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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?

                 

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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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