Suppose an isotropic light source has a power of 59 W and that its light is normally incident on a partially-reflecting surface at a distance of 2.69 m from the source. Calculate the average radiation pressure (in nPa) if the surface reflects 51% of the incident light. (Use c = 2.9979 x 108 m/s)
Suppose an isotropic light source has a power of 59 W and that its light is normally incident on a partially-reflecting surface at a distance of 2.69 m from the source. Calculate the average radiation pressure (in nPa) if the surface reflects 51% of the incident light. (Use c = 2.9979 x 108 m/s)
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![**Problem Statement:**
Suppose an isotropic light source has a power of 59 W and that its light is normally incident on a partially-reflecting surface at a distance of 2.69 m from the source. Calculate the average radiation pressure (in nPa) if the surface reflects 51% of the incident light. (Use \( c = 2.9979 \times 10^8 \) m/s)
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**Explanation of Concepts:**
1. **Isotropic Light Source:** An isotropic light source emits light uniformly in all directions.
2. **Power:** The power of the light source is given as 59 Watts, which is the rate at which energy is emitted from the source.
3. **Radiation Pressure:** This is the pressure exerted by electromagnetic radiation on a surface. It depends on the intensity of the light and the reflectivity of the surface.
4. **Reflectivity:** In this scenario, the surface reflects 51% of the incident light.
5. **Speed of Light:** The speed of light (\( c \)) is provided as \( 2.9979 \times 10^8 \) m/s, which is crucial for calculating the radiation pressure.
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**Solution Approach:**
To solve for the average radiation pressure, use the formula for radiation pressure on a reflecting surface:
\[ P = \frac{2I}{c} \times \text{Reflectivity} \]
Where:
- \( P \) is the radiation pressure.
- \( I \) is the intensity of the light, calculated as \( \frac{\text{Power}}{\text{Area}} \).
- \( c \) is the speed of light.
- Reflectivity is the fraction of light that is reflected.
1. **Calculate Intensity (I):**
The intensity \( I \) at a distance \( r \) from an isotropic source:
\[
I = \frac{P_{\text{source}}}{4\pi r^2}
\]
2. **Calculate Radiation Pressure (P):**
\[
P = \frac{2I}{c} \times \text{Reflectivity}
\]
By plugging in the given values, you can determine the radiation pressure in nanopascals (nPa).
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Transcribed Image Text:**Problem Statement:**
Suppose an isotropic light source has a power of 59 W and that its light is normally incident on a partially-reflecting surface at a distance of 2.69 m from the source. Calculate the average radiation pressure (in nPa) if the surface reflects 51% of the incident light. (Use \( c = 2.9979 \times 10^8 \) m/s)
---
**Explanation of Concepts:**
1. **Isotropic Light Source:** An isotropic light source emits light uniformly in all directions.
2. **Power:** The power of the light source is given as 59 Watts, which is the rate at which energy is emitted from the source.
3. **Radiation Pressure:** This is the pressure exerted by electromagnetic radiation on a surface. It depends on the intensity of the light and the reflectivity of the surface.
4. **Reflectivity:** In this scenario, the surface reflects 51% of the incident light.
5. **Speed of Light:** The speed of light (\( c \)) is provided as \( 2.9979 \times 10^8 \) m/s, which is crucial for calculating the radiation pressure.
---
**Solution Approach:**
To solve for the average radiation pressure, use the formula for radiation pressure on a reflecting surface:
\[ P = \frac{2I}{c} \times \text{Reflectivity} \]
Where:
- \( P \) is the radiation pressure.
- \( I \) is the intensity of the light, calculated as \( \frac{\text{Power}}{\text{Area}} \).
- \( c \) is the speed of light.
- Reflectivity is the fraction of light that is reflected.
1. **Calculate Intensity (I):**
The intensity \( I \) at a distance \( r \) from an isotropic source:
\[
I = \frac{P_{\text{source}}}{4\pi r^2}
\]
2. **Calculate Radiation Pressure (P):**
\[
P = \frac{2I}{c} \times \text{Reflectivity}
\]
By plugging in the given values, you can determine the radiation pressure in nanopascals (nPa).
---
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