Question

A rod made of glass, refractive index $1.5$ and of square cross section, is bent into the shape shown in figure. A parallel beam of light falls normally on the plane flat surface $A$. Referring to the diagram, $d$ is the width of a side and $R$ is the radius of inner semicircle. Find the maximum value of ratio $\frac{d}{R}$ so that all light entering the glass through surface $A$ emerges from the glass through surface $B$.

• A. $\frac{1}{2}$
• B. $\frac{1}{5}$
• C. $\frac{1}{4}$
• D. $\frac{2}{3}$

Answer 1

The correct option is A $\frac{1}{2}$

If all of the light is transmitted to the other end, this means all of the light is undergoing TIR (Total internal reflection).

Drawing the rays incident on the curved glass surface


From the diagram, we can see that the smallest angle of incidence is for the rightmost ray incident on the curved surface. If this angle is greater than the critical angle of the material, then all rays will undergo TIR.

In the triangle,

$sin\theta =\frac{R}{R+d}$

For critical angle

$sin{\theta }_c=\frac{1}{\eta }$ where $\eta$ is the refractive index of the medium (glass).

$\Rightarrow \frac{R}{R+d}>\frac{1}{1.5}$
$\Rightarrow \frac{d}{R}<\frac{1}{2}$