Inteference of incoherent light waves

AI Thread Summary
Interference of incoherent light waves can still occur, but the resulting pattern shifts constantly over time, leading to uniform illumination rather than stable interference fringes. Constructive and destructive interference happen rapidly, making it difficult for the human eye to perceive these changes, resulting in only bright regions being observed. The phase difference between incoherent sources varies continuously, which affects the intensity calculation. The average intensity for incoherent sources is constant, calculated as 2I, where I is the intensity of each slit. Thus, while interference is a phase-independent phenomenon, the rapid fluctuations in incoherent light prevent the detection of distinct patterns.
desmond iking
Messages
284
Reaction score
2

Homework Statement



i know that for intefernece of light waves to occur , the light waves must be coherent. How if two light waves are incoherent?
according to my notes, inteference can still occur if waves are not coherent, but the position of pattern will shift constantly with time to give a uniform illumination.


Homework Equations





The Attempt at a Solution


In my opinion , the constructive intefrence and destructive inteference occur too fast , namely countless times in a second . So our eyes can't detect it. Thus , only 'bright region' will be obersved all the time. please correct me if i am wrong.
 
Physics news on Phys.org
Yes, actually interference is a phase independent phenomena. If you have read YDSE, then you must know of the equation:-
Intensity(central maxima)=4Icos^2(δ/2)
where, δ and I are phase difference b/w the slits and intensity of the slit respectively.

Now, for incoherent sources, δ varies constantly,
So we have to take the average intensity, the average value of cos^2 function is (1/2)
Hence resultant intensity is constant=2I

Your conclusion is correct that our eyes can't detect that rapid change.
 
  • Like
Likes 1 person
I multiplied the values first without the error limit. Got 19.38. rounded it off to 2 significant figures since the given data has 2 significant figures. So = 19. For error I used the above formula. It comes out about 1.48. Now my question is. Should I write the answer as 19±1.5 (rounding 1.48 to 2 significant figures) OR should I write it as 19±1. So in short, should the error have same number of significant figures as the mean value or should it have the same number of decimal places as...
Thread 'A cylinder connected to a hanging mass'
Let's declare that for the cylinder, mass = M = 10 kg Radius = R = 4 m For the wall and the floor, Friction coeff = ##\mu## = 0.5 For the hanging mass, mass = m = 11 kg First, we divide the force according to their respective plane (x and y thing, correct me if I'm wrong) and according to which, cylinder or the hanging mass, they're working on. Force on the hanging mass $$mg - T = ma$$ Force(Cylinder) on y $$N_f + f_w - Mg = 0$$ Force(Cylinder) on x $$T + f_f - N_w = Ma$$ There's also...
Back
Top