Question

A portion of straight glass rod of diameter 4 cm and refractive index 1.5 is bent into an arc of circle of radius R and a parallel beam of light is incident on it as shown in figure. Radius of arc includes outer edge of glass rod. Smallest R in cm which permits all the light to pass around arc is (Write upto two digits after the decimal point.)

Answer 1

For TIR, $\sin \theta =\frac{1}{\mu }=\frac{2}{3}$

From the diagram,
$\sin \theta =\frac{R-d}{R}$
So,
$\frac{R-d}{R}=\frac{2}{3}\Rightarrow 3R-3d=2R$

$R=3d=3\times 4cm=12cm$