Verified Solution • 2025–26 Session moderate

QJEEM22Ph862a: JEE Main 2022 Physics - Class 12 JEEM22Ph862a Solution: The interference pattern is obtained with two coherent light…

The interference pattern is obtained with two coherent light sources of intensity ratio 4 : 1. And the ratio Imax⁡+Imin⁡Imax⁡−Imin⁡{{{I_{\max }} + {I_{\min }}} \over {{I_{\max }} - {I_{\min }}}} is 5x{5 \over x} . Then, the value of x will be equal to :

Multiple Choice Options

  • A 3
  • B 4
  • C 2
  • D 1

Step-by-Step Verified Solution

Imax⁡+Imin⁡Imax⁡−Imin⁡=I1+I2+2I1I2+I1+I2−2I1I2I1+I2+2I1I2−I1−I2+2I1I2{{{I_{\max }} + {I_{\min }}} \over {{I_{\max }} - {I_{\min }}}} = {{{I_1} + {I_2} + 2\sqrt {{I_1}{I_2}} + {I_1} + {I_2} - 2\sqrt {{I_1}{I_2}} } \over {{I_1} + {I_2} + 2\sqrt {{I_1}{I_2}} - {I_1} - {I_2} + 2\sqrt {{I_1}{I_2}} }}

=2(I1+I2)4I1I2= {{2({I_1} + {I_2})} \over {4\sqrt {{I_1}{I_2}} }}

=(I1I2+1)2I1I2=4+12×2=54= {{\left( {{{{I_1}} \over {{I_2}}} + 1} \right)} \over {2\sqrt {{{{I_1}} \over {{I_2}}}} }} = {{4 + 1} \over {2 \times 2}} = {5 \over 4}

So x=4x = 4

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