Question

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

• A. 2 I
• B. 4 I
• C. 5 I
• D. 7 I

Answer 1

Answer: (B)

$I(\phi)=I_1+I_2+2 \sqrt{I_1 I_2} cos \, \phi ...(i)$
Here,$ \, \, \, \, \, I_1=I \, and \, I_2=4I$
At point A, $\phi=\frac{\pi}{2}$
$\therefore \, \, \, \, \, \, \, \, I_A=I+4I=5I$
At point B, $\phi=\pi$
$\therefore \, \, \, \, \, \, \, \, \, I_B=I+4I-4I=I$
$\therefore \, \, \, I_A-I_B=4I$
NOTE Eq. (i) for resultant intensity can be applied only when the
sources are coherent. In the question it is given that the rays
interfere. Interference takes place only when the sources are
coherent. That is why we applied equation number (i). When the
sources are incoherent, the resultant intensity is given by I = $I_1+I_2$