Question

A glass rod of rectangular cross-section is bent into the shape shown in the diagram. A parallel beam of light falls perpendicularly on the flat surface $A$ . Determine the minimum value of the ratio $\frac{R}{d}$ for which all light entering the glass through surface $A$ will emerge from the glass through surface $B$ . The index of refraction of the glass is $1.5$ .

Answer 1

Answer: 2

Consider the representative rays shown in diagram. A ray entering the glass through surface $A$ and passing along the inner side of the rod will be reflected by the outer side with the smallest angle $\alpha$ , at which the reflected ray is tangent to the inner side. We have to consider the conditions under which the ray will undergo total internal reflection before reaching $B$ .

If $\alpha >{\theta }_{\text{c}}$ , the critical angle, at which total internal reflection occurs, all the incident beam will emerge through the surface $B$ . Hence, we require
$\text{sin}\alpha >\frac{1}{\text{n}}$
The geometry gives $\text{sin}\alpha =\frac{\text{R}}{\left(\text{R}+\text{d}\right)}$
Therefore $\frac{R}{R+d}\ge \frac{1}{n}$ ,
or ${\left(\frac{R}{d}\right)}_{min}=\frac{1}{n-1}=\frac{1}{1.5-1}=2$