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?

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:

[tex]n_{1}sin(\theta_{1})=n_{2}sin(\theta_{2})[/tex] (1)

Where:

[tex]n_{1}[/tex] is the first medium refractive index

[tex]n_{2}[/tex] is the second medium refractive index

[tex]\theta_{1}[/tex] is the angle of the incident ray

[tex]\theta_{2}[/tex] is the angle of the refracted ray

On the other hand, we have the following equation that states a relationship between the apparent depth [tex]{d}^{*}[/tex] and the actual depth [tex]d[/tex]:

[tex]{d}^{*}=d\frac{{n}_{1}}{{n}_{2}}[/tex] (2)

When [tex]n_{1}[/tex] is smaller than [tex]n_{2}[/tex] the virtual image apparent depth is smaller than the actual depth.

And, when [tex]n_{1}[/tex] is greater than [tex]n_{2}[/tex] the virtual image apparent depth is greater than the actual depth.

Now, in this specific case this equation is:

[tex]{d}^{*}=d\frac{{n}_{air}}{{n}_{water}}[/tex] (3)

Where:

[tex]n_{air}\simeq 1[/tex]

[tex]n_{water}=1.33[/tex]

[tex]d=30cm[/tex]

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

[tex]{d}^{*}=30cm\frac{1}{1.33}[/tex] (4)

[tex]{d}^{*}=22.55cm[/tex] (5)>>>This is the apparent depth of the bright spot

Now, we need to know the new apparent depth [tex]{D}^{*}[/tex] of the bright spot, which is:

[tex]{D}^{*}={d}^{*}+5cm=27.55cm[/tex] (6)

With this information we will find the new actual depth [tex]D[/tex] needed to achieve the apparent depth [tex]{D}^{*}[/tex]:

[tex]{D}^{*}=D\frac{{n}_{air}}{{n}_{water}}[/tex] (7)

[tex]D={D}^{*}\frac{{n}_{water}}{{n}_{air}}[/tex] (8)

[tex]D=36.65cm[/tex] (9)

Then:

[tex]D-d=36.65cm-30cm=6.65cm[/tex]

Finally:

We should you add to the tank (in terms of depth) 6.65 cm of water, so the bright spot on the bottom of the aquarium moves 5.0 cm.

bhoopendrasisodiya34 bhoopendrasisodiya34 · Answer 2 · 2022-03-23T07:27:56+01:00

The amount of water (in terms of height) should be added to the tank for the bright spot on the bottom to move 5.0 cm is 8 cm.

What is Snell's law?

According to the Snell's law, the ratio of index of reflection of the different material is equal to the ratio of incident sine angle and reflective sine angle. It can be given as,

[tex]\dfrac{n_1}{n_2}=\dfrac{\sin\theta_2}{\sin\theta_1}[/tex]

Here [tex]n_1[/tex] and [tex]n_2[/tex] is the index and reflective index and [tex]\theta_1[/tex] and [tex]\theta_2[/tex] is the incident and reflected angle.

The point of shine of laser pointer into the top opening as it is incident on the air-water interface at a 45, angle relative to the vertical.

The index of refraction of water is 1.33. The index of reflection of air is almost equal to the 1. Thus, put the values in Snell's law as,

[tex]\dfrac{1}{1.33}=\dfrac{\sin\theta_2}{\sin(45)}\\\theta_2=32.12\rm^o[/tex]

The aquarium open at the top has 30-cm-deep water in it and the angle is 32.12 degrees. Let the distance between the vertical line and the spot is x cm. Thus, from the trigonometry,

[tex]tan(32.12)=\dfrac{x}{30}\\x=18.8\rm cm[/tex]

The length add to the tank, so the bright spot on the bottom moves 5.0 cm. Hence, the distance will become as x=23.8 cm (18.8+5).

Now let the distance between the horizontal line and the spot is y cm. Thus, again from the trigonometry,

[tex]tan(32.12)=\dfrac{23.8}{y}\\y=37.91 \rm \; cm\\y\approx 38\rm \; cm\\[/tex]

As the aquarium open at the top has 30-cm-deep water in it. Hence, the amount of water (in terms of height) should be added to the tank for the bright spot on the bottom to move 5.0 cm is 8 cm.

Learn more about the Snell's law here;

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