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Question

A slab of refractive index $$\mu$$ is placed in air and light is incident at maximum angle $$\theta _{0}$$ from vertical. Find minimum value of $$\mu$$ for which total internal reflection takes place at the vertical surface.
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Solution

Using Snell’s law at the surface S1:

$$\mu \sin r=\sin {{\theta }_{0}}............(1)$$

For TIR at vertical surface:

  $$ \mu \sin {{\theta }_{c}}=\sin 90 $$

 $$ \mu \sin {{\theta }_{c}}=1 $$

 $$ \because {{\theta }_{c}}=90-r $$

 $$ \mu \sin (90-r)=1 $$

 $$ \mu \cos r=1......(2) $$

Solving equation (1) and (2)

  $$ {{\mu }^{2}}({{\sin }^{2}}r+{{\cos }^{2}}r)={{\sin }^{2}}{{\theta }_{0}}+1 $$

 $$ {{\mu }^{2}}={{\sin }^{2}}{{\theta }_{0}}+1 $$

 $$ \mu =\sqrt{{{\sin }^{2}}{{\theta }_{0}}+1} $$

 


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Physics

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