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Maurinko [17]
3 years ago
6

The refractive index of a certain glass is 1.66. For what incident angle is light reflected

Physics
1 answer:
MAXImum [283]3 years ago
8 0
If light is moving inside the glass, it will be reflected when the angle of incidence is above the critical angle of the material, which is given by
\theta_c = \arcsin ( \frac{n_a}{n_g} )
where
n_a = 1.00 is the refractive index of air
n_g = 1.66 is the refractive index of this type of glass

If we plug the numbers into the equation, we find
\theta_c = \arcsin ( \frac{1.00}{1.66} ) = \arcsin (0.602) =37.0^{\circ}
So, light will be reflected inside the glass when the angle is equal or greater than this value.
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Therefore the intensity of the two values would be

\frac{I_{27}}{I_{13}} = \frac{(13km)^2}{(27km)^2}

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PART B) Amplitude is inversely proportional to the distance

Amplitude \propto \frac{1}{distance}

\frac{A_{27}}{A_{13}}= \frac{(13km)}{(27km)}

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What component of a longitudinal sound wave is analogous to a trough of a transverse wave?
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Explanation:

There are two components of a longitudinal sound wave which are compression and rarefaction. Similarly, there are two components of the transverse wave, the crest, and trough.

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While compression is that space where the particles are close together while the rarefaction is that space where the particles are far apart from each other.

So, the refraction or the rarefied part of a longitudinal sound wave is analogous to a trough of a transverse wave.  

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In a thunderstorm, electric charge builds up on the water droplets or ice crystals in a cloud. Thus, the charge can be considere
jolli1 [7]

1) 333.6 C

In order to have breakdown, the electric field at the surface of the cloud must be equal to the breakdown electric field:

E=3.00\cdot 10^6 N/C

The electric field strength at the surface of a charged sphere is given by

E=\frac{1}{4\pi \epsilon_0} \frac{Q}{R^2}

where

\epsilon_0 = 8.85\cdot 10^{-12} C^2/(N^2 m^2) is the vacuum permittivity

Q is the charge on the sphere

R is the radius of the sphere

Here we have a cloud of radius

R = 1.00 km = 1000 m

So we can re-arrange the previous equation in order to find the charge on the cloud:

Q=4\pi \epsilon_0 ER^2=4\pi (8.85\cdot 10^{-12})(3.00\cdot 10^6)(1000)^2=333.6 C

2) 2.08\cdot 10^{21} excess electrons

The total charge of the cloud must be (in magnitude)

Q = 333.3 C

We know that one electron carries a charge of

e = 1.6 \cdot 10^{-19}C

The total charge is just given by the charge of each electron multiplied by the number of excess electrons in the cloud:

Q=Ne

where

N is the number of excess electrons

Solving for N, we find:

N=\frac{Q}{e}=\frac{333.3 C}{1.6\cdot 10^{-19} C}=2.08\cdot 10^{21}

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3 years ago
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