Question

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

• A. $\frac{\sqrt{3}+1}{2}$
• B. $\sqrt{\frac{5}{2}}$
• C. $\sqrt{\frac{3}{2}}$
• D. $\sqrt{\frac{7}{2}}$

Answer 1

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.

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