An ideal polarizer with its transmission axis rotated 30° to the vertical is placed in a beam of unpolarized light of intensity 10 W/m². After passing through the polarizer, what is intensity of the beam? a) 10 W/m² b) 5.0 W/m² c) 8.7 W/m² d) 7.5 W/m²
An ideal polarizer with its transmission axis rotated 30° to the vertical is placed in a beam of unpolarized light of intensity 10 W/m². After passing through the polarizer, what is intensity of the beam? a) 10 W/m² b) 5.0 W/m² c) 8.7 W/m² d) 7.5 W/m²
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![**Polarization and Intensity of Light**
**Question:**
An ideal polarizer with its transmission axis rotated 30° to the vertical is placed in a beam of unpolarized light of intensity 10 W/m². After passing through the polarizer, what is the intensity of the beam?
a) 10 W/m²
b) 5.0 W/m²
c) 8.7 W/m²
d) 7.5 W/m²
**Explanation:**
When unpolarized light passes through a polarizer, the intensity of the light is reduced. The intensity \(I\) of the polarized light after passing through an ideal polarizer can be calculated using Malus's Law, which is given by:
\[ I = I_0 \cdot \cos^2(\theta) \]
where:
- \(I_0\) is the initial intensity of the unpolarized light,
- \(\theta\) is the angle between the light's initial polarization direction and the axis of the polarizer.
For unpolarized light, the initial intensity is reduced by half after passing through the first polarizer:
\[ I' = \frac{I_0}{2} \]
Here, \(I_0 = 10 \, \text{W/m}^2\) and \(\theta = 30^\circ\):
\[ I' = \frac{10 \, \text{W/m}^2}{2} = 5 \, \text{W/m}^2 \]
After the polarizer at 30°, the intensity is:
\[ I = 5 \, \text{W/m}^2 \cdot \cos^2(30^\circ) \]
\[ \cos(30^\circ) = \sqrt{3}/2 \]
\[ I = 5 \, \text{W/m}^2 \cdot \left(\frac{\sqrt{3}}{2}\right)^2 \]
\[ I = 5 \, \text{W/m}^2 \cdot \frac{3}{4} \]
\[ I = \frac{15}{4} \, \text{W/m}^2 \]
\[ I = 3.75 \, \text{W/m}^2 \]
Since none of the options directly match this result, further calculations should be double-checked,](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fd9a2776d-a528-4478-be0e-235d689690c2%2Fb63ac85a-c2ff-41ae-bf62-6d74253c69a5%2Fzku015_processed.jpeg&w=3840&q=75)
Transcribed Image Text:**Polarization and Intensity of Light**
**Question:**
An ideal polarizer with its transmission axis rotated 30° to the vertical is placed in a beam of unpolarized light of intensity 10 W/m². After passing through the polarizer, what is the intensity of the beam?
a) 10 W/m²
b) 5.0 W/m²
c) 8.7 W/m²
d) 7.5 W/m²
**Explanation:**
When unpolarized light passes through a polarizer, the intensity of the light is reduced. The intensity \(I\) of the polarized light after passing through an ideal polarizer can be calculated using Malus's Law, which is given by:
\[ I = I_0 \cdot \cos^2(\theta) \]
where:
- \(I_0\) is the initial intensity of the unpolarized light,
- \(\theta\) is the angle between the light's initial polarization direction and the axis of the polarizer.
For unpolarized light, the initial intensity is reduced by half after passing through the first polarizer:
\[ I' = \frac{I_0}{2} \]
Here, \(I_0 = 10 \, \text{W/m}^2\) and \(\theta = 30^\circ\):
\[ I' = \frac{10 \, \text{W/m}^2}{2} = 5 \, \text{W/m}^2 \]
After the polarizer at 30°, the intensity is:
\[ I = 5 \, \text{W/m}^2 \cdot \cos^2(30^\circ) \]
\[ \cos(30^\circ) = \sqrt{3}/2 \]
\[ I = 5 \, \text{W/m}^2 \cdot \left(\frac{\sqrt{3}}{2}\right)^2 \]
\[ I = 5 \, \text{W/m}^2 \cdot \frac{3}{4} \]
\[ I = \frac{15}{4} \, \text{W/m}^2 \]
\[ I = 3.75 \, \text{W/m}^2 \]
Since none of the options directly match this result, further calculations should be double-checked,
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