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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) The following figure shows a cross-section of a ‘light pipe’ made of glass fibre with a refractive index of 1.68. The outer covering of the pipe is made of a material with a refractive index of 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?

                 

Answers (1)

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

Now applying Snell's law on the 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=31^0\

Let,  i  be the  Incident Angle then,

On Applying Snell's law

1sini=\mu_{glass}sinr

sini=1.68sin31=0.8652

i=60^o(approx)

Hence total range of incident angles for which total internal reflection happens is 0<i<60.

Posted by

Pankaj Sanodiya

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