# Black Body Radiation at a Distance from Emitter

**1.**

6. A 100 W light bulb is designed to operate with its filament at 2000K. If the filament

is a perfect cylindrical Black Body and 2 cm long,

6. A 100 W light bulb is designed to operate with its filament at 2000K. If the filament

is a perfect cylindrical Black Body and 2 cm long,

i) What must its diameter be (3 marks)

ii) What will be the wavelength of the intensity peak in its emission. (2 marks)

**iii) You look at this bulb, from a distance of 1 m, through a pair of rose-tinted glasses
that pass only wavelengths 630-650 nm. Assuming that the filament looks like a
uniformly emitting rectangle that is 2cm long, and as wide as your answer to i)
Estimate the optical power entering your pupil if it is 4 mm diameter.
**

**2. Stefan’s Law : P=σAT^4 , Wiens Law : λT=2.879×10^-3**

**3.For part 1 and 2 I have arrived to answers , i.) 1.68×10^-3 (using Stefans Law and rearranging the surface area of the cylinder) and ii.) 1.4485×10^-6 using Wiens Law**

I am confused on part iii. I believe what the process should b is , 1.) Calculate new power as it is no longer a cylinder and is now treated as a rectangle. 2.) Calculate the factor of which the power drops over the distance you are from it . 3.) Then calculate as a proportion of the total area ? The amount of power the pupil itself is receiving . I am not sure how to calculate the power change over the distance . I have drawn a diagram and I think that the light should emit in a circle (as we are only talking 2 dimensions ) Therefore use 1/r^2 law to calculate a proportion ? I’m not entirely sure.

Any help would be appreciated ,

Thanks

Dan !

**1. The problem statement, all variables and given/known data**

**2. Relevant equations**

**3. The attempt at a solution**

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