Light of wavelength 7400 nm passes through a single slit of

Light of wavelength 740.0 nm passes through a single slit of width 2.20 mu m, and a diffraction pattern is observed on a screen 27.5 cm away. Determine the relative intensity of light I/I_max at 17.0 cm away from the central maximum. I/I_max = 8.7e^-03

Solution

wavelength = 740nm

a = 2.20 micro m

D = 27.5 cm

y= 17cm

intensity is

I  =  I₀(sin /)²   where

I  =  I₀(sin /)²   where     = a sin /.

    = a sin /. =   ay/D = 3.14*2.20*10^-6*0.17/(740*10^-9*0.275) = 5.77

I  =  I₀(sin /)²   

I/Io = (sin(5.77)/5.77)^2 =0.01563