Inverse of distance squared? (1 Viewer)

[747captain]

Hey guys, just got some h/w yesterday about "Inverse of distance squared". We have to complete a table which consists of four columns:

• Column 1 - distance (metres); 0.1, 0.2, 0.3, 0.4, 0.5, 0.6, 0.7, 0.8, and 0.9
• Column 2 - time (seconds); (blank to fill in)
• Column 3 - Total area (metres squared); (blank to fill in)
• Column 4 - Intensity; (blank to fill in)

Question: A point source emits 1000J of energy. Calculate the energy per square metre as it moves away from the point source. Use the data in column one and the speed of light (3.0 x 10^8) to complete the table.

Can anyone help me in filling this table out, as my teacher doesn't explain new material such as this very well. Would greatly appreciate it.

Cheers

 

[alcalder]

Hey guys, just got some h/w yesterday about "Inverse of distance squared". We have to complete a table which consists of four columns:

• Column 1 - distance (metres); 0.1, 0.2, 0.3, 0.4, 0.5, 0.6, 0.7, 0.8, and 0.9
• Column 2 - time (seconds); (blank to fill in)
• Column 3 - Total area (metres squared); (blank to fill in)
• Column 4 - Intensity; (blank to fill in)

Question: A point source emits 1000J of energy. Calculate the energy per square metre as it moves away from the point source. Use the data in column one and the speed of light (3.0 x 10^8) to complete the table.

Can anyone help me in filling this table out, as my teacher doesn't explain new material such as this very well. Would greatly appreciate it.

Cheers

No probs. The equation you are using is the following:
I (proportional to) 1/d²

Actually, it is

Brightness (Intensity) = luminosity/Area

Luminosity is the Energy per second reaching that point.
Brightness or Intensity is the Energy per second per square metre reaching a point distance d from a light source.

Column 1 = Just a list of how far away from the point source you are.
Column 2 = How long it would take to get there. v = s/t
velocity = speed of light, distance is in column 1
Column 3 = The area over which the point source has spread. REMEMBER it has spread over the entire surfac area of a sphere with radius d. (ie 4 x pi x d²)
Column 4 = (Initial Energy intensity/time from column 2)/ result in column 3

So the first row of the table would look like this:

Distance (metres) = 0.1
Time (seconds) = 0.1 / 3 x 10⁸ = 3.33 x 10^-10
Area (metres squared) = 4 x pi x 0.1² = 0.12566 m²
Intensity (Joules) = 2.39 x 10¹⁴ Js^-1m^-2


That should be more like it. Now do that graph.

 

[ianc]

Yeah, alcalder is perfectly correct.

Basically the point of the exercise is to show that the energy emitted does not decrease according to the distance but rather the area...hence Intensity is proportional to 1/d²


Hopefully once you see it all laid out in a table it will make sense:

 

[747captain]

Cool. Thankyou so much ianc and alcalder. I really appreciate it, and I now understand how it all works.

However, one last thing. I forgot to mention part 2 of the question:
Draw a graph of I(intensity) against the inverse of distance squared.

What I don't understand is that the inverse of distance squared is intensity. So how can i possibly draw a graph to compare the same figures?

 

[ianc]

no worries, pleased i could help.

bear in mind that the formula for the intensity of light is not necessarily I=1/d², however it is proportional so it could have some sort of constant making it I=k/d² where k is a constant. (thats just so you know, you don't need it for this question)

are you sure you typed the question correctly? because, i agree, as it is written there it doesnt make any sense.

i would assume they mean to plot the intensity against the distance, like so:

(sorry, i couldnt be bothered working out how to axis titles on)

 

[alcalder]

OK, I know where I went wrong.

Go back and look at my original post after yours.

 

[ianc]

ok cool, i didnt think about it being a sphere :S

 

[747captain]

Hey everyone thanks for your help.

However, my teacher went through this homework in class today and it appears that some of it is wrong.

• My teacher says the formula to complete the area column is 4[FONT=&quot]πr[/FONT]^{[FONT=&quot]2[/FONT]}[FONT=&quot]. This doesn't seem right to me.[/FONT]
• [FONT=&quot]Also, the time is actually the distance[/FONT][FONT=&quot] ÷ speed-of-light.[/FONT]

[FONT=&quot]Therefore, because of the above two points, the answers for intensity are also wrong.

Could you guys please check this out? Please refer to my attached word document which contains the table itself.
Cheers.[/FONT]