EM_scattering_homework_1_solution

12 j 1 ka sin s cos s cos s s sin s s w here we have

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Unformatted text preview: � � 0 � a � J � k r � rdr � �� � � � s,z �k 0 0 4� sin � � k s , z � k � h / 2 � � � �k s,� 0 4� sin �� k s , z � k � h / 2 � ks , � a � � � k � k s2,� �uJ � u � � � �� � � � k s ,� a 1 � � k � k s2,� �k s,z J 0 � u � udu 0 4� a sin � � k s , z � k � h / 2 � � � 0 0 s,z � k � ks,� J1 � ks,� a � (1.9) Note that in spherical coordinates k s , z � k cos � s , k s , � � k sin � s . As a result, in spherical coordinates we have the result given by �� � � 0 � 4� a sin �� cos � s � 1� kh / 2 � � � k 2 � cos � s � 1� sin � s J 1 � ka sin � s � (1.10) There is no dependence on �s here but remember that we have to include the vector ˆ ˆ ˆ ˆˆˆ ˆ ˆ (1.11) e � e � k k � x � k k � x � sin � cos � k i � i s � s s ,x s s ss The final result is � � ˆˆ Q� ks , ki � E0 � E0 4� a �� � � 0 � sin � � cos � s � 1� kh / 2 � � � k 2 � cos � s � 1� sin � s 4� a �� � � 0 � sin � � cos � s � 1� kh / 2 � � � k 2 � cos � s � 1� sin � s ˆ ˆ J 1 � ka sin � s � � x � sin � s cos �s ks � � � (1.12) ˆ J 1 � ka sin � s � �cos � s cos �s θs � sin �s ˆ � s� � W here we have used ˆ � cos� s cos �s ,cos � s sin...
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