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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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2 years ago
Two spherical shells have a common center. A -2.1 10-6 C charge is spread uniformly over the inner shell, which has a radius of
julsineya [31]

Answer:

a) E_total = 6,525 10⁴ N /C ,field direction is radial outgoing

b)  E_total = 1.89 10⁶ N / C, field is incoming radial

c) E_total = 0

Explanetion:

For this exercise we can use that the charge in a spherical shell can be considered concentrated at its center and that the electric field inside the shell is zero, since Gauss's law is

                Ф = E .dA = q_{int} /ε₀

inside the spherical shell there are no charges

The electric field is a vector quantity, so we calculate the field created by each shell and add it vectorly.

We have two sphere shells with radii 0.050m and 0.15m respectively

a) point where you want to know the electric field d = 0.20 m

shell 1

the point is on the outside,d>ro,  therefore we can consider the charge to be concentrated in the center

            E₁ = k q₁ / d²

             

shell 2

the point is on the outside,d>ro

             E₂ = k q₂ / d²

the total camp is

              E_total = -E₁ + E₂

              E_total = k ( \frac{-q_1 + q_2}{d^2})

              E_total = 9 10⁹ (-2.1 10⁻⁶+ 5 10⁻⁶ / .2²

              E_total = 6,525 10⁵ N /C

The field direction is radial and outgoing ti the shells

b) the calculation point is d = 0.10m

shell 1

point outside the shell d> ro

                 E₁ = k q₁ / d²

shell 2

the point is inside the shell d <ro

Therefore, according to Gauss's law, since there are no charges in the interior, the electrioc field is zero

                E₂ = 0

               

                 E_total = E₁

                 E_total = k q₁ / d²

                 E_total = 9 10⁹ 2.1 10⁻⁶ / 0.1²

                 E_total = 1.89 10⁶ N / A

As the charge is negative, this field is incoming radial, that is, it is directed towards the shell 1

c) the point of interest d = 0.025 m

shell 1

point  is inside the shell d< ro

                 

as there are no charges inside

                     E₁ = 0

shell 2

point is inside the radius of the shell d <ro

                    E₂ = 0

the total field is

                    E_total = 0

3 0
3 years ago
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klasskru [66]
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3 0
3 years ago
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Elena-2011 [213]

Answer:

B ig lol

Explanation:

6 0
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kap26 [50]

Answer:

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Explanation:

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As Lithium is the left most element in this period. It has the lowest electronegativity value which is equal to 0.98.

6 0
3 years ago
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