Magnitude of a 60 watt light bulb

[Starfleet]

In my mate’s back yard that goes out into an open field there is a 60 watt bulb fitted on a lamp post about 500 metres from his house. At night I could’ve sworn it looks as bright as any bright star in the sky. Now in this scale of some really bright astro objects, I’m thinking where that light must fit in brightness:

App. Mag. Celestial object

?26.73 Sun (449,000 times brighter than full moon)

?20 Sun (As seen from Neptune)

?12.6 Full Moon

?8.0 Maximum brightness of an Iridium (satellite) flare

?6.5 Ahad's constant[4]

?6.0 The Crab Supernova (SN 1054) of AD 1054 (6500 light years away)

?4.7 Maximum brightness of Venus and the International Space Station (when the ISS is at its perigee and fully lit by the sun)[5]

?3.9 Faintest objects observable during the day with naked eye

?3.7 Minimum brightness of Venus

?3.0 Maximum brightness of Mars

?2.8 Maximum brightness of Jupiter

?1.9 Maximum brightness of Mercury

?1.47 Brightest star (except for the sun) at visible wavelengths: Sirius

?0.7 Second-brightest star: Canopus

?0.24 Maximum brightness of Saturn

0 The zero point by definition: This used to be Vega

(see references for modern zero point)

3 Faintest stars visible in an urban neighborhood with naked eye

(List from wikipedia)

Would it be poss to work out this in magnitude? I read in some physics thesis you have the inverse square law but that’s for the stars (not for light bulbs). So can I predict what brightness a lamp would be seen from say 500 metres?

[/]

[EA2007]

I am sure that someone can figure out the correct magnitude for you. There are two types however, absolute and apparent magnitude. You can use the inverse square law though, along with trigonometry and parallax to figure out its true distance!

[narrowbandpaul]

the flux is L/(4pi*r^2); for L=60w and D=500m

F=1.91e-5 W/m2

the flux of vega of visible wavelengths I have found a question sheet stating that the integrated flux of vega over all wavelengths is 2.53e-8 W/m2

I also found the BVR fluxes for vega, adding these up gave a flux of 1.1545e-10W/m2/nm, assuming a constant flux over the vis spectrum, the flux is obtained by multiplying by the wavelength range of the vis spectrum, equal to 300nm.

this yields a flux of 3.4635e-8 w/m2

using pogsons equation, and assuming that the apparant magnitude of vega is 0 for the visible spectrum and using both value for the flux of vega

m= -2.5log(F_bulb/F_vega)

we have two estimates for the apparant magnitude of the bulb.

m= -6.85 and m= -7.2

brighter than venus, but less bright than the moon

Hope this answers the question

[EA2007]

There you go!

[themos]

The 60 Watts is the electric power consumed, the power that is emitted in form of light is a lot less than that, maybe 5%?

[Azelfafage]

Would the Sun REALLY be that bright from Neptune ?

Seems overcooked to me.

A

[thing]

Would the Sun REALLY be that bright from Neptune ?

Seems overcooked to me.

A

I was thinking that.

[Astronut]

Just checked on SN Pro and according to that the Sun's magnitude from Neptune is -19.51 with an angular size of 1.1 arcminutes.

[EA2007]

Round it up to 20, there you go !

[supernova]

the flux is L/(4pi*r^2); for L=60w and D=500m

F=1.91e-5 W/m2

the flux of vega of visible wavelengths I have found a question sheet stating that the integrated flux of vega over all wavelengths is 2.53e-8 W/m2

I also found the BVR fluxes for vega, adding these up gave a flux of 1.1545e-10W/m2/nm, assuming a constant flux over the vis spectrum, the flux is obtained by multiplying by the wavelength range of the vis spectrum, equal to 300nm.

this yields a flux of 3.4635e-8 w/m2

using pogsons equation, and assuming that the apparant magnitude of vega is 0 for the visible spectrum and using both value for the flux of vega

m= -2.5log(F_bulb/F_vega)

we have two estimates for the apparant magnitude of the bulb.

m= -6.85 and m= -7.2

brighter than venus, but less bright than the moon

Hope this answers the question

I cant sleep trying to work that out ahh I get it :idea: brighter than venus, but less bright than the moon

[narrowbandpaul]

interestingly, could a light bulb be used as a reference for photometry?

[EA2007]

Good point, but I assume that the flux of the object would be vastly different given that its electrical and not nuclear. I would point out that if a star could be the size of a light bulb then the light emission would be far greater and far far hotter, wouldn't want to be that one or the one to change it !

[Starfleet]

we have two estimates for the apparant magnitude of the bulb.

m= -6.85 and m= -7.2

brighter than venus, but less bright than the moon

Wow that is some calculation, way above my little pea brain! That's still a very bright star! So if you place the bulb farther into the filed like say 800 metre or 1200 metre can you see it go down to magnitude -1 like Sirius or magnitude 0 like Vega?

This is a way cool formula, so cheers for the brains

[ngc2403]

Good point, but I assume that the flux of the object would be vastly different given that its electrical and not nuclear. I would point out that if a star could be the size of a light bulb then the light emission would be far greater and far far hotter, wouldn't want to be that one or the one to change it !

if the sun was the size of a light bulb then it would be more then 600 million kelvin on the surface :shock:

the photons that it would emitte at the peak of emission would be 100 million electron volts. gamma rays as large as the biggest ever seen :sunny: oh burnny

ally

[narrowbandpaul]

-6.8 is a very bright star.

I cant take all the credit for the equation. i would like to thank max Planck and Pogson, who helped in a small way, this profound discussion!

youre very welcome, these calculations are fun too do.

And its always good to know the flux of vega in different bands.

you asked how far it has to be to be the brightness of vega...ie 0 mag

well, for m=0, the log(F1/Fv)=0 so the fluxes must be the same.

So how far for a 60W bulb (assuming 60W output power...dont know that this is the case)

D= sqrt(L/4pi*F_vega)

=sqrt(60/ 4pi*2.53e-8)

= 13.7 km

thats quite far...

assuming the true output is now 5% of the stated power (ie L=0.05*60)= 3W luminosity

D= 3km

as we can see the distance required to reach zero magnitude is dependant on the efficiency of the bulb...

which you could test..

take an exposure of vega to say 40000 DN on your camera.

take an exposure of bulb to reach same signal level.

the ratio of the two exposure times is the ratio of fluxes.

measure distance to bulb, use F= L*Q/4pi*r^2, where Q is the effiiciency.

we know that the flux of vega is aprox 2.5e-8 W/m2

since cameras are the same the effects of this should cancel.

I think this works?

paul

[EA2007]

you can't take credit for the equation, it was down to max plank.......damn! lol, atleast you understand it Paul, I need to revise my equations somewhat.