| 模型 | 模型答案 | 正确答案 |
|---|---|---|
| 金声玉亮2.0 | Let's denote the intensity of the unpolarized light as $I_0$. When unpolarized light passes through a polarizer, it becomes polarized, and its intensity is halved (since only half of the light's components will align with the polarizer's axis). Thus, the intensity of the light after the first polarizer is $I_1 = \frac{1}{2}I_0$. The light then passes through a second polarizer at an angle of 30 degrees to the first. According to Malus' law, the intensity of the light after this second polarizer is: $$I_2 = I_1 \cdot \cos^2(30^\circ) = \frac{1}{2}I_0 \cdot \left(\frac{\sqrt{3}}{2}\right)^2 = \frac{1}{2}I_0 \cdot \frac{3}{4} = \frac{3}{8}I_0$$ Finally, the light passes through a third polarizer at an angle of 50 degrees to the second. The intensity after the third polarizer is: $$I_3 = I_2 \cdot \cos^2(50^\circ) = \frac{3}{8}I_0 \cdot \cos^2(50^\circ)$$ To find the percentage of the original intensity that comes out of the third polarizer, we divide $I_3$ by $I_0$ and multiply by 100%: $$\text{Percentage} = \left(\frac{I_3}{I_0}\right) \times 100\% = \left(\frac{3}{8} \cdot \cos^2(50^\circ)\right) \times 100\%$$ Using a calculator to find $\cos^2(50^\circ)$: $$\cos^2(50^\circ) \approx 0.3193$$ Thus, $$\text{Percentage} = \left(\frac{3}{8} \cdot 0.3193\right) \times 100\% \approx 12.04\%$$ So, approximately 12.04% of the original intensity of the unpolarized light comes out of the third polarizer. |
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| deepseekr1 | The light passes through three polarizers. Here's the step-by-step solution: 1. **First Polarizer (Unpolarized to Polarized):** Unpolarized light loses half its intensity. \( I_1 = \frac{I_0}{2} \). 2. **Second Polarizer (30° from the first):** Apply Malus' Law: \( I_2 = I_1 \cdot \cos^2(30^\circ) = \frac{I_0}{2} \cdot \left(\frac{\sqrt{3}}{2}\right)^2 = \frac{3I_0}{8} \). 3. **Third Polarizer (50° from the second):** Apply Malus' Law again: \( I_3 = I_2 \cdot \cos^2(50^\circ) = \frac{3I_0}{8} \cdot \cos^2(50^\circ) \). Using \( \cos(50^\circ) \approx 0.6428 \), \( \cos^2(50^\circ) \approx 0.4132 \). \( I_3 \approx \frac{3}{8} \cdot 0.4132 \cdot I_0 \approx 0.1549I_0 \). **Final Intensity Percentage:** \( 0.1549 \times 100 \approx 15.5\% \). **Answer:** Approximately **15.5%** of the original intensity. |
31.0 |
| gpto1 | To solve this problem, we'll use Malus' Law, which relates the intensity of polarized light after passing through a polarizer to the angle between the light's polarization direction and the polarizer's axis. Malus' Law is given by: \[ I = I_0 \cos^2(\theta) \] where: - \( I \) is the transmitted light intensity after the polarizer. - \( I_0 \) is the intensity of the polarized light before the polarizer. - \( \theta \) is the angle between the light's polarization direction and the axis of the polarizer. Additionally, when unpolarized light passes through a polarizer, its intensity is reduced by half, since only the component of the light aligned with the polarizer's axis is transmitted. This can be expressed as: \[ I_{\text{after first polarizer}} = \frac{1}{2} I_{\text{original}} \] Let's solve the problem step by step. **Step 1: First Polarizer** The unpolarized light passes through the first polarizer. The intensity after the first polarizer is: \[ I_1 = \frac{1}{2} I_0 \] where \( I_0 \) is the original intensity of the unpolarized light. **Step 2: Second Polarizer** The light then passes through the second polarizer, which is at an angle of \( 30^\circ \) relative to the first polarizer. Using Malus' Law: \[ I_2 = I_1 \cos^2(30^\circ) \] Substituting \( I_1 \): \[ I_2 = \left( \frac{1}{2} I_0 \right) \cos^2(30^\circ) \] We know that: \[ \cos(30^\circ) = \frac{\sqrt{3}}{2} \] So: \[ \cos^2(30^\circ) = \left( \frac{\sqrt{3}}{2} \right)^2 = \frac{3}{4} \] Substitute back: \[ I_2 = \left( \frac{1}{2} I_0 \right) \left( \frac{3}{4} \right) = \frac{3}{8} I_0 \] **Step 3: Third Polarizer** The light next passes through the third polarizer, which is at an angle of \( 50^\circ \) relative to the second polarizer. The angle between the polarization direction of the light coming from the second polarizer and the axis of the third polarizer is \( 50^\circ \). Using Malus' Law again: \[ I_3 = I_2 \cos^2(50^\circ) \] Substitute \( I_2 \): \[ I_3 = \left( \frac{3}{8} I_0 \right) \cos^2(50^\circ) \] Now, calculate \( \cos^2(50^\circ) \): \[ \cos(50^\circ) \approx 0.6428 \] \[ \cos^2(50^\circ) \approx (0.6428)^2 \approx 0.4132 \] Substitute back: \[ I_3 = \left( \frac{3}{8} I_0 \right) (0.4132) \approx (0.375 \times 0.4132) I_0 \approx 0.15495 I_0 \] **Step 4: Calculate the Percentage** The percentage of the original intensity after the third polarizer is: \[ \text{Percentage} = \left( \frac{I_3}{I_0} \right) \times 100\% \approx 0.15495 \times 100\% \approx 15.5\% \] **Answer:** Approximately **15.5%** of the original intensity was the light coming out of the third polarizer. |
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| GPT-4o | To solve this problem, we need to apply Malus' law iteratively for each polarizer. ### Step-by-Step Solution: 1. **First Polarizer:** - When unpolarized light passes through the first polarizer, the intensity of the light is reduced by half. - Therefore, \( I_1 = \frac{I_0}{2} \). 2. **Second Polarizer:** - The angle between the first and second polarizer is 30 degrees. - Applying Malus' law, the intensity of the light after passing through the second polarizer is: \[ I_2 = I_1 \cos^2(30^\circ) \] - We know that \( \cos(30^\circ) = \frac{\sqrt{3}}{2} \), so: \[ I_2 = \frac{I_0}{2} \left( \frac{\sqrt{3}}{2} \right)^2 = \frac{I_0}{2} \cdot \frac{3}{4} = \frac{3I_0}{8} \] 3. **Third Polarizer:** - The angle between the second and third polarizer is 50 degrees. - Applying Malus' law again, the intensity of the light after passing through the third polarizer is: \[ I_3 = I_2 \cos^2(50^\circ) \] - Therefore: \[ I_3 = \frac{3I_0}{8} \cos^2(50^\circ) \] 4. **Calculating the Percentage:** - To find the percentage of the original intensity \( I_0 \) that comes out of the third polarizer, we need to calculate: \[ \text{Percentage} = \left( \frac{I_3}{I_0} \right) \times 100\% \] - Substituting \( I_3 \) into the equation: \[ \text{Percentage} = \left( \frac{\frac{3I_0}{8} \cos^2(50^\circ)}{I_0} \right) \times 100\% = \frac{3}{8} \cos^2(50^\circ) \times 100\% \] 5. **Evaluating \( \cos^2(50^\circ) \):** - Using a calculator or trigonometric tables, we find that \( \cos(50^\circ) \approx 0.6428 \). - Therefore, \( \cos^2(50^\circ) \approx (0.6428)^2 \approx 0.413 \). 6. **Final Calculation:** - Plugging this value into the percentage calculation: \[ \text{Percentage} = \frac{3}{8} \times 0.413 \times 100\% \approx \frac{1.239}{8} \times 100\% \approx 15.49\% \] ### Conclusion: The percentage of the original intensity that the light coming out of the third polarizer is approximately **15.49%**. |
31.0 |
| llama3.1-70B-Instruct | To solve this problem, we need to apply Malus' law twice, since the light passes through two polarizers after the initial one. ## Step 1: First Polarizer When unpolarized light passes through the first polarizer, the intensity of the light that passes through is half of the original intensity, because unpolarized light can be thought of as having its electric field vectors in all directions equally. Thus, after the first polarizer, the intensity is \(I_1 = \frac{1}{2}I_0\). ## Step 2: Second Polarizer The second polarizer is at an angle of 30 degrees to the first polarizer. According to Malus' law, the intensity of the light that passes through the second polarizer is \(I_2 = I_1 \cos^2(30^\circ)\). Since \(\cos(30^\circ) = \frac{\sqrt{3}}{2}\), we have \(I_2 = \frac{1}{2}I_0 \left(\frac{\sqrt{3}}{2}\right)^2 = \frac{1}{2}I_0 \cdot \frac{3}{4} = \frac{3}{8}I_0\). ## Step 3: Third Polarizer The third polarizer is at an angle of 50 degrees to the second polarizer. Again, applying Malus' law, the intensity of the light that passes through the third polarizer is \(I_3 = I_2 \cos^2(50^\circ)\). Since \(\cos(50^\circ)\) is approximately 0.6428, we have \(I_3 = \frac{3}{8}I_0 \cdot (0.6428)^2\). ## Step 4: Calculate the Percentage To find the percentage of the original intensity that the light coming out of the third polarizer represents, we calculate \(\frac{I_3}{I_0} \times 100\%\). Substituting the value of \(I_3\) from Step 3, we get \(\frac{3}{8} \cdot (0.6428)^2 \times 100\%\). The final answer is: $\boxed{15.6}$ |
31.0 |
| llama3.1-405B-Instruct | To solve this problem, we need to apply Malus' law three times, once for each polarizer. 1. First polarizer: The light is initially unpolarized, so the intensity after passing through the first polarizer is half of the original intensity: I1 = I0 / 2 2. Second polarizer: The angle between the first and second polarizers is 30 degrees. Applying Malus' law, we get: I2 = I1 * cos^2(30) = (I0 / 2) * (sqrt(3) / 2)^2 = (I0 / 2) * (3 / 4) = (3 / 8) * I0 3. Third polarizer: The angle between the second and third polarizers is 50 degrees. Applying Malus' law again, we get: I3 = I2 * cos^2(50) = (3 / 8) * I0 * (cos(50))^2 = (3 / 8) * I0 * (0.6428)^2 = (3 / 8) * I0 * 0.4133 = 0.155 * I0 So, the intensity of the light coming out of the third polarizer is 0.155 times the original intensity. To find the percentage, we multiply by 100: Percentage = 0.155 * 100 = 15.5% Therefore, the light coming out of the third polarizer has 15.5% of the original intensity. |
31.0 |
| DeepSeek-Coder-V2 | To solve the problem step by step, we will use Malus' law, which states that the intensity of polarized light after passing through a polarizer is given by: \[ I = I_0 \cos^2(\theta) \] where \( I \) is the intensity after the polarizer, \( I_0 \) is the intensity before the polarizer, and \( \theta \) is the angle between the polarization direction of the light and the polarizer. ### Step 1: Unpolarized Light through the First Polarizer When unpolarized light passes through the first polarizer, its intensity is reduced by a factor of 1/2. Therefore, the intensity after the first polarizer is: \[ I_1 = \frac{I_0}{2} \] ### Step 2: Light through the Second Polarizer The light then passes through a second polarizer, which is at an angle of 30 degrees to the first polarizer. Using Malus' law, the intensity after the second polarizer is: \[ I_2 = I_1 \cos^2(30^\circ) \] We know that: \[ \cos(30^\circ) = \frac{\sqrt{3}}{2} \] So: \[ \cos^2(30^\circ) = \left(\frac{\sqrt{3}}{2}\right)^2 = \frac{3}{4} \] Thus: \[ I_2 = \frac{I_0}{2} \cdot \frac{3}{4} = \frac{3I_0}{8} \] ### Step 3: Light through the Third Polarizer The light then passes through a third polarizer, which is at an angle of 50 degrees to the second polarizer. Using Malus' law again, the intensity after the third polarizer is: \[ I_3 = I_2 \cos^2(50^\circ) \] We know that: \[ \cos(50^\circ) = \cos(50^\circ) \] So: \[ \cos^2(50^\circ) = \left(\cos(50^\circ)\right)^2 \] Using a calculator: \[ \cos(50^\circ) \approx 0.6428 \] So: \[ \cos^2(50^\circ) \approx (0.6428)^2 \approx 0.413 \] Thus: \[ I_3 = \frac{3I_0}{8} \cdot 0.413 = \frac{3I_0 \cdot 0.413}{8} = \frac{1.239I_0}{8} \approx 0.1549I_0 \] ### Step 4: Calculate the Percentage of the Original Intensity The final intensity \( I_3 \) is approximately \( 0.1549I_0 \). To find the percentage of the original intensity, we calculate: \[ \text{Percentage} = \left(\frac{I_3}{I_0}\right) \times 100\% = 0.1549 \times 100\% \approx 15.49\% \] Thus, the percentage of the original intensity of the light coming out of the third polarizer is approximately: \[ \boxed{15.49\%} \] |
31.0 |
| Qwen2-72B-Instruct | Let's denote the intensity of the unpolarized light as $I_0$. When unpolarized light passes through a polarizer, it becomes polarized, and its intensity is halved (since only half of the light's components will align with the polarizer's axis). Thus, the intensity of the light after the first polarizer is $I_1 = \frac{1}{2}I_0$. The light then passes through a second polarizer at an angle of 30 degrees to the first. According to Malus' law, the intensity of the light after this second polarizer is: $$I_2 = I_1 \cdot \cos^2(30^\circ) = \frac{1}{2}I_0 \cdot \left(\frac{\sqrt{3}}{2}\right)^2 = \frac{1}{2}I_0 \cdot \frac{3}{4} = \frac{3}{8}I_0$$ Finally, the light passes through a third polarizer at an angle of 50 degrees to the second. The angle between the first and third polarizers is the sum of the angles between the first and second, and the second and third, which is $30^\circ + 50^\circ = 80^\circ$. However, we need the angle between the second and third polarizers, which is 50 degrees. The intensity after the third polarizer is: $$I_3 = I_2 \cdot \cos^2(50^\circ) = \frac{3}{8}I_0 \cdot \cos^2(50^\circ)$$ To find the percentage of the original intensity that comes out of the third polarizer, we divide $I_3$ by $I_0$ and multiply by 100%: $$\text{Percentage} = \left(\frac{I_3}{I_0}\right) \times 100\% = \left(\frac{3}{8} \cdot \cos^2(50^\circ)\right) \times 100\%$$ Using a calculator to find $\cos^2(50^\circ)$: $$\cos^2(50^\circ) \approx 0.3193$$ Thus, $$\text{Percentage} = \left(\frac{3}{8} \cdot 0.3193\right) \times 100\% \approx 12.04\%$$ So, approximately 12.04% of the original intensity of the unpolarized light comes out of the third polarizer. |
31.0 |
| baichuan4 | To solve this problem, we will apply Malus' law three times, once for each polarizer the light passes through. 1. When unpolarized light passes through the first polarizer, it becomes polarized. The intensity of the light after the first polarizer is $I_1 = I_0$, because the light is now fully polarized in the direction of the first polarizer. 2. The light then passes through a second polarizer that is at an angle of 30 degrees to the first. Applying Malus' law, the intensity of the light after the second polarizer is: $$I_2 = I_1 \cdot \cos^2(30^\circ)$$ $$I_2 = I_0 \cdot \cos^2(30^\circ)$$ $$I_2 = I_0 \cdot (\sqrt{3}/2)^2$$ $$I_2 = I_0 \cdot 3/4$$ 3. Finally, the light passes through a third polarizer that is at an angle of 50 degrees to the second polarizer. To find the angle between the third polarizer and the original polarized light, we need to find the difference between 50 degrees and 30 degrees, which is 20 degrees. So, the intensity of the light after the third polarizer is: $$I_3 = I_2 \cdot \cos^2(20^\circ)$$ $$I_3 = (I_0 \cdot 3/4) \cdot \cos^2(20^\circ)$$ Now, we need to calculate the percentage of the original intensity that remains after passing through the third polarizer: $$\text{Percentage of original intensity} = \frac{I_3}{I_0} \times 100\%$$ $$\text{Percentage of original intensity} = \frac{3}{4} \cdot \cos^2(20^\circ) \times 100\%$$ We can use a calculator to find the value of $\cos(20^\circ)$ and then square it to get $\cos^2(20^\circ)$. Let's do that: $$\cos(20^\circ) \approx 0.9397$$ $$\cos^2(20^\circ) \approx 0.8825$$ Now, we can calculate the percentage: $$\text{Percentage of original intensity} \approx \frac{3}{4} \cdot 0.8825 \times 100\%$$ $$\text{Percentage of original intensity} \approx 66.19\%$$ So, approximately 66.19% of the original intensity of the light remains after passing through the third polarizer. |
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| doubao | For unpolarized light passing through the first polarizer, the intensity is reduced by half, so $I_1 = \frac{I_0}{2}$. For the light passing through the second polarizer at a 30 degree angle to the first, the intensity is $I_2 = I_1 \cos^2(30^\circ) = \frac{I_0}{2} \times (\frac{\sqrt{3}}{2})^2 = \frac{3I_0}{8}$. For the light passing through the third polarizer at a 50 degree angle to the second, the intensity is $I_3 = I_2 \cos^2(50^\circ) = \frac{3I_0}{8} \times \cos^2(50^\circ)$. To find the percentage of the original intensity, we divide $I_3$ by $I_0$ and multiply by 100: $(\frac{3I_0}{8} \times \cos^2(50^\circ) \div I_0) \times 100 = (\frac{3}{8} \times \cos^2(50^\circ)) \times 100 \approx 14.5\%$. |
31.0 |
| minimax | To solve this problem, we need to apply Malus' law at each polarizer the light passes through. Malus' law states that the intensity of polarized light after passing through a polarizer is given by: \[ I = I_0 \cos^2(\theta) \] where \( I \) is the intensity after the polarizer, \( I_0 \) is the initial intensity before the polarizer, and \( \theta \) is the angle between the polarization direction of the light and the axis of the polarizer. Let's calculate the intensity of the light after each polarizer: 1. After the first polarizer: The light is initially unpolarized, but after passing through the first polarizer, its intensity becomes \( I_1 \), which is a fraction of the original intensity \( I_0 \). Since the first polarizer can only transmit light that is polarized along its axis, the intensity of the light after the first polarizer is half of the original intensity (this is a common result for unpolarized light passing through a polarizer): \[ I_1 = \frac{1}{2} I_0 \] 2. After the second polarizer (at 30 degrees to the first): The second polarizer is at an angle of \( \theta_2 = 30^\circ \) to the first polarizer. Using Malus' law: \[ I_2 = I_1 \cos^2(\theta_2) \] \[ I_2 = \left(\frac{1}{2} I_0\right) \cos^2(30^\circ) \] \[ I_2 = \left(\frac{1}{2} I_0\right) \left(\frac{\sqrt{3}}{2}\right)^2 \] \[ I_2 = \left(\frac{1}{2} I_0\right) \left(\frac{3}{4}\right) \] \[ I_2 = \frac{3}{8} I_0 \] 3. After the third polarizer (at 50 degrees to the second): The third polarizer is at an angle of \( \theta_3 = 50^\circ \) to the second polarizer. Again, using Malus' law: \[ I_3 = I_2 \cos^2(\theta_3) \] \[ I_3 = \left(\frac{3}{8} I_0\right) \cos^2(50^\circ) \] \[ I_3 = \left(\frac{3}{8} I_0\right) \left(\cos(50^\circ)\right)^2 \] Now, we can calculate the percentage of the original intensity that the light coming out of the third polarizer has: \[ \text{Percentage} = \frac{I_3}{I_0} \times 100\% \] \[ \text{Percentage} = \left(\frac{3}{8} \cos^2(50^\circ)\right) \times 100\% \] Using a calculator to find the cosine of 50 degrees and then squaring it: \[ \cos(50^\circ) \approx 0.6428 \] \[ \cos^2(50^\circ) \approx (0.6428)^2 \approx 0.4130 \] Now we can calculate the percentage: \[ \text{Percentage} \approx \left(\frac{3}{8} \times 0.4130\right) \times 100\% \] \[ \text{Percentage} \approx 0.15375 \times 100\% \] \[ \text{Percentage} \approx 15.375\% \] Therefore, the light coming out of the third polarizer has approximately 15.375% of the original intensity. |
31.0 |
| kimi | To solve this problem, we need to apply Malus' law at each stage of the light passing through the polarizers. 1. **First Polarizer:** The light starts as unpolarized, so the intensity after the first polarizer will be reduced to half of its original intensity because unpolarized light has equal intensities in all directions, and only one half of it will pass through the polarizer. Let \( I_{\text{unpol}} \) be the intensity of the unpolarized light. After the first polarizer, the intensity \( I_1 \) will be: \[ I_1 = \frac{1}{2} I_{\text{unpol}} \] 2. **Second Polarizer:** The light now has an intensity \( I_1 \) and is polarized. It passes through a second polarizer at an angle of 30 degrees to the first. Using Malus' law, the intensity after the second polarizer \( I_2 \) will be: \[ I_2 = I_1 \cos^2(30^\circ) \] Since \( \cos(30^\circ) = \frac{\sqrt{3}}{2} \), we get: \[ I_2 = I_1 \left(\frac{\sqrt{3}}{2}\right)^2 = I_1 \frac{3}{4} \] 3. **Third Polarizer:** Now, the light has an intensity \( I_2 \) and passes through a third polarizer at an angle of 50 degrees to the second polarizer. Using Malus' law again, the intensity after the third polarizer \( I_3 \) will be: \[ I_3 = I_2 \cos^2(50^\circ) \] We need to calculate \( \cos(50^\circ) \) to proceed. Using a calculator, we find that \( \cos(50^\circ) \approx 0.6428 \), so: \[ I_3 = I_2 (0.6428)^2 = I_1 \frac{3}{4} (0.6428)^2 \] Now, we can calculate the final intensity \( I_3 \) in terms of the original unpolarized intensity \( I_{\text{unpol}} \): \[ I_3 = \frac{1}{2} I_{\text{unpol}} \frac{3}{4} (0.6428)^2 \] Let's calculate the numerical value: \[ I_3 = \frac{1}{2} \times \frac{3}{4} \times (0.6428)^2 \times I_{\text{unpol}} \] \[ I_3 = 0.25 \times 0.3 \times I_{\text{unpol}} \] \[ I_3 = 0.075 \times I_{\text{unpol}} \] To find the percentage of the original intensity, we multiply by 100: \[ \text{Percentage of original intensity} = 0.075 \times 100\% = 7.5\% \] So, the light coming out of the third polarizer has an intensity that is 7.5% of the original intensity of the unpolarized light. |
31.0 |
| wenxinyiyan | To solve this problem, we will apply Malus' law to each polarizer in sequence. Malus' law states that the intensity of polarized light passing through a polarizer is given by $I = I_0 \cos^2(\theta)$, where $I$ is the intensity of the light after passing through the polarizer, $I_0$ is the intensity of the light before passing through the polarizer, and $\theta$ is the angle between the polarization direction of the light and the axis of the polarizer. 1. **First Polarizer**: The initial light is unpolarized, so it has equal components in all directions. When it passes through the first polarizer, it becomes polarized in the direction of the polarizer's axis. We can consider the intensity of the light after the first polarizer to be $I_1 = I_0 \cos^2(0^\circ) = I_0$ since the angle between the unpolarized light and the polarizer is effectively $0^\circ$. However, since only half of the unpolarized light will align with the polarizer, the intensity after the first polarizer is actually $I_1 = 0.5 \times I_0$. 2. **Second Polarizer**: The light now passes through a second polarizer at an angle of $30^\circ$ to the first. Applying Malus' law, the intensity after the second polarizer is $I_2 = I_1 \cos^2(30^\circ) = 0.5 \times I_0 \times \cos^2(30^\circ)$. 3. **Third Polarizer**: Finally, the light passes through a third polarizer at an angle of $50^\circ$ to the second (and thus $80^\circ$ to the first, but this is not directly relevant). The intensity after the third polarizer is $I_3 = I_2 \cos^2(50^\circ) = 0.5 \times I_0 \times \cos^2(30^\circ) \times \cos^2(50^\circ)$. Now, let's calculate the specific values: * $\cos^2(30^\circ) = \left(\frac{\sqrt{3}}{2}\right)^2 = \frac{3}{4}$ * $\cos^2(50^\circ)$ must be calculated numerically, as it does not simplify to a nice fraction. Using a calculator, we find $\cos^2(50^\circ) \approx 0.6428$ (rounded to four decimal places). Therefore, the intensity after the third polarizer is: $$I_3 = 0.5 \times I_0 \times \frac{3}{4} \times 0.6428 \approx 0.2410 \times I_0$$ To find the percentage of the original intensity, we multiply by 100: $$\text{Percentage of original intensity} = 0.2410 \times 100\% = 24.10\%$$ So, approximately 24.10% of the original intensity of the light emerges from the third polarizer. |
31.0 |
| zhipu | To solve this problem, we need to apply Malus' law at each stage where the light passes through a polarizer. 1. **First Polarizer:** Unpolarized light passes through the first polarizer. When unpolarized light passes through a polarizer, its intensity is reduced by half, regardless of the angle. Therefore, the intensity after the first polarizer, \( I_1 \), is: \[ I_1 = \frac{I_0}{2} \] 2. **Second Polarizer:** The light then passes through a second polarizer at an angle of 30 degrees to the first polarizer. According to Malus' law, the intensity \( I_2 \) after the second polarizer is: \[ I_2 = I_1 \cos^2(30^\circ) \] We know that \( \cos(30^\circ) = \frac{\sqrt{3}}{2} \), so: \[ I_2 = \frac{I_0}{2} \left( \frac{\sqrt{3}}{2} \right)^2 = \frac{I_0}{2} \cdot \frac{3}{4} = \frac{3I_0}{8} \] 3. **Third Polarizer:** Finally, the light passes through a third polarizer at an angle of 50 degrees to the second polarizer. Again, using Malus' law, the intensity \( I_3 \) after the third polarizer is: \[ I_3 = I_2 \cos^2(50^\circ) \] We need to find \( \cos(50^\circ) \). Using a calculator or trigonometric tables, we find: \[ \cos(50^\circ) \approx 0.6428 \] So: \[ I_3 = \frac{3I_0}{8} \cdot (0.6428)^2 \approx \frac{3I_0}{8} \cdot 0.412 \] \[ I_3 \approx \frac{3I_0 \cdot 0.412}{8} = \frac{1.236I_0}{8} \approx 0.1545I_0 \] To find the percentage of the original intensity, we multiply \( I_3 \) by 100%: \[ \text{Percentage} = 0.1545 \times 100\% \approx 15.45\% \] Therefore, the light coming out of the third polarizer is approximately **15.45%** of the original intensity. |
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| gpto1 | To solve this problem, we'll use Malus' Law, which relates the intensity of polarized light after passing through a polarizer to the angle between the light's polarization direction and the polarizer's axis. Malus' Law is given by: \[ I = I_0 \cos^2(\theta) \] where: - \( I \) is the transmitted light intensity after the polarizer. - \( I_0 \) is the intensity of the polarized light before the polarizer. - \( \theta \) is the angle between the light's polarization direction and the axis of the polarizer. Additionally, when unpolarized light passes through a polarizer, its intensity is reduced by half, since only the component of the light aligned with the polarizer's axis is transmitted. This can be expressed as: \[ I_{\text{after first polarizer}} = \frac{1}{2} I_{\text{original}} \] Let's solve the problem step by step. **Step 1: First Polarizer** The unpolarized light passes through the first polarizer. The intensity after the first polarizer is: \[ I_1 = \frac{1}{2} I_0 \] where \( I_0 \) is the original intensity of the unpolarized light. **Step 2: Second Polarizer** The light then passes through the second polarizer, which is at an angle of \( 30^\circ \) relative to the first polarizer. Using Malus' Law: \[ I_2 = I_1 \cos^2(30^\circ) \] Substituting \( I_1 \): \[ I_2 = \left( \frac{1}{2} I_0 \right) \cos^2(30^\circ) \] We know that: \[ \cos(30^\circ) = \frac{\sqrt{3}}{2} \] So: \[ \cos^2(30^\circ) = \left( \frac{\sqrt{3}}{2} \right)^2 = \frac{3}{4} \] Substitute back: \[ I_2 = \left( \frac{1}{2} I_0 \right) \left( \frac{3}{4} \right) = \frac{3}{8} I_0 \] **Step 3: Third Polarizer** The light next passes through the third polarizer, which is at an angle of \( 50^\circ \) relative to the second polarizer. The angle between the polarization direction of the light coming from the second polarizer and the axis of the third polarizer is \( 50^\circ \). Using Malus' Law again: \[ I_3 = I_2 \cos^2(50^\circ) \] Substitute \( I_2 \): \[ I_3 = \left( \frac{3}{8} I_0 \right) \cos^2(50^\circ) \] Now, calculate \( \cos^2(50^\circ) \): \[ \cos(50^\circ) \approx 0.6428 \] \[ \cos^2(50^\circ) \approx (0.6428)^2 \approx 0.4132 \] Substitute back: \[ I_3 = \left( \frac{3}{8} I_0 \right) (0.4132) \approx (0.375 \times 0.4132) I_0 \approx 0.15495 I_0 \] **Step 4: Calculate the Percentage** The percentage of the original intensity after the third polarizer is: \[ \text{Percentage} = \left( \frac{I_3}{I_0} \right) \times 100\% \approx 0.15495 \times 100\% \approx 15.5\% \] **Answer:** Approximately **15.5%** of the original intensity was the light coming out of the third polarizer. |
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| deepseekv3 | To solve this problem, we'll apply **Malus' Law** step by step as the unpolarized light passes through each polarizer. --- ### Step 1: Unpolarized light passes through the first polarizer When unpolarized light passes through a polarizer, its intensity is reduced by half. This is because only the component of the light aligned with the polarizer's axis is transmitted. Thus, the intensity after the first polarizer is: \[ I_1 = \frac{I_0}{2} \] --- ### Step 2: Light passes through the second polarizer at 30° to the first Malus' Law states that the intensity of polarized light after passing through a polarizer is: \[ I = I_0 \cos^2(\theta) \] Here, \(I_0 = I_1\) and \(\theta = 30°\). Thus, the intensity after the second polarizer is: \[ I_2 = I_1 \cos^2(30°) \] Substitute \(I_1 = \frac{I_0}{2}\) and \(\cos(30°) = \frac{\sqrt{3}}{2}\): \[ I_2 = \frac{I_0}{2} \cdot \left(\frac{\sqrt{3}}{2}\right)^2 = \frac{I_0}{2} \cdot \frac{3}{4} = \frac{3I_0}{8} \] --- ### Step 3: Light passes through the third polarizer at 50° to the second Again, apply Malus' Law. Here, \(I_0 = I_2\) and \(\theta = 50°\). Thus, the intensity after the third polarizer is: \[ I_3 = I_2 \cos^2(50°) \] Substitute \(I_2 = \frac{3I_0}{8}\) and \(\cos(50°) \approx 0.6428\): \[ I_3 = \frac{3I_0}{8} \cdot (0.6428)^2 \approx \frac{3I_0}{8} \cdot 0.4132 \approx 0.155I_0 \] --- ### Final Result The intensity of the light after the third polarizer is approximately **15.5%** of the original intensity \(I_0\). \[ \boxed{15.5\%} \] |
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