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2 years ago

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When a flat slab of transparent material is placed under water, the critical angle for light traveling from the slab into water

is found to be 60°. What will the critical angle be if the slab is surrounded by air? Take the index of refraction for water to be 1.33. d) 45.6 c) 44.2 e) 47.3 a) 40.6 b) 42.5

1 answer:

Novosadov [1.4K]2 years ago

3 0

Answer:

(a) 40.6 degree

Explanation:

When refraction takes place from slab to water, the critical angle is 60 degree.

Use Snell's law

refractive index of water with respect to slab

$\mu _{w}^{s}=\frac{Sin60}{Sin90}$

$\frac{\mu _{w}}{\mu _{s}}=0.866$

$\frac{1.33}{\mu _{s}}=0.866$

μs = 1.536

Now for slab air interface, the critical angle is C.

$\mu _{a}^{s}=\frac{SinC}{Sin90}$

$\frac{\mu _{a}}{\mu _{s}}=\frac{SinC}{Sin90}$

1 / 1.536 = Sin C

C = 40.6 degree

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Explanation:

<h3>a) Mass of the continent</h3>

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$m=2.82(10)^{21} kg$ (4) This is the mass of the continent

<h3>b) Kinetic energy of the continent</h3>

Kinetic energy $K$ is given by the following equation:

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Where:

$m=2.82(10)^{21} kg$ is the mass of the continent

$v=4.8 \frac{cm}{year} \frac{1 m}{100 cm} \frac{1 year}{365 days} \frac{1 day}{24 hours} \frac{1 hour}{3600 s}=1(10)^{-9} m/s$ is the velocity of the continent

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Explanation:

Refraction is a phenomenon that occurs when a light rays crosses the boundary between two different mediums.

When this occurs, the light wave changes speed and also direction, according to Snell's law:

$n_1 sin \theta_1 = n_2 sin \theta_2$ (1)

where

$n_1$ is the index of refraction of the first medium

$n_2$ is the index of refraction of the second 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)

The index of refraction is the ratio between the speed of light in a vacuum (c) and the speed of light in the medium (v):

$n=\frac{c}{v}$

Using this definition, we can rewrite eq.(1) as

$\frac{\sin \theta_2}{\sin \theta_1} = \frac{v_2}{v_1}$

Where $v_2$ is the speed of light in the 2nd medium and $v_1$ the speed of light in the 1st medium.

Now let's analyze the situation represented in the figure: we see that as the light ray enters the 2nd medium, it bends towards the normal, this means that the angle of refraction is smaller than the angle of incidence:

$\theta_2 < \theta_1$

This means that

$\frac{\sin \theta_2}{\sin \theta_1}$

And therefore,

$\frac{v_2}{v_1}$

So, the speed of light in the second medium is smaller than the speed of light in the first medium: this occurs only in option C), which is therefore the correct choice.

Learn more about refraction:

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