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4 years ago

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Some unpolarized light has an intensity of 1375 W/m2 before passing through three polarizing filters. The transmission axis of t

he first filter is vertical. The second filter's transmission axis is 31.0° from vertical. The third filter's transmission axis is 41.0° from vertical. What is the intensity of the light that emerges from the three filters?

1 answer:

chubhunter [2.5K]4 years ago

4 0

Answer:

intensity of the light that emerges from the three filters is 560.80 W/m²

Explanation:

Given data

intensity I = 1375 W/m2

angle 1 = 31.0°

angle 2 = 41.0°

to find out

intensity of the light that emerges from the three filters

solution

we know intensity of light pass 1st polarize = I/2 = 1375 / 2 = 687.5 W/m2

so intensity after 2nd polarize pass = I 1st cos²(θ)

I 2nd = 687.5 cos²(31) = 687.5 ( 0.836754) = 575.27 W/m2

and

intensity after 3rd polarize pass = I 2nd cos²(θ)

I 3rd = 575.27 cos²(41) = 575.27 (0.974839) = 560.80 W/m2

so that intensity of the light that emerges from the three filters is 560.80 W/m²

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The question is incomplete. Here is the complete question.

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Part B: How far from Car B's starting line will the cars be when Car B passes Car A? Express your answer in terms of known quantities.

Answer: Part A: $t=\frac{D_{A}}{v_{B}-v_{A}}$

Part B: $x_{B}=\frac{v_{B}D_{A}}{v_{B}-v_{A}}$

Explanation: First, let's write an equation of motion for each car.

Both cars travels with constant speed. So, they are an uniform rectilinear motion and its position equation is of the form:

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v is velocity

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Car A started the race at a distance. So at t = 0, initial position is $D_{A}$ .

The equation will be:

$x_{A}=D_{A}+v_{A}t$

Car B started at the starting line. So, its equation is

$x_{B}=v_{B}t$

Part A: When they meet, both car are at "the same position":

$D_{A}+v_{A}t=v_{B}t$

$v_{B}t-v_{A}t=D_{A}$

$t(v_{B}-v_{A})=D_{A}$

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Car B meet with Car A after $t=\frac{D_{A}}{v_{B}-v_{A}}$ units of time.

Part B: With the meeting time, we can determine the position they will be:

$x_{B}=v_{B}(\frac{D_{A}}{v_{B}-v_{A}} )$

$x_{B}=\frac{v_{B}D_{A}}{v_{B}-v_{A}}$

Since Car B started at the starting line, the distance Car B will be when it passes Car A is $x_{B}=\frac{v_{B}D_{A}}{v_{B}-v_{A}}$ units of distance.

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