In Young's double slit experiment, the intensity of light at a point on the screen where the path difference is is ' I '. The intensity at a point where the path difference is / 6 is [ 6 = 3 2 ][ = wavelength of light ][ =-1]

Step-by-step solution

The intensity of light at any point in Young's double-slit experiment is determined by the phase difference between interfering waves, following the relation I_p = I_ ^2( /2) , where = (2 / ) x relates phase to path difference. Since intensity I occurs when x = , yielding phase = 2 , this corresponds exactly to maximum intensity: I = I_ ^2( ) = I_ . Thus I_ = I . When the path difference becomes x = /6 , the phase becomes = /3 . The intensity at this point is then I₂ = I_ ^2( /6) = I ( 3 /2)^2 = 3I/4 . With ( /6) =