Verified Solution • 2025–26 Session moderate

QJEEM19Ph7da4: JEE Main 2019 Physics - Class 12 JEEM19Ph7da4 Solution: Two coherent sources produce waves of different intensities…

Two coherent sources produce waves of different intensities which interfere. After interference, the ratio of the maximum intensity to the minimum intensity is 16. The intensity of the waves are in the ratio :

Multiple Choice Options

  • A 16:09
  • B 25:09:00
  • C 4:01
  • D 5:03

Step-by-Step Verified Solution

Given that,

Imax⁡Imin=161{{{{\rm I}_{\max }}} \over {{{\mathop{\rm I}\nolimits} _{min}}}} = {{16} \over 1}

We know,

Imax == (I1+I2)2{\left( {\sqrt {{{\rm I}_1}} + \sqrt {{{\rm I}_2}} } \right)^2}

and Imin =(I1−I2)2= {\left( {\sqrt {{{\rm I}_1}} - \sqrt {{{\rm I}_2}} } \right)^2}

∴\therefore     (I1+I2)2(I1−I2)2=161{{{{\left( {\sqrt {{{\rm I}_1}} + \sqrt {{{\rm I}_2}} } \right)}^2}} \over {{{\left( {\sqrt {{{\rm I}_1}} - \sqrt {{{\rm I}_2}} } \right)}^2}}} = {{16} \over 1}

⇒\Rightarrow     I1+I2I1−I2=41{{\sqrt {{{\rm I}_1}} + \sqrt {{{\rm I}_2}} } \over {\sqrt {{{\rm I}_1}} - \sqrt {{{\rm I}_2}} }} = {4 \over 1}

⇒\Rightarrow     4I1−4I2=I1+I24\sqrt {{{\rm I}_1}} - 4\sqrt {{{\rm I}_2}} = \sqrt {{{\rm I}_1}} + \sqrt {{{\rm I}_2}}

⇒\Rightarrow     3I1=5I23\sqrt {{{\rm I}_1}} = 5\sqrt {{{\rm I}_2}}

⇒\Rightarrow     I1I2=53{{\sqrt {{{\rm I}_1}} } \over {\sqrt {{{\rm I}_2}} }} = {5 \over 3}

⇒\Rightarrow     I1I2=259{{{{\rm I}_1}} \over {{{\rm I}_2}}} = {{25} \over 9}

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