Discussion Forum

Interference fringe are produced on a screen by using two light sources of intensities l and 9l. The phase difference between the beams is $\frac{\pi}{2}$  at point P and $\pi$  at point Q on the screen. The difference between the resultant  intensities at point P and Q is 

Answer:

Explanation:

According to question  , resultant  intensity of interferring wave is given by

 $l_{p}=l_{1}+l_{2}+2\sqrt{l_{1}l_{2}\cos \phi}$

 For  $\phi= \frac{\pi}{2},l_{p}=l+9l=10l$

 again at point Q, resultant intensity is given by

$l_{Q}=l_{1}+l_{2}+2\sqrt{l_{1}l_{2}\cos \phi}$

 For ,  $\phi= \pi, l_{Q}=l+9l+(-2\sqrt{9(l)^{2}}$)

 =10l-6l=4l

 Now, difference between the resultant intensity is given by 

$\triangle l=l_{P}-l_{Q}=10l-4l=6l$