A small coin is resting on the bottom of a beaker filled with a liquid. A ray of light from the coin travels u

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?

Accepted Answer

1. **Identify the Phenomenon:** When a ray of light moves along the surface of the liquid after travelling from a denser medium (liquid) to a rarer medium (air), the angle of refraction is $90^\circ$. The angle of incidence in the liquid is called the **Critical Angle** ($i_c$).
2. **Snell's Law:** $\mu \sin i_c = 1 	imes \sin 90^\circ \implies \sin i_c = \frac{1}{\mu}$.
3. **Relation to Speed:** The refractive index $\mu$ is defined as the ratio of the speed of light in vacuum ($c$) to the speed in the medium ($v$): $\mu = \frac{c}{v}$ . Substituting this into the critical angle formula: $\sin i_c = \frac{v}{c}$.
4. **Calculation:** From the standard geometry associated with this AIPMT 2007 problem (typically a triangle with height 4 units and base 3 units, where the ray strikes the surface), the sine of the critical angle is $\frac{3}{\sqrt{3^2+4^2}} = \frac{3}{5} = 0.6$. Using $c \approx 3.0 	imes 10^8 	ext{ m/s}$ :
   $$ v = c 	imes \sin i_c $$
   $$ v = 3.0 	imes 10^8 	imes 0.6 $$
   $$ v = 1.8 	imes 10^8 	ext{ m/s} $$

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