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At what angle should a ray of light be incident on the face of a prism of refracting angle $60^{\circ}$ so that it just suffers total internal reflection at the other face? The refractive index of the material of the prism is $\mathbf{1 . 5 2 4}$.
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Angle of Prism, $\mathrm{A}=60^{\circ}, \mathrm{n}=1.524$
As the ray suffer TIR at the face $A C: r_2=c\left(\begin{array}{c}\text { critical } \\ \text { angle }\end{array}\right)$
As $\sin \mathrm{c}=\frac{1}{\mathrm{n}}=\frac{1}{1.524}=0.6561 \quad \therefore \mathrm{c}=41^{\circ}$
Also, $r_2=41^{\circ}$.
As, $r_1+r_2=A \quad \therefore r_1=A-r_2=60^{\circ}-41^{\circ}=19^{\circ}$
According to Snell's law,
Refractive index, $\mathrm{n}=\frac{\sin \mathrm{i}}{\sin \mathrm{r}_1}$
$\therefore \sin \mathrm{i}=\mathrm{n} \cdot \sin \mathrm{r}_1=1.524 \times \sin 19^{\circ}=1.524 \times 0.325=0.4962$
$\therefore \mathrm{i} \approx 30^{\circ}$.
As the ray suffer TIR at the face $A C: r_2=c\left(\begin{array}{c}\text { critical } \\ \text { angle }\end{array}\right)$
As $\sin \mathrm{c}=\frac{1}{\mathrm{n}}=\frac{1}{1.524}=0.6561 \quad \therefore \mathrm{c}=41^{\circ}$
Also, $r_2=41^{\circ}$.
As, $r_1+r_2=A \quad \therefore r_1=A-r_2=60^{\circ}-41^{\circ}=19^{\circ}$
According to Snell's law,
Refractive index, $\mathrm{n}=\frac{\sin \mathrm{i}}{\sin \mathrm{r}_1}$
$\therefore \sin \mathrm{i}=\mathrm{n} \cdot \sin \mathrm{r}_1=1.524 \times \sin 19^{\circ}=1.524 \times 0.325=0.4962$
$\therefore \mathrm{i} \approx 30^{\circ}$.
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