91Ó°ÊÓ

To find the angle at which sunlight is totally polarized, we will use Brewster's angle formula. Brewster's angle (\theta_B) is given by the following relation: \(\tan(\theta_B) = \frac{n_i}{n_r}\) where \(n_i\) is the refractive index of the medium from which the light is coming (air, in this case) and \(n_r\) is the refractive index of the medium into which the light is reflecting (here, ice). For air, the refractive index is approximately 1, and for ice, it's approximately 1.31. So, \(\tan(\theta_B) = \frac{1}{1.31}\) Now, let's calculate Brewster's angle: \(\theta_B = \arctan\left(\frac{1}{1.31}\right) \approx 37.3^\circ\) So, the sunlight is totally polarized when the incident light is at an angle of \(37.3^\circ\) with respect to the horizontal.

The light reflected from the smooth ice surface at Brewster's angle is polarized parallel to the surface. Therefore, the reflected light is polarized horizontally.

When the light is incident at Brewster's angle, it is still partially transmitted into the ice (even though it's totally polarized). To find the angle at which the transmitted light moves below the horizontal, we will use Snell's law of refraction, given by: \(n_i \sin(\theta_i) = n_r \sin(\theta_t)\) Where \(\theta_i = \theta_B = 37.3^\circ\), \(\theta_t\) is the angle of the transmitted light, and \(n_i\) and \(n_r\) are the refractive indices of air and ice, respectively (1 and 1.31). Plugging in the values, we get: \(1 \times \sin(37.3^\circ) = 1.31 \times \sin(\theta_t)\) Now, solving for \(\theta_t\): \(\theta_t = \arcsin\left(\frac{\sin(37.3^\circ)}{1.31}\right) \approx 27.7^\circ\) Hence, at Brewster's angle, the transmitted light travels at an angle of approximately \(27.7^\circ\) below the horizontal into the ice.