Question

You pass 633 nm laser light through a narrow single slit of width 0.24 mm and observe the diffraction pattern on a screen 6.0 m away. If the intensity at the central bright is Io , what is the intensity at a point on the screen 3.0 mm from the center of the pattern?

Answer 1

The intensity is
`I  = 0.0175I_o`

Step-by-step explanation:

From the question we are told that

The wavelength is
`\lambda = 633 \ nm = 633*10^(-9) \ m`

The width of the slit is
`d = 0.24 \ mm = 0.00024 \ m`

The distance from the screen is
`D = 6.0 \ m`

The distance of the position considered from the center is
`y = 3.0 \ mm = 0.003 \ m`

Generally the intensity from at a point on the screen 3.0 mm from the center of the pattern is

`I = I_o * (sin^2 [(\pi d * y)/( \lambda D ) ])/([(\pi d * y )/(\lambda D) ]^2)`

Here
`I_o` is the intensity of the central bright fringe

=>
`I  = I_o *  (sin^2 [(3.142  * 0.00024 * 0.003)/( 633*10^(-9) * 6 ) ])/([(3.142*0.00024  * 0.003 )/(633*10^(-9) * 6) ]^2)`

=>
`I  = 0.0175I_o`