An isosceles glass prism with base angles 40^ | vedclass

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(B) The base angles of the isosceles prism are $40^{\circ}$. The angle of incidence at the base is determined by the geometry of the prism. Since the

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$\mu > 2.07$

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(B) The base angles of the isosceles prism are $40^{\circ}$. The angle of incidence at the base is determined by the geometry of the prism. Since the light is incident normally on the inclined face,it enters the prism without deviation and strikes the base at an angle of $40^{\circ}$ with respect to the normal.For total internal reflection $(TIR)$ to occur at the base,the angle of incidence $i$ must be greater than the critical angle $	heta_c$ between the glass and water.Given $i = 40^{\circ}$,the condition for $TIR$ is $i > 	heta_c$,which implies $\sin i > \sin 	heta_c$.We know that $\sin 	heta_c = \frac{\mu_w}{\mu_g}$,where $\mu_w = 1.33$ is the refractive index of water and $\mu_g = \mu$ is the refractive index of the glass.Therefore,$\sin 40^{\circ} > \frac{1.33}{\mu}$.Rearranging for $\mu$,we get $\mu > \frac{1.33}{\sin 40^{\circ}}$.Using $\sin 40^{\circ} \approx 0.6428$,we calculate $\mu > \frac{1.33}{0.6428} \approx 2.069$.Rounding to two decimal places,the condition is $\mu > 2.07$.

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