Determine the angle of incidence

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To determine the angle of incidence for light reflecting off a smooth ice surface, the Brewster angle formula is relevant, where tan(θ) = n2/n1. Given that the index of refraction of ice is 1.79, this angle can be calculated. At the Brewster angle, the reflected and refracted rays are at a 90-degree angle, resulting in complete polarization of the reflected light. This polarization occurs because the electromagnetic radiation oscillates perpendicular to the light's direction, and at this angle, only the component parallel to the surface remains. The final calculation will yield the angle of incidence in degrees.
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The index of refraction of ice is 1.79. Light is reflected from a smooth ice surface, and the reflected ray is completely polarized. Determine the angle of incidence. Answer in units of degrees.

Can anyone help me with a formula to set this one up? I'm confused on where to start.
 
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Look up the Brewster angle. Also, at this angle, the angle between the reflected and refracted rays is 90 degrees.
 
<br /> tan(\theta) = \frac{n2}{n1}<br />
This happens when the the angle between the reflected rays and refracted rays is 90 degrees, because in that case, none of the EM radiation oscillating perpendicular to the direction of the light and not parallel to the surface will be able to continue in the new direction. This is because of the fact that the EM radiation of light always oscillates on the plane perpendicular to the direction on the light (this plane can be divided into 2 cooardinites - the x_axis which is parallel to the surface and the y_axis) and never in the same direction. So when the light changes its direction by 90deg. the EM radiation perpendicular to the direction of the lights travel and not parallel to the surface(y_axis), will now have to be in the same direction as the light and so it will disappear from the light in the knew direction leaving only the light that was parallel to the surface(x_axis). So in the end, the light will be polarized with all of the EM radiation parallel to the reflective surface,
 
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