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(D) From the figure,the light ray travels from the coin to the surface at the critical angle $C$. The base of the triangle formed is $3 \, cm$ and the

Vedclass · Accepted Answer

$1.8 	imes 10^8 \, m/s$

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(D) From the figure,the light ray travels from the coin to the surface at the critical angle $C$. The base of the triangle formed is $3 \, cm$ and the height is $4 \, cm$. The hypotenuse is $\sqrt{3^2 + 4^2} = 5 \, cm$.By definition of the critical angle,$\sin C = \frac{	ext{opposite}}{	ext{hypotenuse}} = \frac{3}{5}$.Also,the refractive index of the liquid with respect to air is $\mu = \frac{1}{\sin C} = \frac{1}{3/5} = \frac{5}{3}$.We know that $\mu = \frac{c}{v}$,where $c = 3 	imes 10^8 \, m/s$ is the speed of light in air and $v$ is the speed of light in the liquid.Therefore,$v = \frac{c}{\mu} = \frac{3 	imes 10^8}{5/3} = \frac{9 	imes 10^8}{5} = 1.8 	imes 10^8 \, m/s$.

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

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