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?