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Q 9.17 (b)    What is the answer if there is no outer covering of the pipe?

       

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In the case when there is no outer layer, we apply Snell's law at the glass-air interface (when the ray is emerging out from the pipe),

\mu _{glass} \times sin i' = 1 \times sin 90

sini'=\frac{1}{\mu_{glass}} = \frac{1}{1.68}=0.595

i'=36.5

The refractive angle r corresponding to this is, 90 - 36.5 = 53.5.

 So, from here we can see that the angle r is greater than the critical angle. So for all of the incident angles, the rays will get internally reflected. In other words, rays won't bend in air-glass interference, they would rather hit the glass-air interference and get reflected 

Posted by

Pankaj Sanodiya

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