What Is the Area of the Sky Seen by a Fish?

Notation

\(h=3\) \(\mathrm{m}\)	distance between the eye of the fish and the surface
\(S=\) \(?\)	area in which the fish sees the sky

Hint 1

Draw the light beam travelling from the air to the surface and then to the fish’s eye.

Solution of Hint 1

The beam is incident on the water surface. Part of the beam is refracted and reaches the fish’s eye and the other part is reflected from the surface. Only some of the rays that are incident on the eye are shown in the figure. The other rays incident on the eye are located between the drawn rays.

Hint 2

What is the shape of the area in which the fish sees the sky?

Solution of Hint 2

We assume the fish’s eye as the point in which the rays converge.

The area will be circular with the radius \(r\): \[S=\pi · r^2.\tag{1}\]

Hint 3

In the „Travelling beams“ picture, contained it the section „Solution of Hint \(1\)“, keep only those rays which are refracted to the eye.

Mark the rays in the picture between which the angle is the largest. These rays circumscribe the desired area \(S\). Provide reasoning why.

Solution of Hint 3

The rays which form the largest angle are marked red. The desired area S is marked in orange.

The rays travelling from the sky, which are incident on the desired area of the surface, are then refracted to fish’s eye, thus the fish can see the sky.

Beam cannot be incident on the surface at the angle greater than \(90°.\)

Other beams that could be incident on the eye of the fish would have to travel from the places under the surface, but we are not interested in these beams (they cannot help the fish to see the sky).

Hint 4

Now we focus only on the beams that are refracted at the angle \(\mathrm{α}\), see „Picture to Hint 4“.

We mark the radius of the desired areay \(r\) and the distance from the fish’s eye to the surface \(h\).

1) Write the relation for the angle \(\mathrm{α}\) in accordance with the law of refraction.

2) Deduce the relation between \(\mathrm{α}\), \(r\) and \(h\) from the picture.

Solution of Hint 4

1) We derive the relation for the angle of refraction from Snell’s law: \[\frac{\sin {90°}}{\sin \mathrm{α}}=\frac{n_{2}}{n_{1}},\] where \(n_{1}\) is the refractive index of air and \(n_{2}\) is the refractive index of water.

Since \(\sin {90°}=1\), we can write:

\[\frac{1}{\sin \mathrm{α}}=\frac{n_{2}}{n_{1}},\]

Therefrom we express \(\sin \mathrm{α}\):

\[\sin \mathrm{α}=\frac{n_{1}}{n_{2}},\tag{2}\]

2) Due to the similarity of triangles we can adjust the picture as follows:

We can see from the picture that \[\sin \mathrm{α}=\frac{r}{\sqrt{r^2+h^2}},\tag{3}\] where \(\sqrt{r^2+h^2}\) is a hypotenuse of the triangle and we have determined its length from the Pythagorean theorem.

Hint 5

Express the radius \(r\) from relation (3) and adjust the obtained equation using the relation acquired from the law of refraction (2), so that \(r\) depends only on \(h\), \(n_{1}\) and \(n_{2}\).

Then substitute \(r\) in the relation for calculating area of the circle (1) and compute the value of the desired area numerically.

Solution of Hint 5

(1): \(S=\pi · r^2,\)

(2): \(\sin \mathrm{α}=\frac{n_{1}}{n_{2}},\)

(3): \(\sin \mathrm{α}=\frac{r}{\sqrt{r^2+h^2}}.\)

If we multiply both sides of equation (3) by \(\sqrt{r^2+h^2}\), we will obtain: \[r=\sqrt{r^2+h^2}·\sin \mathrm{α}.\]

According to (2), we substitute \(\sin \mathrm{α}\): \[r= \sqrt{r^2+h^2} · \frac{n_{1}}{n_{2}}.\]

We raise both sides of the equation to the power of two and multiply the brackets:

\[r^2= \left(r^2+h^2 \right)· \left(\frac{n_{1}}{n_{2}}\right)^2=r^2\left(\frac{n_{1}}{n_{2}}\right)^2+h^2\left(\frac{n_{1}}{n_{2}}\right)^2 .\]

Then we subtract \(r^2\left (\frac{n_{1}}{n_{2}}\right )^2\)from both sides of the equation and factor out \(r^2\): \[r^2\left [1 − \left(\frac{n_{1}}{n_{2}}\right)^2\right]= h^2\left (\frac{n_{1}}{n_{2}}\right)^2 .\]

We divide both sides of the equation by \(\left[1 − (\frac{n_{1}}{n_{2}})^2\right]\) and extract the root: \[r= \frac {h\left(\frac{n_{1}}{n_{2}}\right)} {\sqrt{1 − \left(\frac{n_{1}}{n_{2}}\right)^2}}= \frac {h · n_{1}} {n_{2} \sqrt{1 − \left(\frac{n_{1}}{n_{2}}\right)^2}} .\]

We convert the expression inside the square root to the common denominator and adjust the fraction: \[r= \frac {h · n_{1}} {n_{2} \sqrt{\frac{n_{2}^2−n_{1}^2}{n_{2}^2}}} =\frac {h · n_{1}} { \sqrt{n_{2}^2−n_{1}^2}}.\]

We substitute \(r\) in relation (1) and raise the whole equation to the power of two: \[S=\pi r^2=\pi · \left( \frac {h · n_{1}} {\sqrt{n_{2}^2−n_{1}^2}}\right)^2= \pi · \frac {h^2 · n_{1}^2} {n_{2}^2−n_{1}^2} .\]

Numerical solution:

We know from the task assignment:

\(h=3\) \(\mathrm{m}\).

We look up the values of \(\pi\), \(n_{2}\) a \(n_{1}\) in the Tables:

Ludolph’s number			\(\pi\) \(\dot=\) \(3{.}14\)
Refractive index of air			\(n_{1}\) \(\dot=\) \(1{.}00\)
Refractive index of water			\(n_{2}\) \(\dot=\) \(1{.}33\)

We substitute for the values and obtain the size of the desired area:

\(S\) \(\dot=\) \(\frac{3{.}14 · 3^2 ·{1{.}00}^2}{{1{.}33}^2 − {1{.}00}^2}\) \(\mathrm{m^2}=\frac{3{.}14 · 9 ·1}{{1{.}33}^2 − {1}}\) \(\mathrm{m^2}\) \(\dot=\) \(36{.}8\) \(\mathrm{m^2}\)

and its radius:

\(r\) \(\dot=\) \( \frac {3 · 1{.}00} {\sqrt{1{.}33^2 −1{.}00^2}}\) \(\mathrm{m}\) \(=\) \( \frac {3} {\sqrt{1{.}33^2 − 1}}\) \(\mathrm{m}\) \(\dot=\) \(3{.}42\) \(\mathrm{m}\).

Overall Solution

Trajectory of the light beams and the shape of the area in which the fish sees the sky

We assume the fish’s eye as the point in which the rays converge.

The area will be circular with the radius \(r\): \[S=\pi · r^2.\tag{1}\]

The beams which are incident on the surface at the angle of \(\mathrm{90°}\)

Now we focus only on the beams that are refracted at the angle \(\mathrm{α}\) (see „Picture of the beams“).

We derive the relation for the angle of refraction \(\mathrm{α}\) from Snell’s law: \[\frac{\sin {90°}}{\sin \mathrm{α}}=\frac{n_{2}}{n_{1}},\] where \(n_{1}\) is the refractive index of air and \(n_{2}\) is the refractive index of water.

Since \(\sin {90°}=1\), we can write:

\[\frac{1}{\sin \mathrm{α}}=\frac{n_{2}}{n_{1}},\]

Therefrom we express \(\sin \mathrm{α}\):

\[\sin \mathrm{α}=\frac{n_{1}}{n_{2}},\tag{2}\]

Due to the similarity of triangles we can adjust the picture as follows:

We can see from the picture that \[\sin \mathrm{α}=\frac{r}{\sqrt{r^2+h^2}},\tag{3}\] where \(\sqrt{r^2+h^2}\) is a hypotenuse of the triangle and we have determined its length from the Pythagorean theorem.

Size of the desired area

(1): \(S=\pi · r^2,\)

(2): \(\sin \mathrm{α}=\frac{n_{1}}{n_{2}},\)

(3): \(\sin \mathrm{α}=\frac{r}{\sqrt{r^2+h^2}}.\)

If we multiply both sides of the equation (3) by \(\sqrt{r^2+h^2}\), we will obtain: \[r=\sqrt{r^2+h^2}·\sin \mathrm{α}.\]

According to (2)) we substitute \(\sin \mathrm{α}\): \[r= \sqrt{r^2+h^2} · \frac{n_{1}}{n_{2}}.\]

We raise both sides of the equation to the power of two and multiply the brackets:

\[r^2= \left(r^2+h^2 \right)· \left(\frac{n_{1}}{n_{2}}\right)^2=r^2\left(\frac{n_{1}}{n_{2}}\right)^2+h^2\left(\frac{n_{1}}{n_{2}}\right)^2 .\]

Then we subtract \(r^2\left (\frac{n_{1}}{n_{2}}\right )^2\) from both sides of the equation and factor out \(r^2\): \[r^2\left [1 − \left(\frac{n_{1}}{n_{2}}\right)^2\right]= h^2\left (\frac{n_{1}}{n_{2}}\right)^2 .\]

We divide both sides of the equation by \(\left[1 − (\frac{n_{1}}{n_{2}})^2\right]\) and extract the root: \[r= \frac {h\left(\frac{n_{1}}{n_{2}}\right)} {\sqrt{1 − \left(\frac{n_{1}}{n_{2}}\right)^2}}= \frac {h · n_{1}} {n_{2} \sqrt{1 − \left(\frac{n_{1}}{n_{2}}\right)^2}} .\]

We convert the expression inside the square root to the common denominator and adjust the fraction: \[r= \frac {h · n_{1}} {n_{2} \sqrt{\frac{n_{2}^2−n_{1}^2}{n_{2}^2}}} =\frac {h · n_{1}} { \sqrt{n_{2}^2−n_{1}^2}}.\]

We substitute \(r\) in relation (1) and raise the whole equation to the power of two: \[S=\pi r^2=\pi · \left( \frac {h · n_{1}} {\sqrt{n_{2}^2−n_{1}^2}}\right)^2= \pi · \frac {h^2 · n_{1}^2} {n_{2}^2−n_{1}^2} .\]

Numerical solution:

We know from the task assignment:

\(h=3\) \(\mathrm{m}\).

We look up the values of \(\pi\), \(n_{2}\) a \(n_{1}\) in the Tables:

Ludolph’s number			\(\pi\) \(\dot=\) \(3{.}14\)
Refractive index of air			\(n_{1}\) \(\dot=\) \(1{.}00\)
Refractive index of water			\(n_{2}\) \(\dot=\) \(1{.}33\)

We substitute for the values and obtain the size of the desired area:

\(S\) \(\dot=\) \(\frac{3{.}14 · 3^2 ·{1{.}00}^2}{{1{.}33}^2 − {1{.}00}^2}\) \(\mathrm{m^2}=\frac{3{.}14 · 9 ·1}{{1{.}33}^2 − {1}}\) \(\mathrm{m^2}\) \(\dot=\) \(36{.}8\) \(\mathrm{m^2}\)

and its radius:

\(r\) \(\dot=\) \( \frac {3 · 1{.}00} {\sqrt{1{.}33^2 −1{.}00^2}}\) \(\mathrm{m}\) \(=\) \( \frac {3} {\sqrt{1{.}33^2 − 1}}\) \(\mathrm{m}\) \(\dot=\) \(3{.}42\) \(\mathrm{m}\).