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  • Given a slab with index and incident light striking the top horizontal face at an

Given a slab with index \mathrm{n = 1.33} and incident light striking the top horizontal face at angle \mathrm{i} as shown in figure. The maximum value of \mathrm{i} for which total internal reflection occurs is:

Option: 1

\mathrm{\sin ^{-1}(\sqrt{0.77})}


Option: 2

\mathrm{\cos ^{-1}(\sqrt{0.77})}


Option: 3

\mathrm{\sin ^{-1}(0.77)}


Option: 4

\mathrm{\sin ^{-1}(0.38)}


Answers (1)

best_answer

\mathrm{TIR} to take place, \mathrm{90^{\circ}-\mathrm{r}>\theta_{\mathrm{C}} \Rightarrow \mathrm{r}<90^{\circ}-\theta_{\mathrm{C}}}

From the relation, 
\mathrm{\begin{aligned} & \mu=\frac{\sin i}{\sin r} \\ & \sin i=\mu \operatorname{sir} r \end{aligned}}
Since, for \mathrm{TIR} to take place, \mathrm{r<90^{\circ}-\Theta _{c}}
\mathrm{\begin{aligned} \therefore \quad \mathrm{i}= & \sin ^{-1}\{\mu \sin \mathrm{r}\} \\ & \mathrm{i} \leq \sin ^{-1}\left[\left\{\mathrm{n} \sin \left(90^{\circ}-\theta_{\mathrm{C}}\right)\right\}\right] \end{aligned}}
\mathrm{or \quad \mathrm{i} \leq \sin ^{-1}\left\{\mu \cos \theta_{\mathrm{C}}\right\}}\\ \mathrm{or \quad \mathrm{i} \leq \sin ^{-1}\left\{\mu \sqrt{1-\frac{1}{\mu^2}}\right\} \quad\left(\because \sin \theta_{\mathrm{C}}=\frac{1}{\mu_{\mathrm{C}}}\right)}
\begin{array}{ll} \text { or } & i \leq \sin ^{-1}\left\{1.33 \sqrt{1-\frac{1}{(1.33)^2}}\right\} \\ \text { or } & i \leq \sin ^{-1}\{\sqrt{0.77}\} \end{array}
Note: For simplifying the calculation, take 1.33=\frac{4}{3}

Posted by

Deependra Verma

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