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Condition of no emergence

Condition of no emergence :- The light ray entering the prism at interface AB, will not be able to come out from the interface AC, if total internal reflection takes place at this interface. For this the minimum value of angle of incidence at interface AC must be greater than critical angle, i.e.,

$\displaystyle {{({{r}_{2}})}_{min}}>C$ …..(1)

Now as

$\displaystyle A={{r}_{1}}+{{r}_{2}}$ …..(2)

$\displaystyle \Rightarrow {{r}_{2}}=A-{{r}_{1}}$

$\displaystyle \Rightarrow {{\left( {{r}_{2}} \right)}_{\min }}=A-{{\left( {{r}_{1}} \right)}_{\max }}$

But r₁ is maximum when i = i_max = 90°, i.e., incident ray grazes the interface AB.

Applying Snell’s Law at interface AB for i = i_max = 90°, we get

$\displaystyle \frac{\sin i}{\sin {{r}_{1}}}=\mu$

$\displaystyle \Rightarrow \frac{\sin {{90}^{\circ }}}{\sin {{\left( {{r}_{1}} \right)}_{\max }}}=\mu$

$\displaystyle \Rightarrow {{\left( {{r}_{1}} \right)}_{\max }}={{\sin }^{-1}}\left( \frac{1}{\mu } \right)$

$\displaystyle \Rightarrow {{\left( {{r}_{1}} \right)}_{\max }}=C$ …..(3)

From (1), and (3), we get

$\displaystyle {{r}_{1}}+{{r}_{2}}>2C$

Using (2), we can say that

$\displaystyle A>2C$

$\displaystyle \Rightarrow \frac{A}{2}>C$

$\displaystyle \Rightarrow \sin \left( \frac{A}{2} \right)>\sin C$

$\displaystyle \Rightarrow \sin \left( \frac{A}{2} \right)>\frac{1}{\mu }$

$\displaystyle \Rightarrow \mu >\cos \text{ec}\left( \frac{A}{2} \right)$

Hence, a light ray will not emerge out of a prism whatever be the angle of incidence if A > 2C, i.e., if $\displaystyle \mu >\cos \text{ec}\left( \frac{A}{2} \right)$ .

Condition of no emergence

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