Wave Optics

JEE Main level, numerical answer, medium

73 of 191

Question 73 of 191

Two beams of light having intensities II and 4I4I interfere to produce a fringe pattern on a screen. The phase difference between the two beams are π2\dfrac{\pi }{2} and π3\dfrac{\pi }{3} at points AA and BB respectively. The difference between the resultant intensities at the two points is xIxI. The value of xx will be _____ .

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Answer: 2

Given here, phase difference at point AA and BB are ϕA=π2{\phi }_{A}=\dfrac{\pi }{2}and ϕB=π3{\phi }_{B}=\dfrac{\pi }{3}.
Resultant intensity at point AA,
IA=I+4I+2I4Icos⁡(π2){I}_{A}=I+4I+2\sqrt{I}\sqrt{4I}\cos \left(\dfrac{\pi }{2}\right)
⇒IA=5I+4I(0)=5I\Rightarrow {I}_{A}=5I+4I\left(0\right)=5I
Resultant intensity at point BB,
IB=I+4I+2I4Icos⁡(60∘){I}_{B}=I+4I+2\sqrt{I}\sqrt{4I}\cos \left(60^{\circ}\right) =5I+4I×12=7I=5I+4I\times \dfrac{1}{2}=7I
Difference in intensities is IB−IA=7I−5I=2I{I}_{B}-{I}_{A}=7I-5I=2I.
Thus, value of x=2x=2.

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