Optics 6 Question 23
25. Two beams of light having intensities $I$ and $4 I$ 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
(2001, 2M)
(a) $2 I$
(b) $4 I$
(c) $5 I$
(d) 7 I
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Solution:
$$ I(\varphi)=I _1+I _2+2 \sqrt{I _1 I _2} \cos \varphi $$
Here, $\quad I _1=I$ and $I _2=4 I$
At point $A, \varphi=\frac{\pi}{2}$
$\therefore \quad I _A=I+4 I=5 I$
At point $B, \varphi=\pi$
$$ \begin{aligned} & \therefore \quad I _B=I+4 I-4 I=I \\ & \therefore \quad I _A-I _B=4 I \end{aligned} $$
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 $I=I _1+I _2$
