Intensity And Poynting Vector
A laser delivers an average intensity of 6.0 W/m² onto a screen, and a researcher needs the average energy density of the beam in that region.
Select the correct option:
Solution
2.0×10−8J/m3
Building on NCERT Class 12, Chapter 8 (Electromagnetic Waves), the intensity of an electromagnetic wave equals the average energy flowing per unit area per unit time and is related to the average energy density u by I = u·c, because the energy contained in the fields is transported forward at the speed of light. One can picture a column of the wave of cross-section one square metre: in one second it advances a distance c, so all the energy stored in that length of column, namely u·c, crosses the unit area, which is exactly the intensity. Rearranging gives u = I/c = 6.0 / (3 × 10⁸) = 2.0 × 10⁻⁸ J/m³. This small value reflects how thinly energy is spread through the volume even for a visible beam. The value 1.8 × 10⁹ J/m³ is wrong because it multiplies intensity by c rather than dividing, giving an impossibly large density. The value 5.0 × 10⁻⁷ J/m³ uses an incorrect numerical factor unrelated to I/c. The value 3.0 × 10⁻⁹ J/m³ arises from a power-of-ten error while dividing by the speed of light. A dimensional and magnitude check confirms that dividing W/m² by m/s gives J/m³ and yields a plausibly tiny density, validating the correct answer.
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About This Question
- Subject
- physics
- Chapter
- electromagnetic waves
- Topic
- intensity and poynting vector
- Difficulty
- Hard
- Year
- 2025
Solution
Correct Answer:
2.0×10−8J/m3
Building on NCERT Class 12, Chapter 8 (Electromagnetic Waves), the intensity of an electromagnetic wave equals the average energy flowing per unit area per unit time and is related to the average energy density u by I = u·c, because the energy contained in the fields is transported forward at the speed of light. One can picture a column of the wave of cross-section one square metre: in one second it advances a distance c, so all the energy stored in that length of column, namely u·c, crosses the unit area, which is exactly the intensity. Rearranging gives u = I/c = 6.0 / (3 × 10⁸) = 2.0 × 10⁻⁸ J/m³. This small value reflects how thinly energy is spread through the volume even for a visible beam. The value 1.8 × 10⁹ J/m³ is wrong because it multiplies intensity by c rather than dividing, giving an impossibly large density. The value 5.0 × 10⁻⁷ J/m³ uses an incorrect numerical factor unrelated to I/c. The value 3.0 × 10⁻⁹ J/m³ arises from a power-of-ten error while dividing by the speed of light. A dimensional and magnitude check confirms that dividing W/m² by m/s gives J/m³ and yields a plausibly tiny density, validating the correct answer.
This hard difficulty physics question is from the chapter electromagnetic waves, covering the topic of intensity and poynting vector. It appeared in the 2025 exam.
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