Why is the normal used as a reference for angles in physics and math?

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The normal serves as a crucial reference point for measuring angles in physics and math, particularly in the context of light transmission at media interfaces. It is defined as perpendicular to the surface, simplifying calculations related to refraction and reflection. When light beams are parallel to the normal, maximum transmission occurs, which aligns with the principles of optics. Using the normal also ensures that when the incident angle is zero, the refraction angle remains zero, providing a logical framework for analysis. This approach not only streamlines mathematical descriptions but also reflects real-world scenarios effectively.
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Please explain with another simple example
 

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If you didn't, how would you measure i or r?

What would you do in the case of a curved surface?
 
The normal, which is perpendicular to the plane forming the interface between media (e.g. solid and gas), is simply a reference. The maximum transmission of light comes when the beam (photons) is parallel with the normal.
 
The normal is indeed a reference =) We need the normal (as what Astronuc said, perpendicular to the plane) to measure indexes .
 
When we develop the maths to describe the physics we try to make the maths as simple as possible. So the theory dictated that if we use the normal to the interface between the two media the maths will be much easier and natural to the real situation. You can also see that, using the normal as the reference for the angles, when the incident angle is zero the refraction angle will also be zero, which makes the most sense doesn't it? Refraction problems were also initially solved with graphical techniques in which case the normal were used in the diagrams (seee history of Willebrord van Roijen Snell - also called Snellius).
 
The book claims the answer is that all the magnitudes are the same because "the gravitational force on the penguin is the same". I'm having trouble understanding this. I thought the buoyant force was equal to the weight of the fluid displaced. Weight depends on mass which depends on density. Therefore, due to the differing densities the buoyant force will be different in each case? Is this incorrect?

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