(a) Sunlight reflected from the smooth ice surface of a frozen lake is totally polarized when the incident light is at what angle with respect to the horizontal? (b) In what direction is the reflected light polarized? (c) Is any light incident at this angle transmitted into the ice? If so, at what angle below the horizontal does the transmitted light travel?

Short Answer

Expert verified
In which direction is the reflected light polarized, and at what angle does the transmitted light travel below the horizontal into the ice? Answer: The sunlight has to be incident at an angle of approximately 37.3° on the ice's surface to become totally polarized. The reflected light is polarized horizontally, and the transmitted light travels at an angle of approximately 27.7° below the horizontal into the ice.

Step by step solution

01

a) Finding the angle of incidence (Brewster's angle)

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.
02

b) Direction of polarization

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.
03

c) Transmission of light into the ice

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.

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