Final sol 2007

Final sol 2007 - Solution: a). As the aperture is placed at...

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Unformatted text preview: Solution: a). As the aperture is placed at the image plane, from the Gaussian formula, we have: mm s s f n s n s n i i r air i air o air 150 50 1 1 75 1 = ⇒ = + ⇒ = + b). Also from the Gaussian formula: mm s s f n s n s n o o o air i air o air 625 . 10 10 1 10 160 1 1 = ⇒ = + + ⇒ = + c). Case 1. -- = +-- +-- = - ⋅ ⋅ - = 9 . 3 37 . 245 5 . 23 1 245 1 245 1 245 245 1 1 1 1 1 245 1 1 1 1 r o r o r o o r f f f f f f f f M ELEC308 (Spring 2007) Final Exam TA solution Case 2 = - ⋅ ⋅ - ⋅ = 03125 . 37 . 32 1 625 . 10 1 1 10 1 1 1 245 1 1 50 1 1 1 150 1 M d). Case 1 16 625 . 10 170 '- =- =- = o o o s s m 2 75 150 '- =- =- = r r r s s m 32 = ⋅ = r o system m m m Case 2 Since the object plane and the CCD camera are conjugate, B=0, that indicate A is the magnification of the system. 32 = = A m system e). In the CCD camera, the sample will be a pattern of square grids with line to line spacing of 128um, and it locates in a circle with 2mm diameter. Since the diameter of the aperture is larger than the wavelength greatly, we don’t need to consider the diffraction. Solution: Assume the phase shift is δ ,and sin ka δ θ = , where a is the distance between two slits. For the maxima, we have the equation: 1 1 sin 2 k a m θ π = , then 1 1 sin a m θ λ = For the minima, we have the equation: 2 2 1 sin 2( ) 2 k a m θ π =- , then 2 2 1 sin ( ) 2 a m θ λ =- The Fourth minima of 2 λ light coincides with the third maxima of 1 λ light, We get 1 1 2 2 1 ( ) 2 m m λ λ =- , it is known that 1 3 m = , 2 4 m = and 2 436 nm λ = , therefore, 1 508.6 nm λ = Solution:...
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Final sol 2007 - Solution: a). As the aperture is placed at...

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