Verified Solution • 2025–26 Session moderate

QJEEM23Ph6d3f: JEE Main 2023 Physics - Class 12 JEEM23Ph6d3f Solution: In a Young's double slit experiment, the intensities at two…

In a Young's double slit experiment, the intensities at two points, for the path differences λ4\frac{\lambda}{4} and λ3\frac{\lambda}{3} ( λ\lambda being the wavelength of light used) are I1I_{1} and I2I_{2} respectively. If I0I_{0} denotes the intensity produced by each one of the individual slits, then I1+I2I0=\frac{I_{1}+I_{2}}{I_{0}}= __________.

Step-by-Step Verified Solution

I′=Icos⁡2(kΔx2)I' = I{\cos ^2}\left( {{{k\Delta x} \over 2}} \right)

so I1=4I0cos⁡2(2π2λ×λ4){I_1} = 4{I_0}{\cos ^2}\left( {{{2\pi } \over {2\lambda }} \times {\lambda \over 4}} \right)

I1=2I0{I_1} = 2{I_0}

& I2=4I0cos⁡2(2π2λ×λ3){I_2} = 4{I_0}{\cos ^2}\left( {{{2\pi } \over {2\lambda }} \times {\lambda \over 3}} \right)

I2=I0{I_2} = {I_0}

So I1+I2I0=3{{{I_1} + {I_2}} \over {{I_0}}} = 3

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