Question

A small coin is resting on the bottom of a beaker filled with a liquid. A ray of light from the coin travels up to the surface of the liquid and moves along its surface (see figure). How fast is the light traveling in the liquid?

 

\[\text{A}\text{. 1}\text{.2}\times \text{1}{{\text{0}}^{8}}\text{ }m/s\]
\[\text{B}\text{. 1}\text{.8}\times \text{1}{{\text{0}}^{8}}\text{ }m/s\]
\[\text{C}\text{. 2}\text{.4}\times \text{1}{{\text{0}}^{8}}\text{ }m/s\]
\[\text{C}\text{. 3}\text{.0}\times \text{1}{{\text{0}}^{8}}\text{ }m/s\]

Answer 1

Hint: Here, the light ray did not emerge out from water (denser medium) to air (rarer medium) We can see from figure that, the angle of refraction is \[90{}^\circ \]. Then the incident angle is a critical angle. We can find out the velocity of light in water using the equation which relates the refractive index and sin of critical angle.

Formula used:
\[\text{sin C =}\dfrac{1}{\mu }\]
\[\mu =\dfrac{\text{Velocity of light in air}\left( c \right)}{\text{Velocity of light in liquid}\left( v \right)}\]
\[R=h\tan C\]
\[{{\sin }^{2}}a+{{\cos }^{2}}a=1\]
\[\text{sinC=}\dfrac{\text{adjacent side}}{\text{Hypotenuse}}\]
\[\tan C=\dfrac{\operatorname{sin}C}{\operatorname{Cos}C}\]

Complete answer:

As shown in the figure, the light ray from the coin did not emerge. Therefore, the angle of incidence is the critical angle.
(When the light travels from an optically denser medium to an optically less dense (rarer) medium, angle of incidence where the angle of refraction is \[90{}^\circ \] is known as critical angle.)
From the figure,
\[\text{sin C=}\dfrac{3}{\sqrt{{{4}^{2}}+{{3}^{2}}}}=\dfrac{3}{5}\]
Where, \[C\] is the critical angle
We know that,
\[\dfrac{1}{\mu }=\text{ sin C}\]
Where, \[\mu \] is the refractive index.
\[\text{sin C =}\dfrac{1}{\mu }\]
\[\mu =\dfrac{\text{Velocity of light in air}\left( c \right)}{\text{Velocity of light in liquid}\left( v \right)}\]
Therefore,
\[\sin C=\dfrac{v}{c}\]
We have, \[\text{c = 3}\times \text{1}{{\text{0}}^{8}}m/s\]
Then,
\[\sin C=\dfrac{v}{3\times {{10}^{8}}}\]
\[\dfrac{3}{5}=\dfrac{v}{3\times {{10}^{8}}}\]
\[v=\dfrac{3}{5}\times 3\times {{10}^{8}}=1.8\times {{10}^{8}}m/s\]

So, the correct answer is “Option B”.

Note:
This can be solved in another method.

Consider triangle PAB,
We have,
\[\dfrac{R}{h}=\tan C\]
Where,
\[\text{R - circle of radius}\]
\[\text{h- height}\]
\[R=h\tan C\]
\[R=h\times \dfrac{\sin C}{\operatorname{Cos}C}\]
Since, \[\left( {{\sin }^{2}}a+{{\cos }^{2}}a=1 \right)\]
\[R=h\times \dfrac{\sin C}{\sqrt{1-{{\sin }^{2}}C}}\]
We know that,
\[\text{sin C =}\dfrac{1}{\mu }\]
Then,
\[R=\dfrac{h}{\sqrt{{{\mu }^{2}}-1}}\]
\[\dfrac{3}{4}=\dfrac{h}{\sqrt{{{\mu }^{2}}-1}}\]
\[{{\mu }^{2}}=\dfrac{25}{9}\]
\[\mu =\dfrac{5}{3}=1.67\]
We know that,
\[\mu =\dfrac{c}{v}\]
\[v=\dfrac{3\times {{10}^{8}}}{1.66}=1.8\times {{10}^{8}}m/s\]