A glass prism of refracting angle 60^{circ} i | vedclass

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(B) The formula for the refractive index of the prism material $(\mu)$ with respect to the surrounding liquid is given by: $\mu = \frac{\sin((A + \del

Vedclass · Accepted Answer

$45$

Vedclass · Answer

(B) The formula for the refractive index of the prism material $(\mu)$ with respect to the surrounding liquid is given by: $\mu = \frac{\sin((A + \delta_m)/2)}{\sin(A/2)}$, where $A$ is the refracting angle and $\delta_m$ is the angle of minimum deviation.Given $A = 60^{\circ}$ and $\delta_m = 30^{\circ}$, we have:$\mu = \frac{\sin((60^{\circ} + 30^{\circ})/2)}{\sin(60^{\circ}/2)} = \frac{\sin(45^{\circ})}{\sin(30^{\circ})} = \frac{1/\sqrt{2}}{1/2} = \sqrt{2}$.The critical angle $(C)$ is related to the refractive index by the formula: $\sin(C) = 1/\mu$.Substituting the value of $\mu$: $\sin(C) = 1/\sqrt{2}$.Therefore, $C = 45^{\circ}$.

$A$ glass prism of refracting angle $60^{\circ}$ is immersed in a liquid in which the angle of minimum deviation is $30^{\circ}$. The critical angle of glass with respect to the liquid medium is (in $^{\circ}$)

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