Question

The incoming ray suffers total internal reflection (TIR) times as shown. The prism has reflective index $\mu $ and top angle $\theta $ .The emerging ray propagates horizontally. The refractive index $\mu = \dfrac{{\cos \theta }}{{\cos n\theta }}$. Find n.

Answer 1

Hint: We will use Snell’s to find the angle of refraction and incidence
$\dfrac{{\sin {i_p}}}{{\sin r}} = \mu $ ($r$ , is the angle of refraction ) or
${\mu _p}\sin i = {\mu _{air}}\sin r$ ($\mu $ is the refractive index of air and water respectively )
We will also use the relation: $r = {90^0} - {i_p}$

Complete step by step solution:
Let’s discuss the phenomenon of total internal reflection first and then we will calculate the value of n.
Total internal reflection is the internal phenomenon in which the surface of the water in a pool (let’s say an example) when viewed from below the water level, reflects the underwater scene like a mirror without any loss of brightness. In other words TIR occurs when waves in one medium reach the boundary with another medium at a sufficiently slanting angle, providing that the second medium is transparent to the waves and allows them to travel faster than in the first medium.

Now we will calculate the value of n.
According to Snell’s law:
${\mu _p}\sin i = {\mu _{air}}\sin r$.............(1)
We will calculate angle i and r to calculate the refractive index at the point where second total internal reflection takes place.
If we draw the normal at the point of refracted ray then angle of refraction is equal to :
$r = {90^0} - \theta $...................(2)
(Angle of refraction after drawing normal will subtract theta which is an alternate angle)
Angle of incident will be equal to:
$
  i = 90 - 2\theta + 90 - \theta - 90 \\
  i = 90 - 3\theta \\
 $ ...................(3)
(Angle of incident will be calculated after first total internal reflection therefore the angle theta will become twice of it and subtraction will take place to calculate the angle).
We will substitute the equation 2 and 3 in equation 1.
$
   \Rightarrow {\mu _p}\sin i = {\mu _{air}}\sin r \\
   \Rightarrow {\mu _p}\sin (90 - 3\theta ) = 1 \times \sin (90 - \theta ) \\
 $
We have taken refractive index of air as 1:
$
   \Rightarrow {\mu _p}\cos 3\theta = 1 \times \cos \theta \\
   \Rightarrow {\mu _p} = \dfrac{{\cos \theta }}{{\cos 3\theta }} \\
 $............................(4)
If we compare the equation 4 with the given refractive index of question which is given as:
$\mu = \dfrac{{\cos \theta }}{{\cos n\theta }}$

We find that value of n is 3.

Note: There are various uses of total internal reflection of light such as; used in Prisms, binoculars, optical fibres. We have seen various examples of total internal reflection such as mirage formation (road looks like filled with water when seen from a finite distance), fish looks like water above as a mirror.