a) Figure shows a cross - section of a light pipe made of a glass fibre of refractive inde 1.68. The outer covering of the pipe is made of a material of refracitive index 1.44. What is the range of the angles of the incident rays with the axis of the pipe for which total reflections inside the pipe take place, as shown in the figures. b) What is the answer if there is no outer covering of the pipe?

a) Figure shows a cross - section of a light pipe made of a glass fibr

1.	a) Figure shows a cross - section of a light pipe made of a glass fibre of refractive inde 1.68. The outer covering of the pipe is made of a material of refracitive index 1.44. What is the range of the angles of the incident rays with the axis of the pipe for which total reflections inside the pipe take place, as shown in the figures. b) What is the answer if there is no outer covering of the pipe?
Answer» `mu=(1.68)/(1.44)=(1)/(sinC)` `SinC=(1.44)/(1.68)=0.8571` `C=59^(@)` Total internal reflection takes place when `i gt 59^(@)` or angle r may have value between 0 to `31^(@)` `r_("max")=31^(@)` Now `(sin i_("max"))/(sin r_("max"))=1.68` `(sin i_("max"))/(sin 31^(@))=1.68` `Sin i_("max")=0.8562, i_("max")~~60^(@)` Thus all incident rays of angles in the range `0 lt i lt 60^(@)` will suffer total internal reflections in the pipe. b) If there is no outer covering of the pipe `Sin C=(1)/(mu)` `=(1)/(1.68)=0.5962` `sin C=sin 36.5^(@)` `C=36.5^(@)`