = \sqrt{2} 	imes \frac{1}{2} = \frac{1}{\sqrt{2}}$.Thus,$i = \arcsin\left(\frac{1}{\sqrt{2}}ight) = 45^{\circ}$.

$A$ prism of refractive index $\sqrt{2}$ has a refracting angle of $60^{\circ}$. In order for a ray to suffer minimum deviation,it should be incident at an angle of.......$^{\circ}$.

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