Complete Ques
Answer:
The value is [tex]A = 0.00214 \ m^2[/tex]
Explanation:
From the question we are told that
The wavelength is [tex]\lambda = 500 \ nm = 500 *10^{-9} \ m[/tex]
The diameter of the aperture is [tex]d = 30 \mu m = 30*10^{-6} \ m[/tex]
The distance of the screen from the aperture is [tex]D = 350 \ m m = 0.350 \ m[/tex]
Generally the distance from the center the the edge of the central bright fringe is magmatically reparented as
[tex]y = \frac{m * \lambda * D}{d}[/tex]
Generally m = 1 because after the central bright fringe we have the first order fringe
So
[tex]y = \frac{1 * 500 *10^{-9} * 0.350}{30*10^{-6}}[/tex]
=> [tex]y =0.00583 \ m[/tex]
Generally the area of the central bright fringe
[tex]A = 2 \pi * y^2[/tex]
=> [tex]A = 2 * 3.142 * 0.00583^2[/tex]
=> [tex]A = 0.00214 \ m^2[/tex]
