The intensity of light after passing through a polarizer is given by Malus's law, which states that I = I0 * cos²θ, where I0 is the initial intensity, I is the final intensity, and θ is the angle between the light's initial polarization direction and the axis of the polarizer.

In this case, the light passes through two polarizers. The first polarizer is at an angle θ to the initial polarization of the light, and the second polarizer is at an angle of 90° to the initial polarization of the light.

After passing through the first polarizer, the intensity of the light is I1 = I0 * cos²θ.

Then, the light passes through the second polarizer, which is at an angle of 90° to the initial polarization of the light. Since the light's polarization direction was rotated by θ by the first polarizer, the angle between the light's polarization direction and the axis of the second polarizer is 90° - θ.

Therefore, after passing through the second polarizer, the intensity of the light is I = I1 * cos²(90° - θ) = I0 * cos²θ * sin²θ.

Given that the final intensity of the light is 0.093I0, we have:

0.093I0 = I0 * cos²θ * sin²θ.

Dividing both sides by I0 gives:

0.093 = cos²θ * sin²θ.

This is an equation in the variable θ. To solve for θ, we can use the double-angle identity sin²θ = 1/2 - 1/2cos(2θ), which gives:

0.093 = cos²θ * (1/2 - 1/2cos(2θ)).

This is a quadratic equation in cos²θ. Solving this equation for cos²θ gives two solutions, one of which is negative and therefore not physically meaningful. The positive solution gives the value of cos²θ, and taking the square root and then the inverse cosine gives the value of θ.

Solving this equation numerically gives θ ≈ 38.2132°.

Question

The intensity of light after passing through a polarizer is given by Malus's law, which states that I = I0 * cos²θ, where I0 is the initial intensity, I is the final intensity, and θ is the angle between the light's initial polarization direction and the axis of the polarizer.

In this case, the light passes through two polarizers. The first polarizer is at an angle θ to the initial polarization of the light, and the second polarizer is at an angle of 90° to the initial polarization of the light.

After passing through the first polarizer, the intensity of the light is I1 = I0 * cos²θ.

Then, the light passes through the second polarizer, which is at an angle of 90° to the initial polarization of the light. Since the light's polarization direction was rotated by θ by the first polarizer, the angle between the light's polarization direction and the axis of the second polarizer is 90° - θ.

Therefore, after passing through the second polarizer, the intensity of the light is I = I1 * cos²(90° - θ) = I0 * cos²θ * sin²θ.

Given that the final intensity of the light is 0.093I0, we have:

0.093I0 = I0 * cos²θ * sin²θ.

Dividing both sides by I0 gives:

0.093 = cos²θ * sin²θ.

This is an equation in the variable θ. To solve for θ, we can use the double-angle identity sin²θ = 1/2 - 1/2cos(2θ), which gives:

0.093 = cos²θ * (1/2 - 1/2cos(2θ)).

This is a quadratic equation in cos²θ. Solving this equation for cos²θ gives two solutions, one of which is negative and therefore not physically meaningful. The positive solution gives the value of cos²θ, and taking the square root and then the inverse cosine gives the value of θ.

Solving this equation numerically gives θ ≈ 38.2132°.

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