Question

In the figure shown, for an angle of incidence 45° at the top surface, what is the minimum refractive index needed for total internal reflection at vertical surface?

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

Answer 1

The correct option is B $\sqrt{\frac{3}{2}}$

For the total internal reflection from the vertical surface, the angle of incidence on the vertical surface must be greater than critical angle $\left({\theta }_c\right)$.

∴ At surface MR, $sinc=\frac{1}{\mu }$ ....(i)
Also, at surface MN, $\frac{sin{45}^o}{sinr}=\mu$ $\mu =\frac{sin\left(i\right)}{sin\left(r\right)}$
$sinr=\frac{1}{\sqrt{2}\mu }$ ...(ii)
On $DPQ,r+{\theta }_c=$${\theta }_c={90}^{\circ }-r$ ...(iii)
$\sin ({90}^{\circ }-r)=\frac{1}{\mu }$
$\cos r=\frac{1}{\mu }$ ....(iv)
Now, $si{n}^{2}r+co{s}^{2}r=1$
$\frac{1}{{\left(\sqrt{2}\mu \right)}^2}+\frac{1}{{\mu }^2}=1$ [Using (ii) and (iv)]
${\mu }^2=\frac{3}{2}$
$\mu =\sqrt{\frac{3}{2}}$
Hence, the correct answer is option (b).