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a pond with a total depth (ice + water) of 3.25 m is covered by a trans…

Question

a pond with a total depth (ice + water) of 3.25 m is covered by a transparent layer of ice, with a thickness of 0.38 m. find the time required for light to travel vertically from the surface of the ice to the bottom of the pond.

a. 21.3 ns
b. 36.8 ns
c. 12.7 ns
d. 43.1 ns
e. 14.4 ns

Explanation:

Step1: Find time in ice

The speed of light in ice $v_{ice}=\frac{c}{n_{ice}}$, where $c = 3\times10^{8}\ m/s$ is the speed of light in vacuum and $n_{ice}=1.31$. The thickness of ice $d_{ice}=0.38\ m$. Time $t_{ice}=\frac{d_{ice}}{v_{ice}}=\frac{d_{ice}n_{ice}}{c}$.
$t_{ice}=\frac{0.38\times1.31}{3\times 10^{8}}\ s=\frac{0.4978}{3\times 10^{8}}\ s$

Step2: Find time in water

The depth of water $d_{water}=3.25 - 0.38=2.87\ m$. The refractive - index of water $n_{water}=1.33$. The speed of light in water $v_{water}=\frac{c}{n_{water}}$. Time $t_{water}=\frac{d_{water}}{v_{water}}=\frac{d_{water}n_{water}}{c}$.
$t_{water}=\frac{2.87\times1.33}{3\times 10^{8}}\ s=\frac{3.8171}{3\times 10^{8}}\ s$

Step3: Find total time

The total time $t = t_{ice}+t_{water}$.
$t=\frac{0.4978 + 3.8171}{3\times 10^{8}}\ s=\frac{4.3149}{3\times 10^{8}}\ s$.
Converting to nanoseconds, $t=\frac{4.3149}{3\times 10^{8}}\times10^{9}\ ns\approx14.4\ ns$

Answer:

e. 14.4 ns