Information states that "A light beam...encounters a flat surface...such that the smallest angle the beam makes with the surface is 30*."
The question asks "What is the smallest angle the refracted ray makes with the surface?"
The given answer and explanation state that "In this case the incident angle is 60*, since the problem states that the angle with the horizontal is 30*..."
I don't understand how this is computed nor, semantically, what the difference is between the angle of incidence and the "smallest angle the beam makes with the surface."
Is the relationship a minimum to maximum one? Such that the incident angle must always be maximum to the smallest angle the light makes with the horizontal?
Math and Physics CR (Old) Ch. 11 Pr #11

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Re: Math and Physics CR (Old) Ch. 11 Pr #11
Hi jrashaad,
Thanks for the excellent question. Basically, this Q is just asking you to use geometry to distinguish between angle to the horizontal & angle to the normal. I've attached a drawing that should help clarify things.
In Step 1, we take the Q info and need to realize that for classical optics, we always use angles to the normal, not horizontal, so we have to recognize that an incident angle of 30° to the horizontal means that the angle to the normal is 60°, just by simple geometry (because the normal is a vertical line perpendicular to the horizontal). This is Step 1 in the figure. Then, this angle is what we sub into Snell's law to calculate that the refracted angle will be 30° (Step 2). However, again, this is the angle to the normal, and the question asks us about the angle to the horizontal. We have to use geometry again to realize that if the angle to the normal is 30°, the angle to the horizontal is 60° (Step 3). The reason why the question specifies the "smallest" angle to the horizontal is because there's also a 120° angle formed if you start with the refracted ray and then move clockwise to reach the xaxis.
Hope this clarifies things!
Thanks for the excellent question. Basically, this Q is just asking you to use geometry to distinguish between angle to the horizontal & angle to the normal. I've attached a drawing that should help clarify things.
In Step 1, we take the Q info and need to realize that for classical optics, we always use angles to the normal, not horizontal, so we have to recognize that an incident angle of 30° to the horizontal means that the angle to the normal is 60°, just by simple geometry (because the normal is a vertical line perpendicular to the horizontal). This is Step 1 in the figure. Then, this angle is what we sub into Snell's law to calculate that the refracted angle will be 30° (Step 2). However, again, this is the angle to the normal, and the question asks us about the angle to the horizontal. We have to use geometry again to realize that if the angle to the normal is 30°, the angle to the horizontal is 60° (Step 3). The reason why the question specifies the "smallest" angle to the horizontal is because there's also a 120° angle formed if you start with the refracted ray and then move clockwise to reach the xaxis.
Hope this clarifies things!
Andrew D.
Content Manager, Next Step Test Prep.
Content Manager, Next Step Test Prep.
Re: Math and Physics CR (Old) Ch. 11 Pr #11
Yes, that helps. Thank you very much for that explanation.
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