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zubka84 [21]
3 years ago
12

Unpolarized light is incident on a series of polarizing disks. We have three polarizers A, B and C have transmission axes that m

ake angles of 30, 90 and 180 degrees, respectively, measured CW from the vertical y-axis. You want to maximize the relative transmission and can place them in any order. What is the maximum possible relative transmission (is. Final/I0) passing through a series of all three polarizers
Physics
1 answer:
ikadub [295]3 years ago
3 0

Answer:

Explanation:

Formula for intensity of light after transmission for unpolarised light is

I₀ / 2 , We shall take up 180 degree disk first .

For second transmission , the formula is

I = I₀ Cos²θ , θ is angle between axis of polarizer and axis of vibration.

If we take up 30 degree polariser

I = I₀ / 2  Cos²( 180 - 30 )

= .375 I₀

For third transmission

I_ final = .375 I₀  x cos( 90 - 30 )

= .1875  I₀

Final/I₀ = .1875  

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What is radioactive dating? How is it used to determine age of something?
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3 years ago
At a picnic, there is a contest in which hoses are used to shoot water at a beach ball from three different directions. As a res
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Answer:

F₃ = 122.88 N

θ₃ = 20.63°

Explanation:

First we find the components of F₁:

For x-component:

F₁ₓ = F₁ Cos θ₁

F₁ₓ = (50 N) Cos 60°

F₁ₓ = 25 N

For y-component:

F₁y = F₁ Sin θ₁

F₁y = (50 N) Sin 60°

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Now, for F₂. As, F₂ acts along x-axis. Therefore, its y-component will be zero and its x-xomponent will be equal to the magnitude of force itself:

F₂ₓ = F₂ = 90 N

F₂y = 0 N

Now, for the resultant force on ball to be zero, the sum of x-components of the forces and the sum of the y-component of the forces must also be equal to zero:

F₁ₓ + F₂ₓ + F₃ₓ = 0 N

25 N + 90 N + F₃ₓ = 0 N

F₃ₓ = - 115 N

for y-components:

F₁y + F₂y + F₃y = 0 N

43.3 N + 0 N + F₃y = 0 N

F₃y = - 43.3 N

Now, the magnitude of F₃ can be found as:

F₃ = √F₃ₓ² + F₃y²

F₃ = √[(- 115 N)² + (- 43.3 N)²]

<u>F₃ = 122.88 N</u>

and the direction is given as:

θ₃ = tan⁻¹(F₃y/F₃ₓ) = tan⁻¹(-43.3 N/-115 N)

<u>θ₃ = 20.63°</u>

7 0
3 years ago
Suppose that on earth you can throw a ball vertically upward a distance of 1.20 m. Given that the acceleration of gravity on the
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Answer:

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Accelration going up is considered as negetive

Initial Velocity of the ball

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Assuming that the ball is thrown with the same velocity on the Moon, displacement of the ball is

v^2-u^2=2as\\\Rightarrow s=\frac{v^2-u^2}{2a}\\\Rightarrow s=\frac{0^2-4.85^2}{2\times -1.67}\\\Rightarrow s=7.04\ m

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