InterviewSolution
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InterviewSolution
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(a) Figure 9.32 shows a cross-section of a 'light pipe' made of a glass fibre of refractive index 1.68. The outer covering of the pipe is made of a material of refractive 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 figure. (b) What is the answer if there is no outer covering of the pipe? |
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Answer» Solution :`sini._(c)=1.44/1.68" which GIVES "i._(c)=59^(@)`. Total internal reflection takes place when `i gt 59^(@)` or when `r lt r_("max")=31^(@)`. Now, `(sin i_("max")//sinr_("max"))=1.68`, which gives `i_("max")~=60^(@)`. Thus, all incident rays of ANGLES in the range `0ltilt60^(@)` will suffer total internal reflections in the pipe.(If the length of the pipe is finite, which it is in practice, there will be a lower limit on i determined by the ratio of the diameter to the length of the pipe.) (b) If there is no outer coating, `i._(c)=sin^(-1)(1//1.68)=36.5^(@),." Now "i=90^(@)`will have `r = 36.5^(@) and i′_(c) = 53.5^(@)` which is greater than `i′_(c)` . Thus, all incident rays `("in the range "53.5^(@) lt i lt 90^(@))` will suffer total internal reflections. |
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