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- Path: sparky!uunet!srvr1.engin.umich.edu!uwm.edu!rpi!bu.edu!nntp-read!merritt
- From: merritt@macro.bu.edu (Sean Merritt)
- Newsgroups: sci.physics
- Subject: Re: Luminous Intensity
- Message-ID: <MERRITT.93Jan22112710@macro.bu.edu>
- Date: 22 Jan 93 16:27:10 GMT
- References: <4720060@hpcc01.corp.hp.com>
- Sender: news@bu.edu
- Organization: Boston University Physics Department
- Lines: 59
- In-reply-to: selee@hpcc01.corp.hp.com's message of 22 Jan 93 03:18:06 GMT
-
- In article <4720060@hpcc01.corp.hp.com> selee@hpcc01.corp.hp.com (Lee Say Eng) writes:
-
-
-
-
- > If two light rays of luminous intensities I_1 and I_2 are in the
- > same direction and are a small distance x apart, what would be the
- > apparent total intensity to an observer a long distance away?
- >
- > Is it simply I_total = I_1 + I_2 ?
-
-
- (Ignoring losses in intensity as the light rays traverses thru space)
-
- Consider two plane light-waves:
-
- E1 = A1*exp i(k*x-wt)
-
- E2 = A2exp i(k*x-wt+p)
-
- the A's are amplitude
- k is the wave vector propagating in the x dir
- w is 2*pi*frequency
- p is a phase difference
-
- If these waves are to interfere (not clear by the way you state the
- problem , ie how small is a compared to beam width?),
- intensity goes as the square of the fields.
-
-
- I = |E1 + E2|**2 = (E1+E2)^*(E1+E2) where (E1+E2)^* is the complex conj.
-
-
- I = A1**2 + A2**2 + 2*A1*A2* cos(p) = I1 + I2 + 2*I12*cos(p)
-
- The maximum is:
-
- Imax= I1 + I2 + 2*I12 cos(p) = 1 p = 0, or n*2*pi, n=0,1,2...
-
- or Imax = ( A1 + A2 )**2
-
- the minimum is:
-
- Imin = I1 + I2 - 2*I12 cos(p) = -1 p= pi, or m*pi, m =0,1,2...
-
- or Imin = ( A1 - A2 )**2
-
- for cos(p) = 0 p = pi/2 or j*pi/2 (j=m+1)
-
- I = I1 + I2 = A1**2 + A2**2.
-
- your "simple" case.
-
- -sjm
-
- --
- Sean J. Merritt |"Road-kill has it's seasons just like
- Dept of Physics Boston University|anything, there's possums in the autumn
- merritt@macro.bu.edu |and farm cats in the spring." T. Waits
-