Question 333: Ray Optics

Question

A prism of angle $75^{\circ}$ and refractive index $\sqrt{3}$ is coated with thin film of refractive index 1.5 only at the back exit surface. To have total internal reflection at the back exit surface the incident angle must be $\_\_\_\_$
$\left(\sin 15^{\circ}=0.25\right.$ and $\left.\sin 25^{\circ}=0.43\right)$

Accepted Answer

Total internal reflection (TIR) occurs at the interface between the prism ( $n_p=\sqrt{3}$ ) and the film ( $n_f=1.5$ ). $$ \sin C=\frac{n_{ ext {rarer }}}{n_{ ext {denser }}}=\frac{1.5}{\sqrt{3}}=\frac{\sqrt{3}}{2} \Rightarrow C=\sin ^{-1}\left(\frac{\sqrt{3}}{2} ight)=60^{\circ} $$ So, the critical angle is $C=60^{\circ}$. For TIR to occur at the second surface, the angle of incidence $r_2$ must satisfy: $$ r_2>C \Rightarrow r_2>60^{\circ} $$ For a prism with angle $\mathrm{A}=75^{\circ}$ : $$ \mathrm{A}=\mathrm{r}_1+\mathrm{r}_2 \Rightarrow 75^{\circ}=\mathrm{r}_1+\mathrm{r}_2 $$ Substituting the condition $\mathrm{r}_2>60^{\circ}$ : $$ r_1=75^{\circ}-r_2<75^{\circ}-60^{\circ} $$ $$\Rightarrow $$ $$ \mathrm{r}_1<15^{\circ} $$ Apply Snell's Law at the first surface (assuming air to prism) : $$ \mathrm{n}_{ ext {air }} \sin \mathrm{i}=\mathrm{n}_{\mathrm{p}} \sin \mathrm{r}_1 $$ $$\Rightarrow $$ $$ \sin \mathrm{i}=\sqrt{3} \sin \mathrm{r}_1 $$ Since $\mathrm{r}_1<15^{\circ}$, then $\sin \mathrm{i}<\sqrt{3} \sin 15^{\circ}$ : $$ \sin i<1.732 imes 0.25 $$ $$\Rightarrow $$ $$ \sin i<0.433 $$ Given that $\sin 25^{\circ}=0.43$, we can conclude: $$ \mathrm{i}<25^{\circ} $$ For the angle $r_2$ to remain above $60^{\circ}$, i.e., the value of $r_1$ will be between 0 and $15^{\circ}$. The maximum value of $r_2$ can be $75^{\circ}$ for which $\mathrm{r}_1=0$. $$ \sin \mathrm{i}=\sqrt{3} \sin 0^{\circ}=0 $$ So, the value of i will be between $0^{\circ}$ and $25^{\circ}$.

Answer

between $15^{\circ}$ and $20^{\circ}$

Answer

$15^{\circ}$

Answer

$<15^{\circ}$

Answer

$>25^{\circ}$

Ray Optics

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