Intensity of light at a point

1. Jan 12, 2007

Amith2006

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

2. Jan 12, 2007

Mindscrape

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.

3. Jan 13, 2007

Amith2006

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!

4. Jan 13, 2007

arunbg

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 $$\sqrt{x^2+y^2}$$ and the angle wrt the x axis is given by $$tan^{-1}(\frac{y}{x})$$. 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 ?

5. Jan 13, 2007

Amith2006

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]

6. Jan 13, 2007

Mindscrape

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.

7. Jan 13, 2007

Amith2006

Thanks for sharing your knowledge with me.