Ascuba diver is sitting on a boat while waiting to go on a dive and sees light reflected from the water's surface. at what angle of reflection will this light be completely polarized?

θ_p = 53.0º

Explanation:

For reflection polarization occurs when a beam is reflected at the interface between two means, the polarization in total when the angle between the reflected and the transmitted beam is 90º

Let's write the transmission equation

     n1 sin θ₁ = ne sin θ₂

The angle to normal (vertcal) is

    180 = θ2 + 90 + θ_p

    θ₂ = 90 - θ_p

Where θ₂ is the angle of the transmitted ray θ_p is the angle of the reflected polarized ray

We replace

     n1 sin θ_p = n2 sin (90 - θ_p)

Let's use the trigonometry relationship

    Sin (90- θ_p) = sin 90 cos θ_p - cos 90 sin θ_p = cos θ_p

In the law of reflection  incident angle equals reflected angle,  

    ni sin θ_p = ns cos θ_p

    n₂ / n₁ = sin θ_p / cos θ_p

    n₂ / n₁ = tan θ_p

    θ_p = tan⁻¹ (n₂ / n₁)

Now we can calculate it

The refractive index of air is 1 (n1 = 1) the refractive index of seawater varies between 1.33 and 1.40 depending on the amount of salts dissolved in the water

n₂ = 1.33

      θ_p = tan⁻¹ (1.33 / 1)

      θ_p = 53.0º

n₂ = 1.40

      θ_p = tan⁻¹ (1.40 / 1)

      Tep = 54.5º

force is mass*acceleration.

we have a very small mass: 11.4*0.001 kg

if we want 0.035n of force, we need an acceleration:

0.035 = 11.4*0.001*a

a = 0.035/(11.4*0.001) m/s^2

true tan speed = ang vel x radius