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

   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

the answer is a. the yacht traveled from the dock north to the 200-meter marker in the bay in less than 5 minutes, giving the passengers several more hours to fish.

im doing peak rn so sry if its wrong

answer would be letter choice (a),   a ball resting on a shelf. would be another example of potential energy   >   stored energy > not moving.

a ball spinning on a finger   >   kinetic energy > moving energy

a ball rolling on the ground   >   kinetic energy   > moving energy

a   ball flying through the air   > kinetic energy   >   moving energy

hope that                                         : )