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umka21 [38]
3 years ago
8

Unpolarized light with an average intensity of 845 W/m2 moves along the x-axis when it enters a Polarizer A with a vertical tran

smission axis (along the y-axis). The transmitted light then enters a second polarizer, B at an angle in the y-z plane . The light that exits the second polarizer is found to have an average intensity of 275 W/m2. What is the orientation angle of the second polarizer (B) relative to the first one (A)
Physics
1 answer:
horsena [70]3 years ago
8 0

Answer:

θ = 36.2º

Explanation:

When light passes through a polarizer it becomes polarized and if it then passes through a second polarizer, it must comply with Malus's law

         I = I₀ cos² tea

The non-polarized light between the first polarized of this leaves half the intensity, with vertical polarization

          I₁ = I₀ / 2

          I₁ = 845/2

          I₁ = 422.5 W / m²

In this case, the incident light in the second polarizer has an intensity of I₁ = 422.5 W / m² and the light that passes through the polarizer has a value of

I = 275 W / m ²

      Cos² θ = I / I₁

      Cos θ = √ I / I₁

      Cos θ = √ (275 / 422.5)

     Cos θ = 0.80678

     θ = cos⁻¹ 0.80678

     θ = 36.2º

This is the angle between the two polarizers

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Answer:

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Explanation:

Given that,

Velocity of electron, v=5\times 10^7\ m/s

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Magnitude of electric field, E=10^4\ V/m

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From equation (1) and (2) we get:

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B=2\times 10^{-4}\ T

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<h3><u>Explanation:</u></h3>

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We need to calculate the magnitude and direction separately.

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