For the given incident ray as shown in the figure, the condition for total internal reflection will be satisfied if the refractive index of the block is greater than

The correct option is C $\sqrt{\frac{3}{2}}$
Angle of incidence must exceed the critical angle at the interface where total internal reflection has to occur.

Hence,

$\sin \theta >\sin i_C$

From the definition of critical angle

$\sin i_C=\frac{1}{\mu }$

$(\mu$ is the refractive index of the block)

From the geometry of the given figure, a right angled triangle is formed as shown.


Containing angles $\theta$ and $r$, we get

$\therefore \theta ={90}^{\circ }-r$

$\Rightarrow \sin ({90}^{\circ }-r)>\frac{1}{\mu }$

or $\cos r>\frac{1}{\mu }$

or $\mu >\frac{1}{\cos r}.....\left(i\right)$

On applying shell's law,

$\frac{\sin {45}^{\circ }}{\sin r}=\mu$

$\Rightarrow \sin r=\frac{1}{\sqrt{2}\mu }$

or, $\cos r=\sqrt{1-{\sin }^{2}r}=\sqrt{1-\frac{1}{2{\mu }^2}}$

From the limiting condition of total internal reflection, equation $\left(i\right)$ gives,

$\mu =\frac{1}{\cos r}$

$\Rightarrow \mu =\frac{1}{\sqrt{1-\frac{1}{2{\mu }^2}}}$

or, ${\mu }^2=\frac{1}{1-\frac{1}{2{\mu }^2}}$

or, ${\mu }^2-\frac{1}{2}=1$

or, ${\mu }^2=\frac{3}{2}$

$\therefore \mu =\sqrt{\frac{3}{2}}$

Thus, $\mu$ must be greater than $\sqrt{\frac{3}{2}}$ for T.I.R to occur.

$\begin{array}{c}\text{Why this question}?\\ \text{In such problems always demarcate the interface for}\\ \text{T.I.R and apply the condition for critical incidence}\\ \text{along with Snell's law}\end{array}$