Question

A thin beam of light of wavelength 625 nm goes through a thin slit and falls on a screen 3.00 m past the slit. You observe that the first completely dark fringes occur on the screen at distances of ±8.24 mm from the central bright fringe, and that the central bright fringe has an intensity of 2.00 W/m² at its center. What is the intensity of the light at a point on the screen that is one-quarter of the away from the central bright fringe to the first dark fringe?

Answer 1

Answer:

I = 1.62 w/m^2

Explanation:

wavelength = 625 nm

position of screen = 3.00 m past the slit

Determine the intensity of the light

I = $I_{0} ( sin\frac{\beta }{2 } /\frac{\beta }{2} )^2$  

β = 2π*Dsin∅ / л

$I_{0}$  = 2

Attached below is the remaining part of the solution