# Black Body Radiation at a Distance from Emitter

• DanD
In summary, the conversation discusses the calculation of various parameters related to a 100 W light bulb with a filament temperature of 2000K. These include determining the diameter of the filament, the wavelength of the intensity peak in its emission, and the optical power entering the pupil when viewed through rose-tinted glasses. The calculations involve using Stefan's Law and Wien's Law, as well as considering the surface area of the filament and the solid angle of the pupil.
DanD
1.
6. A 100 W light bulb is designed to operate with it’s filament at 2000K. If the filament
is a perfect cylindrical Black Body and 2 cm long,

i) What must it’s 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=2.879x10^-3

3.For part 1 and 2 I have arrived to answers , i.) 1.68x10^-3 (using Stefans Law and rearranging the surface area of the cylinder) and ii.) 1.4485x10^-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 !

No 2D problem, but then: the 1/r2 you mention is a 3D outcome, so applicable (by happenstance?)

Just to help me find my own calculation error, could you show the part i) calculation ? (I end up with 1.75 mm, which I also consider pretty thick)

Also, Wien has 2.898 10-3 as a constant; where does the 2.879 come from ?

Then: for iii you need a ##{dR\over d\lambda }##. Which one do you add under 2. relevant formulas ?

Last edited:
Sorry my mistake I mean 2.898 .

It may have something to do with Planck distribution , calculate B then multiply by some other values to get the right dimensions .. i.e. area , solid angle and frequency.

In response to the diameter question in i.) -->
P=σAT^4

100= 5.67x10^-8 x Surface Area of the Cylinder x (2000)^4

Rearrange to get Surface Area = 100/(5.67x10^-8)x(2000)^4

Now looking at the Surface Area , we know there are two discs and the surface area of the "rectangle" which all together form to make the cylinder . Therefore surface area = (2)(∏d^2)/4 + ∏d(0.02)

Rearranging this you end up with : d^2 +0.04d - 200/(∏x5.67x10^-8)x(2000)^4 = 0

Using the Quadratic Formula you arrive to a positive and negative answer , disregarding the negative answer as you are looking for a length , I get 1.68x10^-3

No, I was the one who ignored the end discs (I am pretty sloppy, and after all, there has to be an entrance and an exit for the current...) which explains the ifference. So I don't think you made a mistake.

You plans look good to me!

As always, include units in your calculations and results.

I managed to do this question via the following , calculating B , Then doing the following Power = B*Area of Emitter*Solid Angle of the Pupil* Range of frequencies.

Thanks guys !

DanD said:
Sorry my mistake I mean 2.898 .

It may have something to do with Planck distribution , calculate B then multiply by some other values to get the right dimensions .. i.e. area , solid angle and frequency.

In response to the diameter question in i.) -->
P=σAT^4

100= 5.67x10^-8 x Surface Area of the Cylinder x (2000)^4

Rearrange to get Surface Area = 100/(5.67x10^-8)x(2000)^4

Now looking at the Surface Area , we know there are two discs and the surface area of the "rectangle" which all together form to make the cylinder . Therefore surface area = (2)(∏d^2)/4 + ∏d(0.02)

Rearranging this you end up with : d^2 +0.04d - 200/(∏x5.67x10^-8)x(2000)^4 = 0

Using the Quadratic Formula you arrive to a positive and negative answer , disregarding the negative answer as you are looking for a length , I get 1.68x10^-3

You are totally correct.

Last edited:

## 1. What is black body radiation at a distance from emitter?

Black body radiation refers to the electromagnetic radiation emitted by a perfect black body at a given temperature. At a distance from the emitter, this radiation can be measured and studied to understand the properties of the emitter.

## 2. How does the distance from the emitter affect black body radiation?

The intensity of black body radiation decreases as the distance from the emitter increases. This is due to the spreading out of the radiation as it travels further away from the source.

## 3. What is the relationship between temperature and black body radiation at a distance from emitter?

The temperature of the emitter directly affects the intensity and wavelength distribution of the black body radiation. As the temperature increases, the radiation shifts to shorter wavelengths and becomes more intense.

## 4. What is the significance of studying black body radiation at a distance from emitter?

Studying black body radiation at a distance from the emitter allows scientists to understand the properties and behavior of the emitter, as well as gather information about the surrounding environment. This can be useful in fields such as astrophysics and materials science.

## 5. How is black body radiation at a distance from emitter measured?

Black body radiation at a distance from the emitter can be measured using specialized instruments such as spectrometers or radiometers. These instruments measure the intensity and wavelength distribution of the radiation and can provide valuable data for analysis and research.

Replies
5
Views
703
Replies
10
Views
1K
Replies
7
Views
1K
Replies
2
Views
2K
Replies
7
Views
4K
Replies
1
Views
2K
Replies
6
Views
1K
Replies
14
Views
1K
Replies
3
Views
2K
Replies
2
Views
1K