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# Two beam of light having intensities $I$ and $4I$ interfere to produce a fringe pattern on a screen . The phase difference between beams is $\large\frac{\pi}{2}$ at point A and n at point B. Then the difference between resultant intensities at A and B is

$(a)\;2I \\ (b)\;4I \\ (c)\;5I \\ (d)\;7I$

$I_R= I+ 4I+2 (I.4I)^{1/2} \cos \phi$
At $A, \phi =\large\frac{\pi}{2}$$=> I_A=5I$
At $B, \phi= \pi$
$I_B=I$
$I_A-I_B=4I$
Hence b is the correct answer.

edited Jul 23, 2014