Critical Angle and Total Internal Reflection

CRITICAL ANGLE AND TOTAL INTERNAL REFLECTION

Consider a ray of light that travels from a denser medium to a rarer medium. The angle of incidence for which the angle of refraction becomes 90° is called critical angle. This is used for constructing totally reflecting prisms.

$\frac{\sin c}{\sin 90}{=}_2{m}_1=\frac{m_2}{m_1}$

$\Rightarrow \sin c=\frac{1}{{}_1{\mu }_2}$

and if i > c

When the angle of incidence of a ray travelling from denser to rarer medium is greater then the critical angle, no refraction occurs. The incident ray is totally reflected back into the same medium. Here the laws of reflection hold good. Some light is also reflected before the critical angle is achieved, but not totally.

Illustration 1: A point source of light a placed at the bottom of a tank containing a liquid of refractive index $\mu$ The level of the liquid is at a height h above the bottom of the tank. A bright circular spot is seen on the surface of the liquid when viewed from above. Find the radius of the spot.

Solution: The rays of light emerging from the source, which are incident at an angle $\theta >{\theta }_c$. (where q_c is the critical angle) will be reflected.

$\text{sin}{\theta }_c=\frac{1}{\mu }...\left(\text{i}\right)$

Further, $\frac{R}{h}=\text{tan}{\theta }_c$

$=\frac{\text{sin}{\theta }_c}{\sqrt{1-\text{si}{\text{n}}^2{\theta }_c}}$

$=\frac{1}{\sqrt{{\mu }^2-1}}$

or, $R=\frac{h}{\sqrt{{\mu }^2-1}}...\left(\text{ii}\right)$