Question
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 $${\frac{\pi }{2}}$$ at point $$A$$ and $$\pi $$ at point $$B.$$ Then the difference between the resultant intensities at $$A$$ and $$B$$ is
A.
$$2\,I$$
B.
$$4\,I$$
C.
$$5\,I$$
D.
$$7\,I$$
Answer :
$$4\,I$$
Solution :
$$I = {I_1} + {I_2} + 2\sqrt {{I_1}} \sqrt {{I_2}} \cos \phi \,\,.....\left( 1 \right)$$
Applying eq. (1) when phase difference is $$\frac{\pi }{2}$$
$$\eqalign{
& {I_{\frac{\pi }{2}}} = I + 4\,I \cr
& \Rightarrow \,\,{I_{\frac{\pi }{2}}} = 5I \cr} $$
Again applying eq. (1) when $$d$$ phase difference is $$\pi $$
$$\eqalign{
& {I_\pi } = I + 4\,I + 2\sqrt I \sqrt {4\,I} \cos \pi \cr
& \therefore \,\,{I_\pi } = I \cr
& \therefore \,\,{I_{\frac{\pi }{2}}} - {I_\pi } = 4\,I \cr} $$