Question

A tank whose bottom is a mirror is filled with water to a depth of 19.4 . A small fish floats motionless 7.10 under the surface of the water.

part A)
What is the apparent depth of the fish when viewed at normal incidence to the water?
Express your answer in centimeters. Use 1.33 for the index of refraction of water.


Part B)
What is the apparent depth of the reflection of the fish in the bottom of the tank when viewed at normal incidence?
Express your answer in centimeters. Use 1.33 for the index of refraction of water.

Answer 1

Concepts and reason

The concept of apparent depth is required to solve the problem.

First, determine the apparent depth of the fish when viewed at normal incidence to the water by using the relationship between actual depth of the fish and apparent depth of the fish. Then determine the real depth of the reflection of the fish in the mirror. Finally determine the apparent depth of the fish’s reflection when viewed at normal incidence by using the relationship between actual depth of the fish’s reflection and apparent depth of the fish’s reflection.

Fundamentals

If viewed at normal incidence to the water, the apparent depth of an object floating in water is determined by using the formula,

${d_2} = \frac{{{n_{{\rm{air}}}}}}{{{n_{{\rm{water}}}}}}{d_1}$

Here, ${n_{{\rm{air}}}}$ is the refractive index of the air, ${n_{{\rm{water}}}}$ is the refractive index of the water, ${d_1}$ is real depth of the object, and ${d_2}$ is apparent depth of the object.

(A)

Calculate the apparent depth of the fish.

The apparent depth of the fish is calculated by using the relation,

${d_2} = \frac{{{n_{{\rm{air}}}}}}{{{n_{{\rm{water}}}}}}{d_1}$

Substitute 7.10 cm for ${d_1}$, 1.00 for ${n_{{\rm{air}}}}$, and 1.33 for ${n_{{\rm{water}}}}$ in the above equation.

$\begin{array}{c}\\{d_2} = \left( {\frac{{1.00}}{{1.33}}} \right)\left( {7.10{\rm{ cm}}} \right)\\\\ = 5.34{\rm{ cm}}\\\end{array}$

Calculate the apparent depth of the reflection of the fish in the bottom of the tank.

The actual depth of the reflection of the fish in the bottom of the tank is given as,

${d_1}' = 2s - {d_1}$

Here, s is the depth of the mirror and ${d_1}$ is the actual depth of the fish in the water.

Substitute 19.4 cm for s and 7.10 cm for ${d_1}$ in the above equation.

$\begin{array}{c}\\{d_1}' = 2\left( {19.4{\rm{ cm}}} \right) - 7.10{\rm{ cm}}\\\\{\rm{ = 31}}{\rm{.7 cm}}\\\end{array}$

The apparent depth of the reflection of the fish in the bottom of the tank is given as,

${d_2}' = \frac{{{n_{{\rm{air}}}}}}{{{n_{{\rm{water}}}}}}{d_1}'$

Substitute 31.7 cm for ${d_1}'$, 1.00 for ${n_{{\rm{air}}}}$, and 1.33 for ${n_{{\rm{water}}}}$ in the equation${d_2}' = \frac{{{n_{{\rm{air}}}}}}{{{n_{{\rm{water}}}}}}{d_1}'$.

$\begin{array}{c}\\{d_2}' = \left( {\frac{{1.00}}{{1.33}}} \right)\left( {31.7{\rm{ cm}}} \right)\\\\ = 23.8{\rm{ cm}}\\\end{array}$

Ans: Part A

The apparent depth of the fish is 5.34 cm.

Part B

The apparent depth of the reflection of the fish in the bottom of the tank is 23.8 cm.