A small spotlight mounted in the bottom of a swimming pool that is 4.3 m deep emits light in all directions. On the surface of the pool, directly above the light, the area illuminated is a circle. Determine the maximum radius of this circle. The index of refraction of water is 1.33.

Radius = 4.9m

Explanation:

The critical angle in optics refers to the angle of incidence, beyond which the total internal reflection of light occurs.

The formula for critical angle is;

θ_crit = sin^(-1)(n_r/n_i)

Where:

θ_crit = The critical angle.

n_r = refraction index.

n_i = incident index.

In this case n_i is refraction index of water which is given 1.33

While n_r is the refraction index of air which has a value of 1.

Thus,

θ_crit = sin^(-1)(1/1.33)

θ_crit = 48.75°

In critical angle statements, If the incident ray is precisely at the critical angle, the refracted ray becomes tangent to the boundary at the incidence point and propagates along the interface surface and there is no reflected ray. Thus;

Radius(r) = d.tan θ_crit

Radius = 4.3tan 48.75

Radius = 4.9m

t2 equal the displacement of the particle over the same time period if t2> t/2? hint: consider a sine wave starting at t1=0, then consider a closing wave starting at t1=0, for your explanatio