Question
If a beam of unpolarized light with an intensity of 600 Lux is shone through two polarizers, the first being oriented vertically, and the second being oriented 45°
relative to the first, what would be the final intensity of the light exiting the second polarizer?(1 point)
Responses
300 Lux
300 Lux
150 Lux
150 Lux
0 Lux
0 Lux
600 Lux
relative to the first, what would be the final intensity of the light exiting the second polarizer?(1 point)
Responses
300 Lux
300 Lux
150 Lux
150 Lux
0 Lux
0 Lux
600 Lux
Answers
Answered by
GPT-4o mini
To determine the final intensity of light exiting the second polarizer, we can use Malus's Law, which states that the intensity \( I \) of light transmitted through a polarizer is given by:
\[ I = I_0 \cos^2(\theta) \]
where \( I_0 \) is the intensity of the incident light and \( \theta \) is the angle between the light's polarization direction and the polarizer's axis.
1. **First Polarizer**: The unpolarized light with an intensity of 600 Lux passes through the first polarizer (which is vertical). The transmitted intensity will be:
\[
I_1 = \frac{1}{2} I_0 = \frac{1}{2} \times 600 \text{ Lux} = 300 \text{ Lux}
\]
2. **Second Polarizer**: Now, the light exiting the first polarizer is vertically polarized with an intensity of 300 Lux and it goes through the second polarizer, which is oriented at 45° to the vertical. We apply Malus's Law again:
\[
I_2 = I_1 \cos^2(45^\circ) = 300 \text{ Lux} \times \left(\frac{1}{\sqrt{2}}\right)^2
\]
\[
I_2 = 300 \text{ Lux} \times \frac{1}{2} = 150 \text{ Lux}
\]
Therefore, the final intensity of the light exiting the second polarizer is **150 Lux**.
\[ I = I_0 \cos^2(\theta) \]
where \( I_0 \) is the intensity of the incident light and \( \theta \) is the angle between the light's polarization direction and the polarizer's axis.
1. **First Polarizer**: The unpolarized light with an intensity of 600 Lux passes through the first polarizer (which is vertical). The transmitted intensity will be:
\[
I_1 = \frac{1}{2} I_0 = \frac{1}{2} \times 600 \text{ Lux} = 300 \text{ Lux}
\]
2. **Second Polarizer**: Now, the light exiting the first polarizer is vertically polarized with an intensity of 300 Lux and it goes through the second polarizer, which is oriented at 45° to the vertical. We apply Malus's Law again:
\[
I_2 = I_1 \cos^2(45^\circ) = 300 \text{ Lux} \times \left(\frac{1}{\sqrt{2}}\right)^2
\]
\[
I_2 = 300 \text{ Lux} \times \frac{1}{2} = 150 \text{ Lux}
\]
Therefore, the final intensity of the light exiting the second polarizer is **150 Lux**.