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QJEEM22Phd69b: JEE Main 2022 Physics - Class 12 JEEM22Phd69b Solution: Two light beams of intensities 4I and 9I interfere on a screen. …

Two light beams of intensities 4I and 9I interfere on a screen. The phase difference between these beams on the screen at point A is zero and at point B is π\pi . The difference of resultant intensities, at the point A and B, will be _________ I.

Step-by-Step Verified Solution

IA=(I1+I2)2=25I{I_A} = {\left( {\sqrt {{I_1}} + \sqrt {{I_2}} } \right)^2} = 25I

IB=(I1−I2)2=I{I_B} = {\left( {\sqrt {{I_1}} - \sqrt {{I_2}} } \right)^2} = I

So, IA−IB=24I{I_A} - {I_B} = 24I

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