Given the following spotlights placed as shown in the diagram. Assume isometric distribution of the light The Area A2 is 12,56m2, The Lamp#2 has a Luminous flux of 600 Lumens and the distance D2 is 2 meters What is the illuminance on the A2 (the Lamp#1 is off) Lamp #1 Lamp #2 Beam Angle (4) D1 D2 Вeam Angle (4) A1 A2 Illuminated Surface Diameter АЗ Light Fixture Distance-
Given the following spotlights placed as shown in the diagram. Assume isometric distribution of the light The Area A2 is 12,56m2, The Lamp#2 has a Luminous flux of 600 Lumens and the distance D2 is 2 meters What is the illuminance on the A2 (the Lamp#1 is off) Lamp #1 Lamp #2 Beam Angle (4) D1 D2 Вeam Angle (4) A1 A2 Illuminated Surface Diameter АЗ Light Fixture Distance-
College Physics
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ISBN:9781305952300
Author:Raymond A. Serway, Chris Vuille
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Chapter1: Units, Trigonometry. And Vectors
Section: Chapter Questions
Problem 1CQ: Estimate the order of magnitude of the length, in meters, of each of the following; (a) a mouse, (b)...
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The Area A2 is 12,56m2, The Lamp#2 has a Luminous flux 600 Lumens and the distance D2 is 2 meters
What is the illuminance on the A2 (the Lamp#1 is off)
![### Transcription for Educational Website
**Problem Statement:**
Given the following spotlights placed as shown in the diagram, assume isometric distribution of the light.
- The Area \(A2\) is \(12.56 \, m^2\).
- Lamp #2 has a luminous flux of 600 Lumens.
- The distance \(D2\) is 2 meters.
What is the illuminance on the area \(A2\) (assuming Lamp #1 is off)?
**Diagram Explanation:**
The diagram shows two spotlights labeled Lamp #1 and Lamp #2. Each lamp is positioned above respective areas on the ground, labeled \(A1\), \(A2\), and \(A3\). For each lamp, a light fixture distance is indicated:
- **Lamp #1:** Illuminates area \(A1\) (This lamp is off in the problem scenario).
- **Lamp #2:** Illuminates area \(A2\) with a luminous flux of 600 Lumens and is 2 meters away from the area.
Both lamps have a specified beam angle \(\Phi\), which describes the spread of the light over the surface areas. The diagram also shows an overlap area \(A3\) where the illuminance from both lamps would converge if Lamp #1 were on.
**Objective:**
Calculate the illuminance on area \(A2\) due to Lamp #2.
**Calculation:**
The formula for illuminance (\(E\)) is given by:
\[ E = \frac{\Phi}{A} \]
Where:
- \(E\) is the illuminance in lux (lumens per square meter),
- \(\Phi\) is the luminous flux in lumens,
- \(A\) is the area in square meters.
For area \(A2\), since \(\Phi = 600 \, \text{lumens}\) and \(A2 = 12.56 \, m^2\),
\[ E = \frac{600}{12.56} \approx 47.77 \, \text{lux} \]
Therefore, the illuminance on area \(A2\) is approximately 47.77 lux.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Ff8e1b55b-e719-4fba-8aef-c002df41f622%2F614f345d-da9b-4acc-9db7-9f8f92493a74%2Fbdn3ut_processed.png&w=3840&q=75)
Transcribed Image Text:### Transcription for Educational Website
**Problem Statement:**
Given the following spotlights placed as shown in the diagram, assume isometric distribution of the light.
- The Area \(A2\) is \(12.56 \, m^2\).
- Lamp #2 has a luminous flux of 600 Lumens.
- The distance \(D2\) is 2 meters.
What is the illuminance on the area \(A2\) (assuming Lamp #1 is off)?
**Diagram Explanation:**
The diagram shows two spotlights labeled Lamp #1 and Lamp #2. Each lamp is positioned above respective areas on the ground, labeled \(A1\), \(A2\), and \(A3\). For each lamp, a light fixture distance is indicated:
- **Lamp #1:** Illuminates area \(A1\) (This lamp is off in the problem scenario).
- **Lamp #2:** Illuminates area \(A2\) with a luminous flux of 600 Lumens and is 2 meters away from the area.
Both lamps have a specified beam angle \(\Phi\), which describes the spread of the light over the surface areas. The diagram also shows an overlap area \(A3\) where the illuminance from both lamps would converge if Lamp #1 were on.
**Objective:**
Calculate the illuminance on area \(A2\) due to Lamp #2.
**Calculation:**
The formula for illuminance (\(E\)) is given by:
\[ E = \frac{\Phi}{A} \]
Where:
- \(E\) is the illuminance in lux (lumens per square meter),
- \(\Phi\) is the luminous flux in lumens,
- \(A\) is the area in square meters.
For area \(A2\), since \(\Phi = 600 \, \text{lumens}\) and \(A2 = 12.56 \, m^2\),
\[ E = \frac{600}{12.56} \approx 47.77 \, \text{lux} \]
Therefore, the illuminance on area \(A2\) is approximately 47.77 lux.
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