InterviewSolution
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InterviewSolution
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A narrow beam of natural light with wavelength lambda = 589nm falls normally on the surface of a Wollaston polarizing prism made of Iceland spar as shown in Fig. The optical axes of the two parts of the prism are mutually perpendicualr. Find the angle delta between the directions of the beams behind the prism if the angle theta is equal to 30^(@). |
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Answer» Solution :In a uniaxial crystal, an unpolarized beam of light (or even a polarized one) splits up into `O` (for ordinary) and `E` (for extraordianary) light waves. The DIRECTION of vibration in the `O` and `E` waves are most easily spectfied in terms of the `O` and `E` PRINCIPLE planes. The principle plane of the ordianry wave is defined as the plane containing the `O` ray and the optic. Similarly the principle plane of the `E` wave is the plane contaning the `E` ray and the optic AXIS. In terms of these plane the following is true: The `O` vibrations are perpendicualr to the principle plane of the `O` ray while the `E` vibrations are in the principle plane of the `E` ray. When we apply this definition to the WOLLASTON prism we find the following: (exaggerated.) When unpolarized light enters from the left the `O` and `E` waves travel in the same direction but with different speeds. The `O` ray on the left its vibrations normal to the plane of the paper and it becomes `E` ray on crossing the diagonal boundary of the two prism similarly the `E` ray on the left becomes `O` ray on the right. In this case Snell's law is applicable only approxilately. The two rays are incident on the boundary at an angle `theta` and in the right prism the ray which we have called `O` ray on the right emerges at `sin^(-1) .(n_(e))/(n_(0))sin theta = sin^(-1) .(1.658)/(1.486) xx (1)/(2) = 33.91^(@)` where we have used `n_(e) = 1.1658, n_(0) = 1.486` and `theta = 30^(@)`. Similarly the `E` ray on the right emerges within the prism at `sin^(-1). (n_(0))/(n_(e)) sin theta = 26.62^(@)` This means that the `O` ray is incident at the bounary between the prism and air at `33.91 - 30^(@) = 3.91^(@)` and will EMERGE into with a deviation of `sin^(-1). n_(0)sin 3.91^(@)` `= sin^(-1).(1.658 sin 3.91^(@)) = 6.49^(@)` The `E` ray will emerge with an oppsite deviation of `sin^(-1). (n_(e)sin(30^(@) - 26.62^(@)))` `= sin^(-1).(1.486 sin 3.38^(@)) = 5.03^(@)` Hence `delta ~~ 6.49^(@) + 5.03^(@) = 11.52^(@)` This result is accurate to first order in `(n_(e) - n_(0))` because Snell's law holds when `n_(e) = n_(0)`. 
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