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23.86. Set Up:
For linearly polarized light the intensity passed by a filter is
where
is the incident
intensity and
I
is the transmitted intensity. After passing through the filter the light is linearly polarized along the
filter axis.
Solve: (a)
Use two filters, with the axis of the second filter at
from the original direction of polarization of the
light.
(b)
Let
be the angle between the original direction of polarization of the light and the axis of the first filter. The
angle between the axes of the two filters therefore is
The intensity passed by the two filters is
so
so
The transmitted intensity has a maximum value of
when
23.87. Set Up:
Let the light initially be in the material with refractive index
and let the final slab have refractive
index
In part (a) let the middle slab have refractive index
Apply Snell’s law to each refraction.
Solve: (a)
1
st
interface:
2
nd
interface:
Combining the two equations gives
This is the equation that would apply if the middle slab were absent.
(b)
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This note was uploaded on 03/06/2009 for the course PHYS 114 taught by Professor Shoberg during the Spring '07 term at Pittsburg State Uiversity.
 Spring '07
 Shoberg
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