How much energy would enter your pupils from an explosion in space?

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SUMMARY

The energy entering the pupils from an explosion in space is calculated to be 1.1 Joules. The discussion highlights the importance of understanding the relationship between intensity, power, and distance in this context. The equation for intensity (S) is crucial, as it connects the power of the light source to the energy received at a given distance. Additionally, accurate unit conversion, such as converting 4 mm to meters, is essential for precise calculations.

PREREQUISITES
  • Understanding of light intensity and its relation to power and distance.
  • Familiarity with basic physics equations related to energy and intensity.
  • Knowledge of unit conversions, specifically between millimeters and meters.
  • Basic grasp of how energy is quantified in Joules.
NEXT STEPS
  • Research the formula for intensity in relation to point sources of light.
  • Study the concept of energy transfer in physics, particularly in explosive events.
  • Explore unit conversion techniques for scientific calculations.
  • Investigate the effects of distance on light intensity and energy distribution.
USEFUL FOR

Students and professionals in physics, engineers working with energy calculations, and anyone interested in the effects of explosions on light intensity and energy transfer.

cestlavie
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Homework Statement
Earth is about 28,000 light years from the center of the Milky Way Galaxy. If a supernova explosion occurred approximately 14,000 light years away from us and released ##2.0 * 10^46 J## of energy in a sudden burst of light, how much energy would enter one of your pupils? Assume the diameter of a pupil to be about 4 mm.
3.0 E 23 J
1.1 J
4.6 J
3.5 J
2.2 J
Relevant Equations
##S_{average}= \frac {energy} {time} * \frac {1} {area}##
The answer is 1.1 J, but I don't know how to get there. The only equation I can think of that might be related to this is Intensity, which I've added above. I could find area, using .0004m as the diameter, and energy using 2.0 E 46 J, but I get stuck on energy.
 
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You need another relation. How does the intensity S of the light depend on the power of the source and the distance from the source?
 
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Also, check your conversion of 4 mm to meters.
 
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