Cambridge International AS & A Level · 25 · structured
Radiant flux decreases with distance
Radiant flux decreases with distance · Original GioPhysics question with a detailed, mark-by-mark answer guide.
- Demand
- demanding
- Marks
- 5
- Topics
- 1
- Answer
- Complete
A star emits a constant total luminosity L=1.00×10²⁸ W isotropically. An observer is 2.00×10¹⁸ m away. Neglect absorption between the star and observer.
- (a)
Recall the inverse-square relation between radiant flux intensity F, luminosity and distance. Explain the spherical-area factor and calculate F.
3 marks - (b)
If the distance doubles without any change in the star, determine the new flux and state whether its luminosity changes.
2 marks
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(a)
Recall the inverse-square relation between radiant flux intensity F, luminosity and distance. Explain the spherical-area factor and calculate F.
List the given quantities with units, identify the required quantity, write the governing relationship before substituting, and keep extra digits until the final line so rounding does not distort the result.
- 1
The same total power crosses each sphere centred on the star; its area is 4πd².
- 2
Therefore F=L/(4πd²).
- 3
F=1.00×10²⁸/[4π(2.00×10¹⁸)²]=1.99×10⁻¹⁰ W m⁻².
(b)
If the distance doubles without any change in the star, determine the new flux and state whether its luminosity changes.
List the given quantities with units, identify the required quantity, write the governing relationship before substituting, and keep extra digits until the final line so rounding does not distort the result.
- 1
Doubling d increases the spreading area by four, so flux becomes one quarter.
- 2
F_new=1.99×10⁻¹⁰/4=4.97×10⁻¹¹ W m⁻²; L remains 1.00×10²⁸ W.
Complete marking guidance
Use these worked steps alongside the part-by-part answers. They explain the method and mark allocation; they do not add extra marks.
- (a) The same total power crosses each sphere centred on the star; its area is 4πd².
- (a) Therefore F=L/(4πd²).
- (a) F=1.00×10²⁸/[4π(2.00×10¹⁸)²]=1.99×10⁻¹⁰ W m⁻².
- (b) Doubling d increases the spreading area by four, so flux becomes one quarter.
- (b) F_new=1.99×10⁻¹⁰/4=4.97×10⁻¹¹ W m⁻²; L remains 1.00×10²⁸ W.
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