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marishachu [46]
3 years ago
15

An example of when total internal reflection occurs is when all the light passing from a region of higher index of refraction to

a region of lower index of refraction
passes through the region of lower index without being affected

passes through the region of lower index but is dispersed

is reflected back into the region of higher index

slows down so that objects appear bent
Physics
1 answer:
Amiraneli [1.4K]3 years ago
8 0

Answer:

is reflected back into the region of higher index

Explanation:

Total internal reflection is a phenomenon that occurs when all the light passing from a region of higher index of refraction to a region of lower index is reflected back into the region of higher index.

According to Snell's law, refraction of ligth is described by the equation

n_1 sin \theta_1 = n_2 sin \theta_2

where

n1 is the refractive index of the first medium

n2 is the refractive index of the second medium

\theta_1 is the angle of incidence (in the first medium)

\theta_2 is the angle of refraction (in the second medium)

Let's now consider a situation in which

n_1 > n_2

so light is moving from a medium with higher index to a medium with lower index. We can re-write the equation as

sin \theta_2 = \frac{n_1}{n_2}sin \theta_1

Where \frac{n_1}{n_2} is a number greater than 1. This means that above a certain value of the angle of incidence \theta_1, the term on the right can become greater than 1. So this would mean

sin \theta_2 > 1

But this is not possible (the sine cannot be larger than 1), so no refraction occurs in this case, and all the light is reflected back into the initial medium (total internal reflection). The value of the angle of incidence above which this phenomen occurs is called critical angle, and it is given by

\theta_c =sin^{-1}(\frac{n_2}{n_1})

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Answer:

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Answer:

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scoundrel [369]

Answer:

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3 years ago
Two identical trucks have mass 5100 kg when empty, and the maximum permissible load for each is 8000 kg. the first truck, carryi
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<span>The 2nd truck was overloaded with a load of 16833 kg instead of the permissible load of 8000 kg. The key here is the conservation of momentum. For the first truck, the momentum is 0(5100 + 4300) The second truck has a starting momentum of 60(5100 + x) And finally, after the collision, the momentum of the whole system is 42(5100 + 4300 + 5100 + x) So let's set the equations for before and after the collision equal to each other. 0(5100 + 4300) + 60(5100 + x) = 42(5100 + 4300 + 5100 + x) And solve for x, first by adding the constant terms 0(5100 + 4300) + 60(5100 + x) = 42(14500 + x) Getting rid of the zero term 60(5100 + x) = 42(14500 + x) Distribute the 60 and the 42. 60*5100 + 60x = 42*14500 + 42x 306000 + 60x = 609000 + 42x Subtract 42x from both sides 306000 + 18x = 609000 Subtract 306000 from both sides 18x = 303000 And divide both sides by 18 x = 16833.33 So we have the 2nd truck with a load of 16833.33 kg, which is well over it's maximum permissible load of 8000 kg. Let's verify the results by plugging that mass into the before and after collision momentums. 60(5100 + 16833.33) = 60(21933.33) = 1316000 42(5100 + 4300 + 5100 + 16833.33) = 42(31333.33) = 1316000 They match. The 2nd truck was definitely over loaded.</span>
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3 years ago
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