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 face AD?

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

Answer 1

Answer: (B)

To find:

Minimum R.I

so that TIR

takes place at AD=?

For TIR

$90-t={\theta }_c$

Applying snesll's Law at Ab

$u_1\sin i=u_2\sin r$

$1\times \sin 45=u\sin r$

$u\sin t=\frac{1}{\sqrt{2}}$............(1)

Applying snell's law at AD

$u\sin (90-t)=1\times \sin 90$

$u=\frac{1}{\cos r}$............(2)

From eq (1) and (2)

$\tan r=\frac{1}{\sqrt{2}}$

from eq (1)

$u.\frac{1}{\sqrt{3}}=\frac{1}{\sqrt{2}}\Rightarrow u=\sqrt{\frac{3}{2}}$