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Unformatted text preview: Problem 1 Problem 2 Please go through page 250 and 251 of the textbook. Problem 3 (a). (b). From part a, we know cm f cm f 50 , 25 2 1 = = , this is the focal length of the equivalent thin lens of the corresponding thick lens in air. Using the matrix of thick lens but the thickness is zero. For the first thin lens: ELEC308 (Spring 2008) Assignment 3 TA solution cm R R R R P P P C cm f P P P n n d n P d n n n P n n d n P d L L L L 25 5 . 1 1 1 5 . 1 1 1 1 1 25 , 1 , 1 , , ) 1 ( 1 2 1 2 1 2 1 2 1 ' 2 ' ' 1 = = ⇒ + = + = = = = + = = = =  = For the second thin lens cm R R cm f 275 55 . 1 1 25 1 55 . 1 1 50 3 3 = ⇒ + = = (c). For a doublet, its power 2 2 1 1 2 1 ) 1 ( ) 1 ( K n K n P P P + = + = , where 22 21 2 12 11 1 1 1 , 1 1 r r k r r k = = chromatic aberration is eliminated when: 1 2 1 1 = ∂ ∂ + ∂ ∂ = ∂ ∂ λ λ λ n K n K P since < ∂ ∂ λ n , that means to eliminate chromatic aberration, K1 and K2 must have opposite signs. That requires combination of a positive lens made K1 and K2 must have opposite signs....
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This note was uploaded on 10/16/2011 for the course ELEC 308,315,10 taught by Professor Prof.shenghuisong during the Spring '11 term at CUHK.
 Spring '11
 Prof.ShenghuiSong

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