Question

In a double slit experiment, the intensity of light at the center of the central bright fringe is measured to be 6.2 µW/m2. What is the intensity half

Answer 1

Answer:

   I_FWHW = 3.2 μW / m²

Explanation:

In the analysis of optics and electricity a very useful magnitude is the width at half height (FWHW) and the intensity at this height, which is given by

               I_FWHW = I₀ / 2

corresponds to the width of the line for this intensity.

In this case they give the maximum intensity for which

               I_FWHW = 6.2 / 2

               I_FWHW = 3.2 μW / m²

You do not give more data in your exercise, but the most interesting calculation is to find the angle values ​​for which you have this intensity since it is this range is 50% of the energy of the system, have I write the equation for this calculation

             I = Io cos² x₁   (sin x / x)²

             x₁ = π d sin θ /λ

             x = π b sin θ /λ

where d is the separation of the slits and b the width of each slit