• Support PF! Buy your school textbooks, materials and every day products Here!

Intensity of light at a point

  • Thread starter Amith2006
  • Start date
427
2
1. Homework Statement
1) At a given point in space, the total light wave is composed of 3 phasors P1 = a, P2 = (a/2)e^(i(theta)), P3 = (a/2)e^(-i(theta)). What is the intensity of light at this point?



2. Homework Equations



3. The Attempt at a Solution

I solved it in the following way:
P = P1 + P2 + P3
= a + (a/2)e^(i(theta)) + (a/2)e^(-i(theta))
= a + (a/2)[cos(theta) + isin(theta) + cos(theta) - isin(theta)]
= a + acos(theta)
= 2acos^2(theta/2)
I = P^2
= 4a^2cos^4(theta/2)

But if P3 = ae^(-i(theta)) then this method won’t work, isn’t it? So could someone suggest a more general way to solve such problems?
 

Answers and Replies

1,860
0
I don't know what you did for the fourth step, the cos^2 one. Seems okay if that fourth step is legitimate. Why doesn't the method work when you change P3?

The general method to adding sinusoids is exactly what you did. You want to convert them to phasors, as was done for you, and then add in their cartesian coordinates. Then, usually it is nice to take your summation from cartesian back to polar.
 
427
2
I don't know what you did for the fourth step, the cos^2 one. Seems okay if that fourth step is legitimate. Why doesn't the method work when you change P3?

The general method to adding sinusoids is exactly what you did. You want to convert them to phasors, as was done for you, and then add in their cartesian coordinates. Then, usually it is nice to take your summation from cartesian back to polar.
Suppose P1 = a, P2 = (a/2)e^(i(theta)), P3 = (a)e^(-i(theta))
P = a + (a/2)e^(i(theta)) + (a)e^(-i(theta))
= a + (a/2)[cos(theta) + isin(theta)] + a[cos(theta) - isin(theta)]
= a + (a/2)[3cos(theta) - isin(theta)]
How do u proceed further? In the previous case, the imaginary part got cancelled, but that is not the case here.Please help!:confused:
 
587
0
Note that waves don't actually have imaginary parts. The i seen in the phasor representation is actually the rectangular form for representation of the phasor.
This representation is adopted to make addition of phasors easier. After operating, it is better to convert to polar form which gives you magnitude and relative angle. In essence these are mathematical tools that make such operations easier.

To give you an idea, use what you have learnt in complex nos. The absolute value of the complex no is given by [tex]\sqrt{x^2+y^2}[/tex] and the angle wrt the x axis is given by [tex]tan^{-1}(\frac{y}{x})[/tex]. Similarly, in the case of phasors the first former gives the magnitude and the latter, the phasor angle relative to other phasors. Can you do your problem now ?
 
427
2
Note that waves don't actually have imaginary parts. The i seen in the phasor representation is actually the rectangular form for representation of the phasor.
This representation is adopted to make addition of phasors easier. After operating, it is better to convert to polar form which gives you magnitude and relative angle. In essence these are mathematical tools that make such operations easier.

To give you an idea, use what you have learnt in complex nos. The absolute value of the complex no is given by [tex]\sqrt{x^2+y^2}[/tex] and the angle wrt the x axis is given by [tex]tan^{-1}(\frac{y}{x})[/tex]. Similarly, in the case of phasors the first former gives the magnitude and the latter, the phasor angle relative to other phasors. Can you do your problem now ?
I have solved it with whatever I could understand from you. Please see if it is right. I have assumed a to be amplitude.
P = (a/2)[(2 + 3cos(theta)) - isin(theta)]
|P| = (a/2)[(2 + 3cos(theta))^2 + sin^2(theta)]^(1/2)
= (a/2)[4 + 9cos^2(theta) + 6cos(theta) + sin^2(theta)]^(1/2)
= (a/2)[8cos^2(theta) + 6cos(theta) + 5]^(1/2)
Intensity = |P|^2
= ((a^2)/4)[ 8cos^2(theta) + 6cos(theta) + 5]
 
1,860
0
Yes, that would be the intensity, but you don't want to lose the phase information either. I would keep it until you directly report your answer.
 
427
2
Thanks for sharing your knowledge with me.
 

Related Threads for: Intensity of light at a point

  • Last Post
Replies
8
Views
3K
  • Last Post
Replies
1
Views
1K
  • Last Post
Replies
2
Views
5K
  • Last Post
Replies
1
Views
3K
  • Last Post
Replies
9
Views
3K
Replies
1
Views
2K
Replies
2
Views
30K
  • Last Post
Replies
3
Views
680
Top