Question

A fish looking up through the water sees the outside world contained in a circular horizon. If the refractive index of water is 4/3 and the fish is 12cm below the surface, the radius of this circle (in cm) is:
A. \[36\sqrt{7}\]
B. $36/\sqrt{7}$
C. $36\sqrt{5}$
D. $4\sqrt{5}$

Answer 1

Hint: Total internal reflection occurs if angle of incidence is greater than the critical angle. The sine of the critical angle is inversely proportional to the refractive index of the medium. The circular horizon will form the base of a cone with half angle equal to the critical angle. If the critical angle is known, then the radius of the circular horizon can be found using basic trigonometry.

Formula used: $\sin {{i}_{c}}=\dfrac{1}{n}$
Here,
     $n$ is the refractive index of denser medium
     ${{i}_{c}}$ is the critical angle of incidence

Complete step by step solution:
Light rays travelling from denser to rarer medium experience total internal reflection. If the angle of incidence is greater than critical angle, then the light rays will get reflected from the interface back into the denser medium. Thus, the fish living under water will only see the outside world restricted to a circular horizon when the angle of incidence is less than the critical angle.

The refractive index of water is $n=\dfrac{4}{3}$

The critical angle is
${{i}_{c}}={{\sin }^{-}}^{1}\dfrac{1}{n}={{\sin }^{-}}^{1}\dfrac{3}{4}$

If $r$ is the radius of the circular horizon and $h$ is the distance of the fish from the surface, then form geometry
$\cot {{i}_{c}}=\dfrac{r}{h}$

From trigonometry
\[\begin{align}
  & 1+{{\cot }^{2}}{{i}_{c}}={{\csc }^{2}}{{i}_{c}} \\
 & 1+\dfrac{{{h}^{2}}}{{{r}^{2}}}=\dfrac{1}{si{{n}^{2}}{{i}_{c}}} \\
 & 1+\dfrac{{{h}^{2}}}{{{r}^{2}}}=\dfrac{{{4}^{2}}}{{{3}^{2}}} \\
 & r=h\sqrt{\dfrac{9}{7}}=12\times \sqrt{\dfrac{9}{7}}\text{ cm = }\dfrac{36}{\sqrt{7}}\text{ cm} \\
\end{align}\]

The correct answer is option B.

Additional information: Optical fibres are based on the principle of total internal reflection. They are used to transmit signals over large distances with very little attenuation. It is also used in optical fingerprinting and modulation.
The knowledge of critical angle is utilised in cutting of gemstones for maximum brilliance.

Note: The critical angle is the angle of incidence when the angle of refraction is $90^o$. Total internal reflection also occurs in other types of waves such as sound and water waves.