As shown in the figure, a ray AB of unpolarized light enters from water of refractive index n_w = 4/3 into a medium of refractive index n_p = 4/ 3 after passing through a glass plate of refractive index n_g = 1.5 and a layer of water. At a particular incident angle i the reflected ray CD is polarized in the direction as shown in the figure. The value of i (in degrees) is:
A. A
Let the angle of incidence at the first interface (water to glass) be i. By applying Snell's law at the successive parallel interfaces, we can find the angle of incidence at the interface where the reflection occurs (water to medium n_p). For the first interface (water to glass): n_w i = n_g r_1 For the second interface (glass to water): n_g r_1 = n_w r_2 From the above two equations, we get: n_w i = n_w r_2 r_2 = i Thus, the angle of incidence at the third interface (water to medium n_p) is also i. The problem states that the reflected ray is completely polarized with its electric field perpendicular to the plane of incidence (indicated by the dots on ray CD). This complete polarization upon reflection occurs when the light is incident at Brewster's angle _B. According to Brewster's law for the interface between water and medium n_p: _B = n_p n_w Since the angle of incidence at this interface is i, we have i = _B. Substituting the given refractive indices: i = 4/ 3 4/3 i = 3 3 = 3 Therefore, the angle of incidence is: i = 60^ Answer: 60
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