How Much of a Glass Cube's Surface Must Be Covered to Hide a Central Spot?

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SUMMARY

The problem involves determining the fraction of a glass cube's surface that must be covered to obscure a central spot from all viewing angles. Given a cube with an edge length of 1 cm and a refractive index of 1.50, the critical angle is calculated using Snell's Law. The solution requires blocking rays that do not undergo total internal reflection, leading to the conclusion that a circular area on the side face of the cube must be covered to prevent visibility of the spot.

PREREQUISITES
  • Understanding of Snell's Law and critical angle calculations
  • Basic knowledge of optics and light refraction
  • Familiarity with geometric representations of light paths
  • Concept of total internal reflection in optics
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  • Study Snell's Law in detail, focusing on applications in optics
  • Explore the concept of critical angles and total internal reflection
  • Learn about geometric optics and ray diagrams for light behavior
  • Investigate practical applications of refractive indices in materials
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Students studying optics, physics enthusiasts, and anyone interested in understanding light behavior in transparent materials.

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Homework Statement



A glass cube has a small spot at its center. What parts of the cube face must be covered to prevent the spot from being seen, no matter what the direction of viewing? What fraction of the cube face must be covered? Assume a cube edge of 1 cm and a refractive index of 1.50. (Neglect subsequent behaviour of an internally reflected ray.)



Homework Equations



I think the only relevant equations are snells law and the critical angle equation:

[tex]n_1 sin( \theta_1)=n_2 sin( \theta_2)[/tex]

[tex]\theta_c=\frac{n_2}{n_1},,,,(n_2<n_1)[/tex]


The Attempt at a Solution



I have no clue how to even start this. I need some hints first.

-Josh
 
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The rays from the spot that do not suffer total internal reflection need to be blocked.

Draw a little square with a dot in the middle (a top view of the situation). All rays from the one straight towards the front of the cube up to first ray that will be refracted along the face of the cube (the "last ray") need to blocked off. This region will then form a circle on the side face. Notice that the incident angle of the "last ray" will be the critical incident angle.
 

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