In young's double slit experiment, light of wavelength 480 ~nm is incident on two slits separated by a distance of 4 10⁻⁴ ~m . If a thin plate of thickness 1.4 10⁻⁶ ~m and refractive index 13 7 is placed between one of the slits and screen, the phase difference introduced at the position of central maxima is

Step-by-step solution

The YDSE setup is shown below Additional path difference introduced by medium plate, L=( -1) t As, a path difference of one wavelength is equals to a phase difference = L 2 so, = ( -1) t 2 Here, = 13 7 , t=1.4 10⁻⁶, =480 10⁻⁹ ~m = ( 13 7 -1 ) 1.4 10⁻⁶ 2 480 10⁻⁹ = 6 1.4 2 10⁻⁶ 7 480 10⁻⁹ =5 rad