The minimum deviation produced by a hollow pr | vedclass

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Question

(A) For a prism,at the condition of minimum deviation,the angle of refraction $r$ is related to the prism angle $A$ by $A = 2r$. Given $r = 30^{\circ}

Vedclass · Accepted Answer

$\sqrt{2}$

Vedclass · Answer

(A) For a prism,at the condition of minimum deviation,the angle of refraction $r$ is related to the prism angle $A$ by $A = 2r$. Given $r = 30^{\circ}$,we have $A = 2 	imes 30^{\circ} = 60^{\circ}$. The formula for the refractive index $\mu$ of the prism material is $\mu = \frac{\sin((A + \delta_m)/2)}{\sin(A/2)}$. Substituting the given values $\delta_m = 30^{\circ}$ and $A = 60^{\circ}$: $\mu = \frac{\sin((60^{\circ} + 30^{\circ})/2)}{\sin(60^{\circ}/2)} = \frac{\sin(45^{\circ})}{\sin(30^{\circ})}$. Since $\sin(45^{\circ}) = \frac{1}{\sqrt{2}}$ and $\sin(30^{\circ}) = \frac{1}{2}$,we get $\mu = \frac{1/\sqrt{2}}{1/2} = \frac{2}{\sqrt{2}} = \sqrt{2}$.

The minimum deviation produced by a hollow prism filled with a certain liquid is found to be $30^{\circ}$. The light ray is also found to be refracted at an angle of $30^{\circ}$. Then the refractive index of the liquid is

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