Verified Solution • 2025–26 Session moderate

QJEEM19Phb732: JEE Main 2019 Physics - Class 12 JEEM19Phb732 Solution: In a Young's double slit experiment, the ratio of the slit's…

In a Young's double slit experiment, the ratio of the slit's width is 4 : 1. The ratio of the intensity of maxima to minima, close to the central fringe on the screen, will be :

Multiple Choice Options

  • A 25:09:00
  • B 4:01
  • C (3+1)4:16{\left( {\sqrt 3 + 1} \right)^4}:16
  • D 9:01

Step-by-Step Verified Solution

I1=4I0{I_1} = 4{I_0}

I2=I0{I_2} = {I_0}

Imax⁡=(I0+I2)2{I_{\max }} = {\left( {\sqrt {{I_0}} + \sqrt {{I_2}} } \right)^2}

=(2I0+I0)2=9I0= {\left( {2\sqrt {{I_0}} + \sqrt {{I_0}} } \right)^2} = 9{I_0}

Imin⁡=(I1−I2)2{I_{\min }} = {\left( {\sqrt {{I_1}} - \sqrt {{I_2}} } \right)^2}

=(2I0−I0)2=I0= {\left( {2\sqrt {{I_0}} - \sqrt {{I_0}} } \right)^2} = {I_0}

∴\therefore Imax⁡Imin⁡=9I0I0=91{{{{\mathop{\rm I}\nolimits} _{\max }}} \over {{I_{\min }}}} = {{9{I_0}} \over {{I_0}}} = {9 \over 1}

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