How Do You Calculate the Index of Refraction in a Light Refraction Problem?

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Homework Help Overview

The problem involves calculating the index of refraction of a plastic block based on the angles of incidence and refraction of a light ray entering and exiting the material. The context is within the subject area of optics, specifically focusing on light refraction and Snell's law.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • Participants discuss applying Snell's law to determine the index of refraction, with some attempting to relate the angles of incidence and refraction. Questions arise regarding the correct application of trigonometric relationships and the interpretation of angles at different interfaces.

Discussion Status

The discussion is ongoing, with participants offering hints and clarifications about the relationships between angles at the interfaces. There is an acknowledgment of misunderstandings regarding the equality of angles and a suggestion to explore geometric relationships in the context of the problem.

Contextual Notes

Participants express confusion over the application of Snell's law and the relationships between the angles involved, indicating a need for further exploration of the problem setup and assumptions regarding the angles of incidence and refraction.

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Homework Statement



A light ray enters a rectangular block of plastic at an angle 1 = 44.0° and emerges at an angle 2 = 79.0°, as shown in Figure P22.61.
p35-71.gif


Figure P22.61
(a) Determine the index of refraction of the plastic.
wrong check mark
(b) If the light ray enters the plastic at a point L = 50.0 cm from the bottom edge, how long does it take the light ray to travel through the plastic?
ns


Homework Equations


n1sin(1)=n2sin(2) or sin(critical)=n2/n1

The Attempt at a Solution


i attempted arcsin(11/44) and arcsine(79/44) but both were wrong. Can someone please help me get on the right path?
 
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Treat this as two refractions and apply Snell's law at each interface. Hint: How does the second angle (the angle of refraction) at the first interface relate to the first angle (angle of incidence) at the second interface?
 
I have sin(44)=nsin(theta2) and sin(79)=nsin(theta2) but that cannot be correct. I know that the angle of incidence and angle of refraction are equal. Please help I am mising something here.
Thanks
 
floridianfisher said:
I have sin(44)=nsin(theta2) and sin(79)=nsin(theta2) but that cannot be correct.
True, that cannot be correct, since \sin 44 \neq \sin 79.
I know that the angle of incidence and angle of refraction are equal.
They are related, but not equal.

For clarity, I'll rewrite the equations as:
\sin 44 = n \sin \theta_a
\sin 79 = n \sin \theta_b

Find the simple trignometric relationship between \theta_a & \theta_b by examining the diagram. Look for triangles.
 

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