Glass cube refraction

  • Thread starter kreil
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kreil
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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
 

Answers and Replies

  • #2
andrevdh
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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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