Physics - Geometrical Optics Question with Solution | TestHub

For an equilateral prism with $A=60^{\circ }$ and $\mu =2$. For P: Minimum deviation ${\delta }_m$. $\mu =\frac{\sin ((A+{\delta }_m)/2)}{\sin (A/2)}⟹2=\frac{\sin ((60^{\circ }+{\delta }_m)/2)}{\sin (30^{\circ })}=\frac{\sin ((60^{\circ }+{\delta }_m)/2)}{1/2}$. So $\sin ((60^{\circ }+{\delta }_m)/2)=1$, which means $(60^{\circ }+{\delta }_m)/2=90^{\circ }$. Thus ${\delta }_m=120^{\circ }$. (P)-(3) For Q: Critical angle ${\theta }_c$. $\sin ({\theta }_c)=1/\mu =1/2$. So ${\theta }_c=30^{\circ }$. (Q)-(1) For R: Grazing emergence ($e=90^{\circ }$). From Snell's law at the second surface, $\mu \sin (r_2)=1\sin (90^{\circ })=1$. So $\sin (r_2)=1/2$, which means $r_2=30^{\circ }$. Since $r_1+r_2=A$, $r_1=60^{\circ }-30^{\circ }=30^{\circ }$. From Snell's law at the first surface, $1\sin (i)=\mu \sin (r_1)=2\sin (30^{\circ })=1$. So $i=90^{\circ }$. (R)-(4)

For S: If the ray enters normally, $i=0^{\circ }$. Then $r_1=0^{\circ }$. Since $r_1+r_2=A$, $r_2=60^{\circ }$. As $r_2=60^{\circ }>{\theta }_c=30^{\circ }$, total internal reflection occurs at the second surface. (S)-(5)