Question
As shown in Fig. 37-10, a ray enters the flat end of a long rectangular block of glass that has a refractive index of $n_2$. Show that all entering rays can be totally internally reflected only if $n_2>1.414$
Step 1
We have a ray entering a rectangular glass block with refractive index nâ‚‚ through its flat end. We need to determine the condition for total internal reflection to occur for all entering rays at the side faces of the block. Show more…
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As shown in Fig. $37-10$, a ray enters the flat end of a long rectangular block of glass that has a refractive index of $n_{2}$. Show that all entering rays can be totally internally reflected only if $n_{2}>1.414 .$ The larger $\theta_{1}$ is, the larger $\theta_{2}$ will be, and the smaller $\theta_{3}$ will be. Therefore, the ray is most likely to leak out through the side of the block if $\theta_{1}=90^{\circ} .$ In that case, $$ n_{1} \sin \theta_{1}=n_{2} \sin \theta_{2} \quad \text { becomes } \quad(1)(1)=n_{2} \sin \theta_{2} $$ For the ray to just escape, $\theta_{4}=90^{\circ} .$ Then $$ n_{2} \sin \theta_{3}=n_{1} \sin \theta_{4} \quad \text { becomes } \quad n_{2} \sin \theta_{3}=(1)(1) $$ We thus have two conditions to satisfy: $n_{2} \sin \theta_{2}=1$ and $n_{2} \sin \theta_{3}=1$. Their ratio gives $$ \frac{\sin \theta_{2}}{\sin \theta_{3}}=1 $$ But we see from the figure that $\sin \theta_{3}=\cos \theta_{2}$, and so this becomes $$ \tan \theta_{2}=1 \quad \text { or } \quad \theta_{2}=45.00^{\circ} $$ Then, because $n_{2} \sin \theta_{2}=1$, we have $$ n_{2}=\frac{1}{\sin 45.00^{\circ}}=1.414 $$ This is the smallest possible value the index can have for total internal reflection of all rays that enter the end of the block. It is possible to obtain this answer by inspection. How?
As shown in Fig. $37-10$, a ray enters the flat end of a long rectangular block of glass that has a refractive index of $n_{2}>1.414$. The larger $\theta_{1}$ is the larger $\theta_{2}$ will be, and the smaller $\theta_{3}$ will be. Therefore, the ray is most likely to leak out through the side of the block if $\theta_{1}=90^{\circ} .$ In that case, $$ \mathrm{n}_{1} \sin \theta_{1}=\mathrm{n}_{2} \sin \theta_{2} \text { becomes }(1)(1)=\mathrm{n}_{2} \sin \theta_{2} $$ For the ray to just escape, $\theta_{4}=90^{\circ}$. Then $$ \mathrm{n}_{2} \sin \theta_{3}=\mathrm{n}_{1} \sin \theta_{4} \text { becomes } \mathrm{n}_{2} \sin \theta_{3}=(1)(1) $$ We thus have two conditions to satisfy: $\mathrm{n}_{2} \sin \theta_{3}=1$ and $\mathrm{n}_{2} \sin \theta_{3}=$ 1. Their ratio gives $$ \frac{\sin \theta_{2}}{\sin \theta_{3}}=1 $$ But we see from the figure that $\sin \theta_{3}=\cos \theta_{2}$, and so this becomes $$ \tan \theta_{2}=1 \text { or } \theta_{2}=45.00^{\circ} $$ Then, because $n_{2} \sin \theta_{2}=1$, we have $$ n_{2}=\frac{1}{\sin 45.00^{\circ}}=1.414 $$ This is the smallest possible value the index can have for total internal reflection of all rays that enter the end of the block. It is possible to obtain this answer by inspection. How?
In Fig. 35-40 a light ray enters a glass slab at point $A$ and then undergoes total internal reflection at point $B$. What minimum value for the index of refraction of the glass can be inferred from this information?
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