In Young's double slit experiment, at two points P and Q on screen, waves from slits s₁ and s₂ have a path difference of 0 and 4 respectively. The ratio of intensities at point P to that at Q will be ( 0^ =1, 45^ = 1 2 )

Step-by-step solution

Intensity in Young's double-slit experiment follows I = I_ ^2 ( 2 ) , where = 2 x relates phase difference to path difference. At point P where x_P = 0 , the phase difference is _P = 0 , yielding intensity I_P = I_ ^2(0) = I_ . At point Q with path difference x_Q = 4 , the phase difference becomes _Q = 2 4 = 2 . The intensity is I_Q = I_ ^2 ( 4 ) = I_ ( 1 2 )^2 = I_ 2 . The intensity ratio is I_P I_Q = I_ I_ /2 = 2 , giving I_P : I_Q = 2:1 . 2:1