At a beach the light is generally partially polarized due to reflections off sand and water. At a particular beach on a particular day near sundown, the horizontal component of the electric field vector is 2.8 times the vertical component. A standing sunbather puts on polarizing sunglasses; the glasses eliminate the horizontal field component. (a) What fraction of the light intensity received before the glasses were put on now reaches the sunbather's eyes

the glasses received 0.1131 of the total intensity

Explanation:

Given the data in the question;

the intensity is proportional to the electric field amplitude;

now, we assign the electric field of 1 to the vertical component and an electric field of 2.8 to the horizontal component

so;

Horizontal intensity =  1² = 1

Vertical intensity = 2.8² = 7.84

so total intensity = 1 + 7.84 = 8.84

fraction of the light intensity received before the glasses were put on now reaches the sunbather's eyes,

since the sunglasses blocks the horizontal intensity so only the the vertical intensity goes through;

hence

⇒ 1 / 8.84 = 0.1131 of the total intensity

Therefore, the glasses received 0.1131 of the total intensity

talk back to him about the sport and show him your expertise or leave the room

speed of light in vacuum is 300,000 km/s

speed of light in other mediums is given by the expression v = c/n, where c is the speed of light in vacuum and n is the refractive index of medium.

for the first medium v = 200,000 km/s

  so, refractive index of medium = 300000/200000 = 1.5

    1.5 refractive index is usually for glass. so first medium is glass.

for the second medium v = 250,00 km/s

so, refractive index of medium = 300000/250000 = 1.2

    1.2 refractive index is usually for water. so second medium is water.

for the third medium v = 0 km/s

so, refractive index of medium = 300000/0 = ∞

    ∞ refractive index is usually for opaque mediums. so third medium is opaque medium.