$(a)\;1.3 \\ (b)\;1.4 \\ (c)\;1.5 \\ (d)\;1.6 $

According to the Snell's law $n_1 \sin \theta_1=n_2 \sin \theta_2$

Given $n_1=n; \sin \theta_1=\sin 45^{\circ}=\large\frac{1}{\sqrt 2}$

$n_2 \geq 1; \theta \geq 90^{\circ}$

$\therefore \sin \theta_2 =\large\frac{n_1}{n_2} $

$\sin \theta_2 =\large\frac{n_1}{1} \bigg( \large\frac{1}{\sqrt 2}\bigg) $

Since for internal reflection $\theta_2 \geq 90^{\circ}$

limiting value of $\sin \theta_2 = \large\frac{n}{\sqrt 2}$$=1$ or $n= \sqrt 2$

Thus ray will undergo total internal reflection if $ n \geq \sqrt 2$

Hence b is the correct answer.

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