Questions & Answers

Question

Answers

a)$1.3$

b)$1.4$

c)$1.5$

d)$1.2$

Answer
Verified

$i>{{\sin }^{-1}}(\dfrac{1}{\mu })$

For total internal reflection to occur, the angle of incidence must be greater than the critical angle.

Let the refractive index be assumed as $\mu $and the angle of incidence as $i$,

The critical angle is calculated as,

${{\sin }^{-1}}(\dfrac{1}{\mu })$

Therefore, the formula turns into,

$\begin{align}

& i > {{\sin }^{-1}}(\dfrac{1}{\mu }) \\

& \sin (i) > \dfrac{1}{\mu } \\

& \sin ({{45}^{0}}) > \dfrac{1}{\mu } \\

& \mu > \sqrt{2} \\

& \mu > 1.44 \\

& \\

\end{align}$

The total internal reflection occurs if the angle of incidence is greater than the critical angle. It generally takes place at the boundary between two transparent mediums. The indices of refraction depend on wavelength, the critical angle will vary slightly with wavelength with color. If the angle of reflection is less than the critical angle, both refraction and reflection occur at different proportions. Critical angle is the angle at which the light ray grazes through the meeting point of two different mediums. The incident ray will get reflected if the incident angle is greater than the critical angle.

The total internal reflection occurs when the angle of incidence is greater than critical angle. Also, the refraction phenomenon occurs only when this condition is satisfied. In all other cases, both refraction and reflection occur at different proportions.