Optics Problem: Find Smallest Refractive Index of Slab

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Homework Help Overview

The problem involves determining the smallest refractive index of a rectangular slab such that all incident light from air emerges from the opposite face. The context is within the subject area of optics, specifically focusing on the principles of refraction and total internal reflection.

Discussion Character

  • Exploratory, Assumption checking

Approaches and Questions Raised

  • Participants discuss the conditions for total internal reflection and the relationship between angles of incidence and critical angles. There is an attempt to express the maximum angle of incidence in terms of trigonometric functions and critical angles.

Discussion Status

The discussion includes attempts to derive relationships involving the refractive index and angles, with some participants suggesting different approaches to express these relationships mathematically. There is an indication that one participant believes they have arrived at an answer, but no consensus or resolution is presented.

Contextual Notes

Participants note the importance of considering the refractive index of air, which is close to 1, and the implications this has on the calculations involved in the problem.

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Homework Statement


Light falls on the surface AB of a rectangular slab from air. Determine the smallest refractive index n that the material of the slab can have so that all incident light emerges from the opposite face CD.
upload_2015-6-9_20-39-0.png


Homework Equations

The Attempt at a Solution


Let's think about this case:
upload_2015-6-9_20-49-45.png

There must be total internal reflection at Q and S.
That means, ##90-r>\theta _c## [##\theta _c## is the critical angle]
## sin(90-r)>sin \theta _c##
##cos r > \frac{1}{n}## [## sin \theta _c = \frac {1}{n}##]
##n>sec(r)##
But, the maximum value of ##sec(r)## is infinite.
That means n should be infinite.
But that is not the answer.
 
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Try it again. Write down the maximum ##r## in terms of arc-sin of something. Write down ##\theta_C## in terms of arc-sin of something. Then write down ##90-r > \theta_C## in terms of these arc-sin formulas. Don't forget that air has an index of refraction very close to 1. See what you get.
 
I have got the answer. :)
 
..
 
Last edited:

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