*21. ssm In Figure 26.6, suppose that the angle of incidence is q1 = 30.0°, the thickness of the glass pane is 6.00 mm, and the refractive index of the glass is n2 = 1.52. Find the amount (in mm) by which the emergent ray is displaced relative to the incident ray.

 

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Below is another sketch which will be helpful.

Since the actual exiting ray and the original ray both make 30° angles with the normal to the top surface, they are parallel.  If we find any separation between them we know the distance the ray was shifted.  So in the diagram x, is the distance shifted.

 

are baselines for right triangles made by either the refracted ray or the original ray.  Both have an adjacent side of length t, the thickness of the glass slab.  So the differences between the two baselines are due to the two different angles with the normal at the bottom surface.  So we have the shifted distance is given by

 

And we get the two baselines by

 

And

So

 

We know , and we find  from Snell’s law

 

 

 

Now we find the separation x

 

 

 

 

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This page last updated on July 1, 2018