Graduate Students do all problems. Undergraduate students choose three problems.

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1 OPTI 45/55 Midterm Due: Februr, Grdute Students do ll problems. Undergrdute students choose three problems.. Google Erth is improving the resolution of its globl mps with dt from the SPOT5 stellite. The stellite orbits t n ltitude 8 km bove the Erth s surfce. Assume its cmer contins / CCD with sensor size of mm nd piel dimensions of Ech piel on the sensor corresponds to.5 m ptch on the ground. You cn ssume thin lens sstem. Answer the following: ht re the dimensions of the piels? The piels re 6.4 mm / 48 b 4.8 mm / 56, which is equivlent to. μm squre piels. ht mgnifiction is needed? The.5 m ptch on the ground corresponds to the. μm on the sensor, so the mgnifiction is given b /, where is the imge height nd is the object height. In this cse, the mgnifiction. -6 / ht is the imge distnce for the stellite cmer? The mgnifiction is lso equl to the rtio of the imge distnce so the object distnce, so the imge distnce is given b.4-6 (8, m). m. ht is the focl length of the stellite cmer? The Gussin imging equtions tells us tht Effectivel, the object is t infinit. f.m.. 8 f

2 If we wnt to mke the diffrction spot dimeter mtch the piel size, wht is the size of the entrnce pupil? (Assume λ.5 μm) The diffrction limited spot size is given b Spot Size.44λ f D.μm Solving for D gives 4 mm. Technicll, the working F/# should be used here, but the difference between the F/# nd orking F/# is miniscule here since m is so smll.. Suppose we mesured n unknown thick lens sstem with the reciprocl mgnifiction technique. Using the technique, we determined tht the distnce from the rer principl plne to the imge plne ws mm. e lso determined tht the distnce from the front principl plne to the object plne ws -5 mm. Answer the following questions: ht is the mgnifiction of the lens? The object nd imge distnces re -5 mm nd mm respectivel. The mgnifiction is the rtio of the imge to object distnce or -4. ht is the focl length of the lens? The Gussin imging equtions tells us tht f mm. 5 f How fr do we need to move the lens to hve reciprocl mgnifiction for the sme object nd imge plnes?

3 Here, there re multiple routes to get the nswer, but bsicll use one of the equtions on pge 5 of the notes nd solve for the distnce d the lens needs to be moved. In terms of the object distnce l : d l d 5 m ( 4) 75mm here re the nodl points locted? Since the lens is in ir, the nodl points coincide with the principl plnes.. The wvefront coefficients for n opticl sstem with n eit pupil dimeter is mm nd the reference sphere rdius is mm re 4 μm, 6.6 μm,.4 μm. Do the following: rite n epression for the wvefront error. 4 ( h,, ψ).. 66h. 4h cos ( ψ) rite n epression for the trnsverse r error. (.( ).h ) (. ( ). h. 68h ) Sketch the trnsverse r error in the cse where h (i.e. on-is). ( ). ( ).

4 Sme for direction. Sketch the trnsverse r error in the cse where h (i.e. full field) long the nd directions. ( ).. ( ) In clss, we sid for non-rottionll smmetric sstems, the Seidel berrtions generlize to terms on is. Suppose we hve wvefront (, θ) θ cos Rewrite this wvefront in terms of Zernike polnomils such tht (, θ) Z (, θ) Z (, θ) Z (, θ) Z (, θ) Z (, θ ) ht re the coefficients,,,, in terms of nd?

5 There re two ws to solve for this problem. First, equte the epressions nd replce cos θ [ cos θ]. cosθ Z (, θ) Z (, θ) Z (, θ) Z (, θ) Z (, θ) Substitute the definitions of ech of the Zernike terms into the right hnd side.. cos θ cos θ ( ) 6 cos θ 8( ) cos θ Collect like terms ( ) ( 8 ) cosθ 6 cosθ 8 θ cosθ cos Compring both sides gives series of simultneous equtions ( ) ( 8 ) 6 8 From inspection, this gives 6

6 Alterntivel, we cn use the orthogonl properties of the Zernike polnomils to clculte ech of the coefficients vi nm m (, θ) Z (, θ) ddθ n So, 8 ( cos ) ( cos ) 6 ( cos θ)( ) ( cos ) θ ddθ 4 ( cos θ)( ) cosθddθ θ cosθddθ ddθ θ cosθddθ 6 For wht vlues of θ is (,θ) mimum when nd >? For nd >, the peks occurs when cos θ. This occurs when θ, ± Note, the following trig reltionship m be useful: cos θ [ cos θ]. Also, the first few Zernike polnomils re given b

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β 1 = 2 π and the path length difference is δ 1 = λ. The small angle approximation gives us y 1 L = tanθ 1 θ 1 sin θ 1 = δ 1 y 1

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