Question

A diffraction pattern is formed on a screen 150 cm away from a 0.500-mm-wide slit. Monochromatic 546.1-nm light is used. Calculate the fractional intensity I/Imax at a point on the screen 4.10 mm from the center of the principal maximum.

Answer 1

The expression for the intensity of light is given by :

`$I=I_(max)\left((\sin (\pi a \sin \theta)/(\lambda))/((\pi a \sin \theta)/(\lambda) )\right)^2$`

For a small angle, θ

sin θ = tan θ

`$=(y)/(L)$`

Therefore the above equation becomes,

`$I=I_(max)\left((\sin (\pi a y)/(\lambda L))/((\pi a y)/(\lambda L) )\right)^2 $`

The given data is

λ = 546.1 nm

L = distance between the slit and the screen = 140 cm

= 1.40 m

a = width of the slit

=
`$0.50 * 10^(-3) \ m$`

Therefore,

`$I=I_(max)\left((\sin (\pi * 0.50 * 10^(-3) * 4.10 * 10^(-3))/(546.1 * 10^(-9) * 1.20))/((\pi * 0.50 * 10^(-3) * 4.10 * 10^(-3))/(546.1 * 10^(-9) * 1.20) )\right)^2 $`

`$=\left((0.170)/(9.82)\right)^2$`

`$= 2.89 * 10^(-4) \ I_(max)$`

Therefore the fractional intensity is
`$(I)/(I_(max))= 2.89 * 10^(-4) $`