Question

The intensity of illumination, I, produced by a light of power P at a point distant x m from the light is given by \[I =frac{ P }{ x^2 }\] Two lights, of powers P and 2P are 20 metres apart. Find to the nearest metre the distance from the less powerful light at which the intensity of illumination is least.

Answer 1

\[I=\frac{ P }{ x^2 }\]

Answer 2

Okay Hero.

Answer 3

Thanks for that.

Answer 4

Is there some assumption not stated about where the measurement point might be? For example, are we looking for a spot somewhere on the line between the two sources?

Answer 5

Yeah I think so. I'm not particularly sure about the wording of the question. That's why I'm asking. But I think it's a spot between the two light sources.

Answer 6

what happens when two lights illuminate the same spot? do we take the average of the intensity at that spot?

Answer 7

do we add them together

Answer 8

i would assume we would add them but this is a physics problem and there is probably something i'm missing

Answer 9

if its just the summation then i would suggest minimizing the equation
\[\frac{ 2P }{ (20 - x)^{2} } + \frac{ P }{ x^{2} }\]

Answer 10

It's not a Physics question. It looks like one, but it's not.

Answer 11

i would assume then that the intensity at any given point is the summation of the two intensities. if so then

\[I _{1} = \frac{ 2P }{ (20 - x )^{2}}\]

that is the brightest intensity bulb

and the other would would be

\[I _{2} = \frac{ P }{ x^{2} }\]

where x is the distance from the bulb

I assume then that the combined intensity at any point is just I1 + I2

I suggest to minimize I1 + I2 using differentiation

Answer 12

So we add the two and then differentiate? Because if so, don't think that's going to work.

Answer 13

yeah you'll get

\[-\frac{ 2P }{ x^{3} } - \frac{ 4P }{ (x - 20)^{3} }\]

then you can set this = 0

Answer 14

Both aren't negative. One's got to be positive, if you move them to one side.

Answer 15

But I don't think that's how you do it. I will wait for whpalmer's reply.

Answer 16

Isn't the spot where the minimum illumination occurs going to be the spot where the illumination from each source is equal?

Answer 17

Nope.

Answer 18

the one with the less power which is the one that's P.

Answer 19

oh, well if you solve for the derivative = 0 you will get x = about 8.84989 meters from the less intense bulb

Answer 20

That's right!!!!!

Answer 21

and this is a verified minimum because values close to that result in a higher intensity

Answer 22

WTF...It's 9m

Answer 23

Can you get it to exxactly 9?

Answer 24

nearest meter

Answer 25

exactly*

Answer 26

oh, just round up then

Answer 27

Don't think the answers rounded up. Not sure, but unlikely it will need rounding up. Because if it did, the question would tell you.

Answer 28

question asked for "to the nearest metre"

Answer 29

idk, that's what i get

Answer 30

In my previous question, I had $403, but the answers had $400. So I did it again, and then found out something about the question, and redid it using something else. Got it to $400.

Answer 31

Wait hold on for a minute a two there.

Answer 32

Ah okay yeah you're right.

Answer 33

Sorry about that guys. Thanks mate.

Answer 34

Thanks both of yous. can't give you guys both medals. Lolz....

Answer 35

Fun problem!

Answer 36

@whpalmer4 @binarymimic Holy cow, I couldn't get it. Could you show me the working?

Answer 37

How the hell do you factor x^6? That's where I ended when I made the derivative=0

Answer 38

The derivative of P/x^2 is

\[-\frac{ 2P }{ x^3 }\]

and the derivative of
2P/(20 - x)^2 is

\[\frac{ 4P }{ (20 - x)^3 }\]

Answer 39

You forgot the minus.

Answer 40

I'm stuck at when you make the whole thing equal to zero,

Answer 41

minus?

Answer 42

\[-\frac{ 4P }{ (20-x)^3 }\]

Answer 43

oh wait nvm

Answer 44

You're correct. But I'm stuck at when you make them equal to zero.

Answer 45

And isn't it 4Px*

Answer 46

not 4P?

Answer 47

wait nvm that either lolz. I'm at a total mess here.

Answer 48

we want to take their sum and equal it to 0 so

\[\frac{ 4P }{ (20 - x)^{3} } -\frac{ 2P }{ x^{3} } = 0\]

Answer 49

Yeah I did that, Do you combine it?

Answer 50

You could set up the equality

4P * x^3 = 2P * (20 - x)^3

Answer 51

Yeah but how do you factorise a cubic polynomial?

Answer 52

I can't seem to factorise the cubic polynomial*

Answer 53

we dont have to factor anything really

first cancel the Ps and divide by 2 to get

\[2x^{3} = (20 - x)^{3}\]

Answer 54

see where we're going ?

Answer 55

Yeah you cube root everything. So you get cube root2x which equals 20-x?

Answer 56

yep

Answer 57

Ah okay, so you get 8.84?

Answer 58

8.85*

Answer 59

yes thats what i get

Answer 60

How do you know it's the minimum distance? Would you have to differentiate the derivative?

Answer 61

i just tested values on the original equation close to that point and found that if i used a value smaller than 8.85 then the combined intensity was larger, and if i used a value larger than 8.85, the combined intensity was still larger

thats why i concluded it was a minimum

Answer 62

Ah okay.

Answer 63

If you want to take a second derivative then .. . >_>

Answer 64

lolz, no way man. That's gonna take a trillion years.

Answer 65

i used wolfram to do it, and then evaluated it at 8.849866680, and it returned a positive value. that means it is concave up at that point, so still a minimum :D

Answer 66

ah okay. COuld you check this closed question how the answers got when f increases x<0?

Answer 67

http://openstudy.com/study#/updates/50eab348e4b0d4a537cc3994