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4 years ago

A. How long does it take light to travel through a 3.0-mm-thick piece of window glass?

Physics

1 answer:

hodyreva [135]4 years ago

6 0

Answer:

a) $1.517\times10^{-11}$ s

b) 3.41 mm

Explanation:

We take the speed of light, c = $3.0\times10^8$ m/s and the refractive index of glass as 1.517.

Speed = distance/time

Time = distance/speed

Refractive index, n = speed of light in vacuum / speed of light in medium

$n=\dfrac{c}{s}$

$s=\dfrac{c}{n}$

$t=\dfrac{d}{c/n}$

$t=\dfrac{dn}{c}$

$t=\dfrac{3\times10^{-3}\times1.517}{3.0\times10^8}$

$t=1.517\times10^{-11}$

We take the refractive index of water as 1.333.

Speed in water = speed in vacuum / refractive index of water

Distance = speed * time

$d=s\times t$

$d=\dfrac{c}{n_w}\times \dfrac{3\times10^{-3}\times1.517}{c}$

$d=\dfrac{3\times10^{-3}\times 1.517}{1.333}$

d = 3.41 mm

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The options are not shown, so it is kinda hard to give you the exact answer.

But i can try to give a general explanation on the topic.

Interference happens when two waves interact between them. We usually use waves with the same wavelength and same phase.

There are two "main" types of interference:

Destructive interference: Happens when the peak of one wave and the trough of the other wave happen in the same place, then when we add these waves, we have a subtraction, and then the amplitude of the resultant wave is smaller than the amplitudes of the original waves (if both initial waves have the same amplitude, then the amplitude of the resultant wave in that position is zero).

Constructive interference: Similar to the previous case, but now the two waves have the peaks or toughs in the same position, then when we add the waves, we have an addition in the amplitude, then the amplitude of the resultant wave will be larger than the amplitudes of the initial waves.

Then if you are looking for constructive interference, you need to look for an option where the peaks of the waves coincide. For example, in the top image we can see, at the top, a wave with an amplitude of 2 units.

And at the bottom a wave with an amplitude of 1 unit.

We also can see that the peaks (and the tough) are in the same place, then we will have constructive interference. Then the amplitude of the resultant wave will be equal to the sum of the amplitudes of the initial waves (2 and 1)

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3 years ago

Two point charges are 10.0cm apart and have charges of 2.0uC and -2.0uC, respectively. What is the magnitude of the electric fie

elena-14-01-66 [18.8K]

The electric field generated by a point charge is given by:
$E= k_e \frac{Q}{r^2}$
where
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Q is the charge
r is the distance from the charge

We want to know the net electric field at the midpoint between the two charges, so at a distance of r=5.0 cm=0.05 m from each of them.

Let's calculate first the electric field generated by the positive charge at that point:
$E_1=k_e \frac{Q_1}{r^2}=(8.99 \cdot 10^9 Nm^2C^{-2}) \frac{(2.0 \cdot 10^{-6} C)}{(0.05 m)^2} =+7.19 \cdot 10^6 N/C$
where the positive sign means its direction is away from the charge.

while the electric field generated by the negative charge is:
$E_2=k_e \frac{Q_1}{r^2}=(8.99 \cdot 10^9 Nm^2C^{-2}) \frac{(-2.0 \cdot 10^{-6} C)}{(0.05 m)^2} =-7.19 \cdot 10^6 N/C$
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If we assume that the positive charge is on the left and the negative charge is on the right, we see that E1 is directed to the right, and E2 is directed to the right as well. This means that the net electric field at the midpoint between the two charges is just the sum of the two fields:
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From the equation we can say that,

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