Question

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?

Answer 1

Answer:

Height of water should be 10 cm.

Answer 2

Answer:

The answer is 8cm

Explanation:

We will use Snell Refraction law to solve this exercise.

First we will need to note that in both scenarios we are using an air-water interface, so the refraction angle doesnt change

Second we need to stablish some parameters, such as the air index of refraction which will be 1. We already know the index of refraction of the water is 1. 33

Let´s get to the action!

According to the Snell Refracton law we have the next equation:

(n1)(sin(theta1))=(n2)(sin(theta2))

Where n1 is the index of refraction of the air

And n2 is the index of refraction of the water

theta1 is the incidence angle relative to the vertical

theta2 is the refraction angle relative to the vertical

Replacing we will find the refraction angle this way:

$(1)(sin(45))=(1.33)(sin(theta2))$

$(1/1.33)(sen(45))=sin(theta2)$

Finally we get that theta2=32.12°  

With this angle we will trace a triangle, where the vertical line will be 30cm (or 0.3m), with this triangle we will find the distance between the vertical and the spot where the beam hits the bottom, this way:

$tan(theta2)=(X/0.3m)$e

We get that X is 0.19m (or 19cm).  

After it, we will increase X by adding 5cm to it, it means the new X will be 19cm+5cm, 24cm (or 0.24m)

As we know how much the new X is and we also know what´s the refraction angle (theta2) all we do is find how much the new height is, this way:

$tan(theta2)=(0.24m/Y)$

We get that the new Y is 0.38m (or 38cm)

It means, the amount of water you need to add to the tank is 8cm