What is the Distance d in Snell's Law and the Pythagorean Theorem?

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Homework Help Overview

The discussion revolves around finding the distance d in the context of Snell's Law and the Pythagorean Theorem, specifically involving a scenario with angles and distances related to light refraction.

Discussion Character

  • Exploratory, Conceptual clarification, Problem interpretation

Approaches and Questions Raised

  • Participants discuss the application of Snell's Law and the Pythagorean Theorem to determine the distance d, with some questioning the correctness of their angle calculations and others seeking clarification on how to visualize the problem geometrically.

Discussion Status

The conversation indicates that participants are actively engaging with the problem, with some providing guidance on using geometry and diagrams to approach the solution. However, there remains uncertainty among participants about the steps needed to find the distance d.

Contextual Notes

There is mention of a diagram that is necessary for understanding the problem, and participants express confusion about the setup and calculations involved.

h20proof
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In the following diagram find the distance d if a=4.0 mm, Θ=30°

n_asinΘ_a=n_bsinΘ_b: snells law
a^2+b^2=c^2: pythagorean theorem

I think I got the angle to the problem correct. I am not sure if this is correct. Is this correct Θ_b=sin^-1(1sin30/1.52)?
 

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That will give you the internal angle.

You are apparently required to find the offset distance, d, of the emergent ray.
 
thank you. How do you find it?
 
h20proof said:
thank you. How do you find it?
Using geometry. Specifically, the geometry of right-angled triangles.

You start by drawing a large clear diagram, and using a straight-edge to make all your straight lines look straight. Any lines you draw can be made into a side of a right-angled triangle.
 
okay, I still don't see it.
 
h20proof said:
okay, I still don't see it.
Where's your diagram?
 

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