Physics 107 HOMEWORK ASSIGNMENT #20

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1 Physcs 107 HOMEWORK ASSIGNMENT #0 Cutnell & Jhnsn, 7 th etn Chapter 6: Prblems 5, 7, 74, 104, 114 *5 Cncept Smulatn 6.4 prves the ptn f explrng the ray agram that apples t ths prblem. The stance between an bject an ts mage frme by a vergng lens s 49.0 cm. The fcal length f the lens s.0 cm. Fn (a) the mage stance an (b) the bject stance. 7 A farsghte wman breaks her current eyeglasses an s usng an l par whse refractve pwer s pters. Snce these eyeglasses nt cmpletely crrect her vsn, she must hl a newspaper 4.00 cm frm her eyes n rer t rea t. She wears the eyeglasses.00 cm frm her eyes. Hw far s her near pnt frm her eyes? **74 The cntacts wrn by a farsghte persn allw her t see bjects clearly that are as clse as 5.0 cm, even thugh her uncrrecte near pnt s 79.0 cm frm her eyes. When she s lkng at a pster, the cntacts frm an mage f the pster at a stance f 17 cm frm her eyes. (a) Hw far away s the pster actually lcate? (b) If the pster s 0.50 m tall, hw tall s the mage frme by the cntacts? 104 A sle prjectr has a cnvergng lens whse fcal length s mm. (a) Hw far (n meters) frm the lens must the screen be lcate f a sle s place mm frm the lens? (b) If the sle measures, what are the mensns (n mm) f ts mage? *114 An bject s n frnt f a cnvergng lens (f = 0.0 m). The magnfcatn f the lens s m = 4.0. (a) Relatve t the lens, n what rectn shul the bject be mve s that the magnfcatn changes t m = 4.0? (b) Thrugh what stance shul the bject be mve?

2 5. REASONING A vergng lens always pruces a vrtual mage, s that the mage stance s negatve. Mrever, the bject stance s pstve. Therefre, the stance between the bject an the mage s + = 49.0 cm, rather than = 49.0 cm. The equatn + = 49.0 cm an the thn-lens equatn cnsttute tw equatns n tw unknwns, an we wll slve them smultaneusly t btan values fr an. a. Slvng the equatn + = 49.0 cm fr, substtutng the result nt the thn-lens equatn, an suppressng the unts gve 1 r + = f =.0 (1) Grupng the terms n the left f Equatn (1) ver a cmmn enmnatr, we have = = ().0 ( 49.0 ) ( 49.0 ) Crss-multplyng an rearrangng n Equatn () gves 49.0 = r = 0 () Usng the quaratc frmula t slve Equatn (), we btan ± ( ) ( 1.00) = = 85.1 cm We have scare the pstve rt, because we knw that must be negatve fr the vrtual mage. b. Usng the fact that + = 49.0 cm, we fn that the bject stance s = 49.0 cm = 49.0 cm 85.1 cm = 14.1 cm 7. REASONING The eyeglasses frm a vrtual mage f the newspaper page at her near pnt. The mage stance s negatve, an we can calculate t by usng the thn-lens equatn. In ng s, we nee t keep tw thngs n mn. Frst, we must take nt accunt the fact that she wears her eyeglasses.00 cm frm her eyes. Ths s because the stances that appear n the thn-lens equatn are measure wth respect t the lens f the eyeglasses, nt wth respect t her eyes. Thus, the bject stance (the lcatn f the newspaper page) s = 4.00 cm.00 cm. Secn, n usng the thn-lens equatn we must have a value fr

3 the fcal length f, an ths s nt rectly gven. Hwever, we knw the refractve pwer f the lens, whch s the recprcal f the fcal length, accrng t Equatn 6.8. After calculatng the mage stance, we can btan her near pnt wth respect t the eyeglasses by takng the magntue f ths negatve number. Fnally, we wll accunt fr the fact that the eyeglasses are.00 cm frm her eyes. Accrng t the thn-lens equatn, the recprcal f the mage stance s = f T use ths expressn, we nee a value fr the fcal length, an we use Equatn 6.8 fr the refractve pwer t btan t. Ths equatn ncates that the refractve pwer n pters s the recprcal f the fcal length n meters, s we have Refractve pwer = r f = = = m 60.4 cm f Refractve pwer pters Rememberng that the eyeglasses are wrn.00 cm frm the eyes, we can nw apply the thnlens equatn: -1 = = = cm r = 119 cm f 60.4 cm 4.00 cm.00 cm The magntue f ths value fr s 119 cm an gves the lcatn f the wman s near pnt wth respect t the eyeglasses. Snce the eyeglasses are wrn.00 cm frm the eyes, the near pnt s lcate at a stance f 11 cm frm the eyes. 74. REASONING AND If the near pnt s 79.0 cm, then = 79.0 cm, an = 5.0 cm. Usng the thn-lens equatn, we fn that the fcal length f the crrectng lens s (5.0 cm)( 79.0 cm) f = = = cm cm + ( 79.0 cm) a. The stance t the pster can be btane as fllws: = = f 6.6 cm 17 cm b. The mage sze s 17 cm h = h = (0.50 m) =.4 m 1. cm r = 1. cm

4 104. REASONING The stance frm the lens t the screen, the mage stance, can be btane rectly frm the thn-lens equatn, Equatn 6.6, snce the bject stance an fcal length are knwn. The wth an heght f the mage n the screen can be etermne by usng Equatn 6.7, the magnfcatn equatn. a. The stance t the screen s s that = = = mm f mm mm = mm =.78 m. 4 1 b. Accrng t the magnfcatn equatn, the wth an heght f the mage n the screen are mm 108 mm Wth h h = = 4.0 mm = mm The wth s mm mm 108 mm Heght h h = = 6.0 mm = mm The heght s mm REASONING T fn the stance thrugh whch the bject must be mve, we must btan the bject stances fr the tw stuatns escrbe n the prblem. T ths, we cmbne the thn-lens equatn an the magnfcatn equatn, snce ata fr the magnfcatn s gven. a. Snce the magnfcatn s pstve, the mage s uprght, an the bject must be lcate wthn the fcal pnt f the lens, as n Fgure 6.8. When the magnfcatn s negatve an has a magntue greater than ne, the bject must be lcate between the fcal pnt an the pnt that s at a stance f twce the fcal length frm the lens, as n Fgure 6.7. Therefre, the bject shul be mve away frm the lens. b. Accrng t the thn-lens equatn, we have = 1 f r = f f (1)

5 Accrng t the magnfcatn equatn, wth expresse as n Equatn (), we have m = = 1 f = f f r f = f m 1 m () Applyng Equatn () t the tw cases escrbe n the prblem, we have ( ) + m ( + ) f m 1 f f = = = m () ( ) m f m 1 f f = = = (4) m Subtractng Equatn () frm Equatn (4), we fn that the bject must be mve away frm the lens by an atnal stance f 5.0 f.0 f.0 f 0.0 m = = = = 0.15 m m + m