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Alona [7]
3 years ago
8

The top of a swimming pool is at ground level. If the pool is 3.00 m deep, how far below ground level does the bottom of the poo

l appear to be located for the following conditions? (The index of refraction of water is 1.333.)
(a) The pool is completely filled with water.
______m below ground level

(b) The pool is filled halfway with water.
______m below ground level
Physics
1 answer:
Yuri [45]3 years ago
4 0

Answer:

a) 2.25 m

b) 2.625 m

Explanation:

Refraction is the name given to the phenomenon of the speed of light changing the the boundary when it moves from one physical medium to the other.

Refractive index is the ratio of the speed of light in empty vacuum (air is an appropriate substitution) to the speed of light in the medium under consideration.

In terms of real and apparent depth, the refractive index is given by

η = (real depth)/(apparent depth)

a) Real depth = 3.00 m

Apparent depth = ?

Refractive index, η = 1.333

1.333 = 3/(apparent depth)

Apparent depth = 3/1.3333 = 2.25 m.

Hence the bottom of the pool appears to be 2.25 m below the ground level.

b) Real depth = 1.5 m

Apparent depth = ?

Refractive index, η = 1.333

1.3333 = 1.5/(apparent depth)

Apparent depth = 1.5/1.3333 = 1.125 m

But the pool is half filled with water, there is a 1.5 m depth on top of the pool before refraction starts.

So, apparent depth of the pool = 1.5 + 1.125 = 2.625 m below the ground level

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Answer:

5 hours

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<em>D=S×T </em>

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~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

A second method:

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A ray of light passes from air into a block of clear plastic. How does the angle of incidence in the air compare to the angle of
andre [41]

Answer:

The angle of incidence is greater than the angle of refraction

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When a ray of light is refracted, it changes speed and direction, according to Snell's Law:

n_1 sin \theta_1 = n_2 sin \theta_2

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n_2 is the index of refraction of the 2nd medium

\theta_1 is the angle of incidence (the angle between the incident ray and the normal to the boundary)

\theta_2 is the angle of refraction (the angle between the refracted ray and the normal to the boundary)

In this problem, we have a ray of light passing from air into clear plastic. We have:

n_1=1.00 (index of refraction of air)

n_2=1.50 approx. (index of refraction in clear plastic)

Snell's Law can be rewritten as

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\frac{n_1}{n_2}

And so

\theta_2

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