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A small coin is resting on the bottom of a beaker filled with liquid. A ray of light from the coin travels upto the surface of the liquid and moves along its surface. How fast is the light travelling in the liquid?
(The ray meets the surface at a horizontal distance of 3 cm from the point directly above the coin; the depth of the liquid is 4 cm.)
(The ray meets the surface at a horizontal distance of 3 cm from the point directly above the coin; the depth of the liquid is 4 cm.)
A
$2.4 \times 10^{8}$ m/s
B
$3.0 \times 10^{8}$ m/s
C
$1.2 \times 10^{8}$ m/s
D
$1.8 \times 10^{8}$ m/s
Detailed Solution
The refracted ray moves along the surface, so the angle of refraction is $r = 90^\circ$ and the angle of incidence $i$ is the critical angle.
In the right triangle formed, the vertical side is 4 cm and the horizontal side is 3 cm, so the hypotenuse comes out to be $\sqrt{3^2 + 4^2} = 5$ cm.
$\sin i = \dfrac{3}{5}$
By Snell's law for light going from the liquid to air: $\dfrac{1}{\mu} = \dfrac{\sin i}{\sin 90^\circ}$
$\mu = \dfrac{1}{\sin i} = \dfrac{5}{3}$
Speed of light in the liquid: $v = \dfrac{c}{\mu} = \dfrac{3 \times 10^{8}}{5/3}$
$v = 1.8 \times 10^{8}$ m/s