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
