Physics, 30.03.2020 16:37 darius12318

Unpolarized light of intensity I0=950 W/m2 is incident upon three polarizers. The axis of the first polarizer is vertical. The axis of the second polarzer is rotated at an angle of 55 degrees from the vertical. The axis of the third polarizer is horizontal.

Write an equation for the intensity of light after it has passed through all three polarizers in terms of the intensity of unpolarized light entering the first polarizer I0 and the angle of the second polarizer relative to the first, given that the first and third polarizers are crossed (90° between them). Use trigonometric identities to simplify and give the results in terms of a single trigonometric function of φ = 2θ.

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Answer from: hey840

The equation for the intensity of light after it has passed through all three polarizers is:

$I_{3}=\frac{1}{8}I_{0}sin^{2}(\phi)$

Explanation:

When unpolarized light goes through the first polarizer, whose axis of transmission is vertical, the intensity is cut by half:

$I_{1}=\frac{1}{2} I_{0}$

When the light passes through the second polarizer, whose axis of transmission is rotated an angle $\theta=55^{o}$ from the axis of transmission of the first polarizer, using Malus's law we have that:

$I_{2}=I_{1}cos^{2}(\theta)=\frac{1}{2}I_{0}cos^{2}(\theta)$

When the light passes through the third polarizer, whose axis of transmission is horizontal, using once again Malus's law we have that:

$I_{3}=I_{2}cos^{2}(\frac{\pi}{2}-\theta)=\frac{1}{2}I_{0}cos^{2}(\theta)cos^{2}(\frac{\pi}{2}-\theta)$

We use the next trigonometric identities:

$cos^{2}(\frac{\pi}{2}-\theta)=sin^{2}(\theta)$ $sin^{2}(2\theta)=4sin^{2}(\theta)cos^{2}(\theta)$

Manipulating the last equation we get that:

$I_{3}=\frac{1}{8}I_{0}sin^{2}(2\theta)$

Replacing $\phi=2\theta$ we have that:

$I_{3}=\frac{1}{8}I_{0}sin^{2}(\phi)$

Answer from: Quest

i think its

Answer from: Quest

most voltage drop calculators want the distance between the source and destination. then internally, they double it so both the source and return wires are calculated. but you need to know which distance the calculator uses so you're not half of what you should be or double what you should be.

most circuit breakers will not take a #6 directly, so the electrician will have to splice on a smaller pigtail. you also have to upsize the grounding wire when you upsized the ungrounded (hot) wire (see note in 250.122). this will preclude using most cable assemblies, as a cable with #6 copper will most likely only have a #10 grounding conductor. you need a #6 grounding conductor if the circuit is 20a. a #8 cable is no better, as it will have a #10 grounding conductor too. but most 20a breakers will take a #8 directly.

in your related post, you didn't say what type of load this is. if it is a light or heater, voltage drop won't be a big issue. if this is a motor, the current needs are huge at startup and the voltage drop can pose a problem. if this is a motor, knowing if the motor is under much load during startup will answer the voltage drop effect (a saw is easier to start than a compressor).

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Unpolarized light of intensity I0=950 W/m2 is incident upon three polarizers. The axis of the first...