In a Young's double slit experiment, the separation between the slits = 2.0 mm, the wavelength of the light = 600 nm and the distance of the screen from the slits = 2.0 m. If the intensity at the centre of the central maximum is `0.20 W m^-2`, what will be the intensity at a point 0.5 cm away from this centre along the width of the fringes ?

Answer

B

Explanation

Given that: `d=2mm=2xx10^-3m`
`l=600nm`
`=6xx10^-7m`
`I_(max)=0.20W/m^2,D`
=2m
for the point y=0.5cm
We know path difference `=x=(yd)/D`
`=0.5xx10^I-2xx2xx10^-3/2`
`=5xx10^-5m`
So the corresponding phase difference is
`phi=(2pix)/lamda=(2pixx5xx10^-6)/(6xx10^-7)`
`=(50pi)/3=(3xx16pi+2pi)/3`
`rarr phi=(2pi)/3`
so the amplitude of the resulting wave at the point y=0.5 cm is
`A=sqrt(r^2+r^2+2r^2cos(2pi)/3)0`
`sqrt(r^2+r^2-r^2)=r`
`Since I/I_(max)-A^2/(2r^2)`
[since maximum amplitude -2r]
`rarr I/0.2=A^2/(4r^2)=r^2/(4r^2)`
`rarr I=.2/4=0.05W/m^2`