Question

In the figure for an angle of incidence $45^{\circ},$ at the top surface, what is the minimum refractive index needed for total internal reflection at vertical face:

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

Answer 1

Answer: (B)

At point A by snell's law
$\mu=\frac{\sin 45^{\circ}}{\sin r }$
$ \Rightarrow \sin r =\frac{\sin 45^{\circ}}{\mu}$
$\therefore \sin r =\frac{1}{\mu \sqrt{2}}$
At point B for total internal reflection
$\sin i =\frac{1}{\mu}$
$i=90-r$
$\ldots \ldots$
$\therefore \sin \left(90^{\circ}- r \right)=\frac{1}{\mu}=\cos r =\frac{1}{\mu} \dots$(1)
Now $\cos r=\sqrt{1-\sin ^{2} r}=\sqrt{1-\frac{1}{2 \mu^{2}}}$
$=\sqrt{\frac{2 \mu^{2}-1}{2 \mu}} \cdots \cdots$ (2)
Putting the value of $\cos r$ in equation (2) and squaring both sides
$\frac{1}{\mu}=\sqrt{\frac{2 \mu^{2}-1}{2 \mu^{2}}}$
$=\frac{1}{\mu^{2}}=\frac{2 \mu^{2}-1}{2 \mu^{2}}$
$2=2 \mu^{2}-1=\frac{3}{2}=\mu^{2} $
$\Rightarrow \mu=\sqrt{\frac{3}{2}}$