Question

For the given incident ray as shown in figure, the condition of total refraction of this ray the minimum refractive index of prism will be :

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

Answer 1

The correct option is C $\frac{\sqrt{5}}{2}$

From $\Delta BCD$ we have

$i_c+r+{90}^{\circ }={180}^{\circ }$

$\Rightarrow r={90}^{\circ }-i_c$

$\mu =\frac{\sin i}{\sin r}=\frac{\sin {30}^{\circ }}{\sin r}=\frac{\sin {30}^{\circ }}{\sin \left({90}^{\circ }-i_c\right)}=\frac{1/2}{\cos i_c}$

We know that

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

$\therefore \frac{1/2}{\cos i_c}=\frac{1}{\sin i_c}$

$\Rightarrow \tan i_c=2$

Placing this in a triangle we get $p=2,b=1$

Hence $h=\sqrt{2^2+1}=\sqrt{5}$

$\therefore \sin i_c=\frac{2}{\sqrt{5}}$

Hence, $\mu =\frac{\sqrt{5}}{2}$

Therefore the correct answer is option (C).