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

real, smaller and upside-down.   so the answer should be b.