.70 Optc Hmewrk # February 8, 04 Prblem. Reractng Surace (Me rm Pertt -) Part (a) Fermat prncple requre that every ray that emanate rm the bject an pae thrugh the mage pnt mut be chrnu (.e., have equal ptcal path length). The cmmn ptcal path length (OP cmmn ) can be ealy un ung the ray that pae alng the ptcal a rm the bject t the mage pnt, whle takng nt accunt the reractve nee ar (n ) an gla (n ): OP cmmn n + n et an y be the hrzntal an vertcal crnate, repectvely, wth the rgn at the bject pnt. The ptcal path length (OP arbtrary ) any arbtrary ray that emanate rm the bject at (0, 0), enter the gla at crnate (, y), an then pae thrugh the mage pnt at ( +, 0), can be un rm the n-ar prpagatn tance ( ) an n-gla prpagatn tance ( ), whle takng nt accunt the nee reractn: + y ( + ) + y OP n + n OP arbtrary arbtrary n + y + n ( + ) + y The crrect reractng urace crrepn t all (, y) uch that OP arbtrary equal OP cmmn. OP arbtrary OP cmmn n + y + n ( + ) + y n + n Pluggng n n, n.5, 0 cm, an 0 cm: + y +.5 (30 ) + y 35 Nte: Th equatn nly -mennal. The 3-mennal lutn wul be a raalymmetrc urace, where each cr-ectn (thrugh the verte) llw the D equatn. Page 0
.70 Optc Hmewrk # February 8, 04 Part (b) Belw a plt the reractng urace ung label rm Part (a). I un (, y) crnate n the urace by ubttutng back nt the urace equatn, a llw: + y +.5 (30 ) + y + y +.5 900 60 + ( + y 35 ) 35 +.5 900 60 + 35.5 (900 60 + ) (35 ) In th rm, can ealy be un rm, an y can be un rm an :.5 (900 + ) (35 ) 60.5 y By chng a ere value r, I can calculate (, y) crnate that aty the rgnal urace equatn, a llw: y 0 0 0 0.90 ±.07.8 ±.85 3.75 ±3.38 Ater calculatng many clely-pace (, y) crnate n th way, I wa able t plt the reractng urace abve wth Matlab. Page 0
.70 Optc Hmewrk # February 8, 04 Prblem. Ice-cream Spn Frt, let ene me varable: et an be the ntal an nal bject tance, repectvely. et an be the ntal an nal mage tance, repectvely. et m an m be the ntal an nal magncatn, repectvely. et be the tance mve by the mrrr (ntal t nal) The ecrbe prblem can be quante by 5 equatn wth 5 unknwn,,,, ) : ( Mrrr equatn at the ntal bject tance Mrrr equatn at the nal bject tance Magncatn at the ntal bject tance Magncatn at the nal bject tance Derence between ntal an nal tance + Eq. + Eq. m Eq. 3 m Eq. 4 + Eq. 5 The ytem equatn can be lve r the unknwn a llw: Cmbne Eq. an Eq. + + Page 3 0
.70 Optc Hmewrk # February 8, 04 Ue Eq. 3 an Eq. 4 t remve an m m Ue Eq. 5 t remve m + m ( + ) Slve r ( m m m ) m m Slve r remanng unknwn + m m + In th prblem, we are gven.5 cm, m 0.5, an m /3, but we are nt gven ther gn (+ r -). Frtunately, we can aely (but arbtrarly) et +.5 cm. Ater all, the gn hul be ncnequental, nce the lutn (t the ytem equatn) r ±.5 wll mply be the revere ne anther (let becme rght, rght becme let). Hwever, the lutn t the ytem equatn wll, nee, epen n the gn m an m (.e., whether the mage are nverte r nt) The table belw hw the lutn t the ytem equatn r all 4 cmbnatn m (+ r -) an m (+ r -). The table reveal that, epenng n the gn m an m, the mrrr may take n a cncave r cnve gemetry r the ntal an/r nal bject-t-mrrr tance. A hghlghte belw, nly the lat cmbnatn m (-0.5) an m (-/3) cnrm t the preme the prblem (.e., that the ce-cream pn beng ue a a cncave mrrr). Knwn Slutn t Sytem Equatn (n centmeter) Mrrr Gemetry (cm) m m. ' '.5 0.5 /3.5 5 -.5 -.67 -.5 Cnve mrrr r ntal an nal tance.5 0.5 -/3-0.5 0.5 0.67 0.5 Cnve r ntal; cncave r nal.5-0.5 /3 -.5-0.75-0.33-0.5 Cncave r ntal; cnve r nal.5-0.5 -/3 7.5 0 3.75 3.33.5 Cncave r ntal an nal tance The ptn (p) the mage, relatve t the bject, at each bject-t-mrrr tance : p 7.5 cm 3.75 cm 3.75 cm p 0 cm 3.33 cm 6.67 cm Thu, the anwer t the pecc quetn are: When the pn mve.5 cm (.e., ) urther away rm the bject, the mage mve.9 cm (.e., p p ) urther away rm the bject. The cal length the pn cncave mrrr.5 cm. Page 4 0
.70 Optc Hmewrk # February 8, 04 Prblem 3. Hre an Camera Part (a) The mage tance the hre ne ( ne ) can be un wth the thn-len equatn, gven the bject tance the hre ne ( ne 5 m) an the cal length the thn len ( 3 m): e 5 3 45 + ne n 3.75 n ne e ne 5 3 The heght the mage the hre ne (h ne ) can be un wth the magncatn equatn, gven the heght the hre ne (h ne.5 m): n e 3. 75 m 0.5 ne 5 m e h n e mh ne 0.5.5 0.565 h ' n h ne Thu, the mage the hre ne lcate 3.75 m t the rght the len an 56.5 cm belw the ptcal a. Part (b) The magncatn m -0.5. In the vertcal rectn, the mage nverte, a ncate by the negatve magncatn. In ther wr, the hre appear upe-wn. In the hrzntal rectn (a calculate n Part ()), the hre tal cme t cu cler t the len than the hre ne e. In ther wr, the mage the hre tal t the let the mage the hre ne. The graph abve hw the rentatn the mage (but nt t cale), a ecrbe here. Page 5 0
.70 Optc Hmewrk # February 8, 04 Part (c) The mage 56.5 cm tall (but nverte), a calculate n Part (a). Part () The mage tance the hre tal ( tal ) can be un wth the thn-len equatn, gven the bject tance the hre tal ( tal 7.5 m) an the cal length the thn len ( 3 m): tal tal 7.5 3 5.5 + tal 3. 6 7.5 3 4.5 tal tal The length the mage () can be calculate a llw: ne tal 3.75 3.6 0.3 Thu, the mage the hre 3 cm lng (rm ne t tal). Page 6 0
.70 Optc Hmewrk # February 8, 04 Prblem 4. Prm Par Part (a) At the rt urace the prm, the angle ncence zer, n reractn ccur. At the ecn urace, a ncate n the rawng, the angle ncence α an the angle reractn (α + θ). Snell law then gve the relatn amng θ, α, an n (aumng the ne reractn ar ): n n( α ) n( α + θ ) r α < n n I α n, then ttal nternal relectn (TIR) ccur an the ray mut cr the bttm n nterace the prm t ecape. Thu, the gemetry the rgnal prm par becme nval. Part (b) A ncate n the rawng, the length () ver whch the beam nterect the ecn urace the prm : h n c(α) The wth the beam ater the prm : Page 7 0
.70 Optc Hmewrk # February 8, 04 h c( α + θ ) c( α + θ ) c( α) h n Thu, the wth the beam reuce by a actr (r) : r c( α + θ ) n ( α + θ ) c( α) c( α) T remve θ rm the equatn, we can ubttute n the relatn rm Part (a): r n n ( α) c( α) The ecn prm mply repeat the ame prce, gvng a ttal reuctn actr r. Thu, the reuctn actr (r an r ) ater the rt an ecn prm, relatve t the ntal beam wth, are: r r n n ( α) r r α < n c( α) n n n ( α) r r α < n c ( α) n Nte: Fr ucently mall α, then n(α) α, n(α + θ) α + θ, an c(α). The anwer t the prblem cul then pbly be apprmate a llw: n α α + θ θ (n ) α r r n α r r n α Page 8 0
.70 Optc Hmewrk # February 8, 04 Page 9 0 Prblem 5. Tw Thn ene The cal length a cmbnatn tw thn lene ( an ), eparate by tance, : + + Pluggng n 0 cm, -8 cm, an 5 cm: 5 8 0 8) ( 0 53.33 cm The tance rm the negatve len t the lm plane hul equal the tance pat the negatve len at whch a hrzntal, cllmate beam wul cme t cu. Th can be etermne a llw. The matr (M ytem ) repreentatn the tw-len ytem 0 0 0 M ytem The vectr ( n ) repreentatn a ray that ha a heght an parallel t the ptcal a (.e., crrepnng t a hrzntal cllmate beam) : 0 n I n ent nt the tw-len ytem, the utput ray ( ut ) wul be: M n ytem ut I ut allwe t prpagate a tance () pat the negatve len, the reultng ray ( cu ) wul be: + 0 ut cu The tance rm the negatve len t the lm plane hul equal the at whch the heght cu zer (.e., cu () 0):
.70 Optc Hmewrk # February 8, 04 Page 0 0 0 ) ( cu ) ( ) ( + Pluggng n 0 cm, -8 cm, an 5 cm: 8) (0 5 0) (5 8 3.33 cm I a tant bject ubten egree at the camera, then that mean: ) tan( h The magncatn rmula ay: h h m h h Pluggng n 53.33 cm (nce the lm plane at the cu the ytem): h -tan( º ) 53.33 -.86 Thu, the mage a egree tant bject.86 cm tall.
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