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
$\sin i=\frac{1}{\mu }$
$i=90-r$
$\dots \dots$
$\therefore \sin \left(90^{\circ }-r\right)=\frac{1}{\mu }=\cos r=\frac{1}{\mu }\dots$(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
$\frac{1}{\mu }=\sqrt{\frac{2{\mu }^2-1}{2{\mu }^2}}$
$=\frac{1}{{\mu }^2}=\frac{2{\mu }^2-1}{2{\mu }^2}$
$2=2{\mu }^2-1=\frac{3}{2}={\mu }^2$
$\Rightarrow \mu =\sqrt{\frac{3}{2}}$