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QJEEM22Phbf58: JEE Main 2022 Physics - Class 12 JEEM22Phbf58 Solution: Two beams of light having intensities I and 4I interfere to…

Two beams of light having intensities I and 4I interfere to produce a fringe pattern on a screen. The phase difference between the two beams are π/2\pi / 2 and π/3\pi / 3 at points A\mathrm{A} and B\mathrm{B} respectively. The difference between the resultant intensities at the two points is xIx I . The value of xx will be ________.

Step-by-Step Verified Solution

IR1=I1+I2+2I1I2cos⁡ϕ{I_{{R_1}}} = {I_1} + {I_2} + 2\sqrt {{I_1}{I_2}} \cos \phi

IA=I+4I+2I.4Icos⁡90∘{I_A} = I + 4I + 2\sqrt {I.4I} \cos 90^\circ

=5I= 5I

IB=I+4I+2I.4Icos⁡60∘{I_B} = I + 4I + 2\sqrt {I.4I} \cos 60^\circ

=7I= 7I

IB−IA=2I{I_B} - {I_A} = 2I

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