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  • In the following figure 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.

Q 9.17 (a)    In the following figure 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.

                       

Answers (1)

best_answer

We are given,

Refractive index of glass(\mu _{glass}) and outer covering(\mu _{outerlayer}) is 1.68 and 1.44 respectively.

Now applying snell's law on upper glass - outer layer, 

\mu _{glass}sini'=\mu_{outerlayer}sin90

i' = the angle from where total Internal reflection starts

sini'=\frac{\mu_{outerlayer}}{\mu_{glass}} = \frac{1.44}{1.68}=0.8571

i' = 59^0

At this angle, in the air-glass interface

Refraction angle r = 90 - 59 = 31 degree

let Incident Angle be  i .

Applying Snell's law

1sini=\mu_{glass}sinr

sini=1.68sin31=0.8652

i=60(approx)

Hence total range of incident angle for which total internal reflection happen is 0<i<60.

Posted by

Pankaj Sanodiya

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