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alina1380 [7]
3 years ago
15

Vertically polarized light with an intensity of 36.8 lux passes through a polarizer whose transmission axis is an angle of 51.0°

with the vertical. What is the intensity and direction of transmitted light? If the second polarizer whose transmission axis is an angle of 88.0° with the vertical is placed after first polarizer, what is the intensity of transmitted light?
Physics
1 answer:
Annette [7]3 years ago
5 0

Answer:

0.01774 lux

Explanation:

I_0 = Polarized light intensity = 36.8 lux

\theta = Angle

Through first filter

I_1=I_0cos^2\theta\\\Rightarrow I_1=36.8cos^2{51}\\\Rightarrow I_1=14.57\ lux

Through second filter

I_2=I_1cos^2\theta\\\Rightarrow I_2=14.57cos^2{88}\\\Rightarrow I_2=0.01774\ lux

intensity of transmitted light is 0.01774 lux

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The ball has height <em>y</em> and velocity <em>v</em> at time <em>t</em> according to

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and

<em>v</em> = <em>v</em>₀ - <em>g t</em>

where <em>v</em>₀ is its initial speed and <em>g</em> = 9.80 m/s² is the magnitude of the acceleration due to gravity.

The ball is falling with a velocity of 7.94 m/s when it's 2.72 m below the release point, which at time <em>t </em>such that

-2.72 m = <em>v</em>₀ <em>t</em> - 1/2 <em>g</em> <em>t</em> ²

-7.94 m/s = <em>v</em>₀ - <em>g t</em>

Solve for <em>t</em> in the second equation:

<em>t </em>= (<em>v</em>₀ + 7.94 m/s)/<em>g</em>

Substitute this into the first equation and solve for <em>v</em>₀ :

-2.72 m = <em>v</em>₀ (<em>v</em>₀ + 7.94 m/s) /<em>g</em> - 1/2 <em>g</em> ((<em>v</em>₀ + 7.94 m/s)/<em>g</em>)²

-2.72 m = <em>v</em>₀²/<em>g</em> + (7.94 m/s) <em>v</em>₀/<em>g</em> - 1/2 (<em>v</em>₀ + 7.94 m/s)²/<em>g</em>

2 (-2.72 m) <em>g</em> = 2<em>v</em>₀² + 2 (7.94 m/s) <em>v</em>₀ - (<em>v</em>₀ + 7.94 m/s)²

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-53.3 m²/s² = <em>v</em>₀² - 63.0 m²/s²

<em>v</em>₀² = 9.73 m²/s²

<em>v</em>₀ = 3.12 m/s

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This means that the gravity force decreases when the distance between these two bodies increases.

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Substituting this distance in (1):

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