Total Internal Reflection - Practice Questions & MCQ

Total Internal Reflection:

When a ray of light goes from denser to rarer medium it bends away from the normal and as the angle of incidence in denser medium increases, the angle of refraction in rarer medium also increases and at a certain angle, angle of refraction becomes $90^{\circ}$ this angle of incidence is called critical angle (C).

When Angle of incidence exceeds the critical angle than light ray comes back into the same medium after reflection from interface. This phenomenon is called Total internal reflection (TIR).

Using snell's law :
$\mu_2 \sin C=\mu_1 \sin r$

$
\begin{aligned}
& \Longrightarrow \mu_2 \sin C=\mu_1 \text { since, } \sin r=1 \\
& \Longrightarrow \sin C=\frac{\mu_1}{\mu_2}=\frac{\text { R.I of rarer medium }}{\text { R.I of denser medium }}
\end{aligned}
$

$
\mu=\frac{1}{\sin C}
$

when $\mu_{1=1}$ for air and $\mu_2=\mu$.

 Conditions for TIR : 

 (i) The ray must travel from denser medium to rarer medium.

 (ii) The angle of incidence 'i' must be greater than critical angle 'C' i.e  i>c.

Circle of illuminance:

In the figure, ray 1 strikes the surface at an angle less than critical angle C and gets refracted in rarer medium. Ray 2 strikes the
surface at critical angle and grazes the interface. Ray 3 strikes the surface making an angle greater than the critical angle and gets
internally reflected. The locus of points where ray strikes at critical angle is a circle, called circle of illuminance. All light rays striking inside the circle of illuminance get refracted in the rarer medium. If an observer is in the rarer medium, the observer will see light coming out only from within the circle of illuminance. If a circular opaque plate covers the circle of illuminance, no light will get refracted in the rarer medium and then the object cannot be seen from the rarer medium. The radius of C.O.I. can be easily found.

From the figure:

$\tan \left(\theta_c\right)=\frac{R}{h} \quad$ where R is radius of $\mathrm{C} . \mathrm{O} . \mathrm{I}$
$\Longrightarrow R=h \tan \left(\theta_c\right)$
Also, $\sin \left(\theta_c\right)=\frac{1}{\mu}$

Therefore using trigonometry

$
\tan \left(\theta_c\right)=\frac{1}{\sqrt{\mu^2-1}}
$

So, the radius of the circle of illuminance, $\quad R=\frac{h}{\sqrt{\mu^2-1}}$