Apparent magnitude of a light bulb

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SUMMARY

The apparent magnitude of a 100W light bulb at a distance of 3 meters can be calculated using the formulas m-M=5log(d/10) and m=-2.5log(l/l0). The solar luminosity is approximately 4 × 1026 W, and the relative magnitude of the Sun is -27. To eliminate the unknown constant in the expression relating flux density to apparent magnitude, one must compare the light bulb's intensity to that of the Sun.

PREREQUISITES
  • Understanding of apparent magnitude and flux density in astronomy
  • Familiarity with logarithmic equations in physics
  • Knowledge of solar luminosity and its significance
  • Basic principles of light intensity and distance calculations
NEXT STEPS
  • Calculate the apparent magnitude of the 100W light bulb using the provided formulas
  • Research the concept of flux density and its application in astronomy
  • Explore the relationship between intensity and magnitude in astronomical contexts
  • Study the significance of the solar luminosity in astrophysics
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Students in physics, astronomy enthusiasts, and anyone interested in understanding the calculations related to light intensity and apparent magnitude.

icedragon
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Homework Statement



What is the apparent magnitude of a 100W light bulb at a distance of 3 m?

[Hint: Compare with the Sun to eliminate the unknown constant in the expression relating flux density to apparent magnitude.]

. The solar luminosity is L \approx 4 \times 10^{26} W.
. The Earth-Sun distance is approximately 1.5 \times 10^8 km.

Homework Equations



m-M=5log \frac{d}{10}
m=-2.5log \frac{l}{l_{0}}

The Attempt at a Solution



I have attempted to substitute m into the equation for m-M but I do not see how that helps me eliminate the unknown constant (which I assume is l_{0}. Any help would be appreciated!
 
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Hello icedragon,

Welcome to Physics Forums (PF)!
icedragon said:

Homework Statement



What is the apparent magnitude of a 100W light bulb at a distance of 3 m?

[Hint: Compare with the Sun to eliminate the unknown constant in the expression relating flux density to apparent magnitude.]

. The solar luminosity is L \approx 4 \times 10^{26} W.
. The Earth-Sun distance is approximately 1.5 \times 10^8 km.

Homework Equations



m-M=5log \frac{d}{10}
Where did that formula come from? What's d and where did the '10' come from?
m=-2.5log \frac{l}{l_{0}}
(In your equation you are using l's as in 'lesson'. But they should be capitol I's as in 'Intensity'.)

That looks right if the relative magnitude of I_0 is 0 (which is just fine for what we're doing in this problem).

The more general formula is

m_1 - m_0 = -2.5 \log \frac{I_1}{I_0}
which you can then set m_0 = 0 if you want m_1 to be relative to 0 magnitude. But you'll still need to solve for I_0 if you wish to do that.

The Attempt at a Solution



I have attempted to substitute m into the equation for m-M but I do not see how that helps me eliminate the unknown constant (which I assume is l_{0}. Any help would be appreciated!

If it helps, the relative magnitude of the sun is -27 (relative to magnitude 0)*. Your first step is to use that and solve for I_0.

*(See http://en.wikipedia.org/wiki/Magnitude_%28astronomy%29)
 

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