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: $\mu = \frac{\sin i}{\sin(A/2)}$,where $i$ is the angle of incidence and $A$ is th

Vedclass · Accepted Answer

$45$

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(A) For a prism,the refractive index $\mu$ is given by the formula: $\mu = \frac{\sin i}{\sin(A/2)}$,where $i$ is the angle of incidence and $A$ is the refracting angle of the prism at the condition of minimum deviation.Given: $\mu = \sqrt{2}$ and $A = 60^{\circ}$.Substituting the values into the formula:$\sqrt{2} = \frac{\sin i}{\sin(60^{\circ}/2)}$$\sqrt{2} = \frac{\sin i}{\sin(30^{\circ})}$Since $\sin(30^{\circ}) = 0.5$ or $1/2$:$\sin i = \sqrt{2} 	imes \frac{1}{2} = \frac{1}{\sqrt{2}}$Therefore,$i = \arcsin(1/\sqrt{2}) = 45^{\circ}$.

$A$ 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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