Solved: PHYSICS Question for NEET

NEET PHYSICSMedium

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?

$1.8 \times 10^8 \text{ m/s}$

$2.4 \times 10^8 \text{ m/s}$

$3.0 \times 10^8 \text{ m/s}$

$1.2 \times 10^8 \text{ m/s}$

Step-by-Step Solution

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 ( $i_c$ ).
Snell's Law: $\mu \sin i_c = 1 \times \sin 90^\circ \implies \sin i_c = \frac{1}{\mu}$ .
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: $\sin i_c = \frac{v}{c}$ .
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 $c \approx 3.0 \times 10^8 \text{ m/s}$ : $v = c \times \sin i_c$ $v = 3.0 \times 10^8 \times 0.6$ $v = 1.8 \times 10^8 \text{ m/s}$

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