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.part ahow much water (in terms of height) should you add to the tank so the bright spot on the bottom moves 5.0 cm?

This situation is shown in the image attached and is known as Refraction.  

Refraction is a phenomenon in which a wave (the light in this case) bends or changes it direction when passing through a medium with a refractive index different from the other medium.  

In this context, the Refractive index is a number that describes how fast light propagates through a medium or material.  

According to Snell’s Law:  

(1)  

Where:  

is the first medium refractive index  

is the second medium refractive index  

is the angle of the incident ray  

is the angle of the refracted ray

On the other hand, we have the following equation that states a relationship between the apparent depth and the actual depth :  

(2)  

When is smaller than the virtual image apparent depth is smaller than the actual depth.  

And, when is greater than the virtual image apparent depth is greater than the actual depth.

Now, in this specific case this equation is:  

(3)  

Where:  

Knowing this, let’s find the apparent depth of the bright spot from (3):  

(4)  

(5)>>>This is the apparent depth of the bright spot  

Now, we need to know the new apparent depth of the bright spot, which is:  

(6)

With this information we will find the new actual depth needed to achieve the apparent depth :  

(7)  

(8)  

(9)  

Then:  

Finally:

answer: 0.20

explanation: this can only be the answew because all the other measuremnets are not close

hèllo, sorry i really need some more points so if you need this qustion answered than retype it