An aquarium open at the top has 30-cm-deep water in it. You shine a laser pointer into the top opening so it is incident on the air-water interface at a 45∘ angle relative to the vertical. You see a bright spot where the beam hits the bottom of the aquarium. The index of refraction of water is 1.33. How much water (in terms of height) should you add to the tank so the bright spot on the bottom moves 5.0 cm? Express your answer with the appropriate units.

This law relates the index of refraction of the water (n2), the index of refraction of the air (n1), the incidence angle relative to the vertical (theta1) and the refraction angle relative to the vertical (theta2) by using the next equation:

[tex](n1)*(sin(theta1))=(n2)*(sin(theta2))[/tex]

Then we will find the refraction angle relative to the vertical this way:

[tex](n1/n2)*(sin(theta1))=sin(theta2)[/tex]

[tex](1/1.33)*(sin(45))=sin(theta2)[/tex]

Then, theta2=32.12°

Now that we have this information we can imagine a triangle with a 30cm height and a 32.12° angle. This way we can find how much X is, this X will be the distance between the vertical line and the spot the beam hits the bottom, so we can use some trigonometry to find it, this way:

[tex]tan(32.12)=(X/30cm)[/tex]

[tex]X=(tan(32.12))*(30cm)[/tex]

Then, X=18.8cm, we can approximate it to 19cm

Once we have X we will add 5cm to it which is how much the beam needs to be moved, then the new X will be 24cm

Now, with the new horizontal distance we will find the new vertical distance, let´s call it Y, this way we will know how much water we must add to move the beam, then we will have a triangle with a vertical distance called Y, the same 32.12° angle will be used as we are still working with the air-water interface and a 19cm horizontal distance, then:

[tex]tan(32.12)=(24cm/Y)[/tex]

[tex]Y=(24cm/tan(32.12))[/tex]

Then, Y=38cm

In this case, you must add 8cm of water to the tank to move the beam on the bottom 5cm