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Assignment3_solution_updata

# Assignment3_solution_updata - Assignment3 TAs Solution...

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Assignment3 TAs’ Solution Problem 1 6.20 1 5 . 0 / ) 1 5 . 1 ( 1 = - = P and 2 ) 25 . 0 /( ) 1 5 . 1 ( 2 = - - - = P - = 6 . 0 2 . 0 , 6 . 2 , 8 . 0 A det(A)=0.8*0.6-0.2*(-2.6)=1 6.23 - 385 . 0 6 . 2 ) 5 . 1 / 3 . 0 * 2 * 1 2 1 ( ) / ( / 1 2 1 2 1 = - = - + - = - + - = f n d P P P P f l 23 . 0 6 . 2 6 . 0 1 1 - = - = = C D F V , 31 . 0 6 . 2 8 . 0 2 2 = - - = - = C A F V Problem 2 Please go through example 5 of tutorial 4 , Problem 3 Choose BK7 and F2 as the material. 36 1 64 1 2 2 7 1 - - = = - - = = C F D F C F D BK n n n V V n n n V V n 1 =1.52, n 2 =1.63 (1) If the doublet is consisted of two thin lenses, the principle plane is in the center of the doublet and the focal plane is in the focal point.

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45 1 0 2 1 2 1 1 2 = + = = + P P P V P V P So, f 1 =19.7mm, f 2 =-35mm, according to the lens makers formula r 11 =-r 12 =20.5mm, r 22 =10.6mm (2) If the doublet is consisted of two thick lenses, then For simplicity, r22= , then, r12=-20.1mm, if d1=d2=10mm, then r11=17.3mm. The doublet is like this = = = - = = - = - = - = 1 10 , 0 , 1 63 . 1 0 , 0 , 1 1 2 0 , 3 1 2 , 1 933 . 0 0 , 0034 . 0
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