Physics 11 HW #10 Solutions

Transcription

1 Physics HW #0 Slutins Chapter 5: Fcus On Cncepts: 4,, 3, 5 Prblems: 3, 5,, 9, 33, 37, 4, 44 Fcus On Cncepts 5-4 (c) The ray f light strikes the mirrr fur units dwn frm the tp f the mirrr with a 45 angle f incidence. The ray reflects frm the mirrr at an angle f 45 and passes thrugh pint C. Fcus On Cncepts 5- (c) Accrng t the scussin in Sectin 5.5, a cncave mirrr can prduce an enlarged image, prvided the bject stance is less than the raus f curvature. A cnvex mirrr cannt prduce an enlarged image, regardless f where the bject is lcated. Fcus On Cncepts 5-3 (b) A cnvex mirrr always prduces a virtual, upright image (see Sectin 5.5). Fcus On Cncepts 5-5 (c) A cnvex lens always prduces an upright image that is smaller than the bject. Prblem 5-3 REASONING The drawing shws the image f the bird in the plane mirrr, as seen by the camera. Nte that the image is as far behind the mirrr as the bird is in frnt f it. We can find the stance d between the camera and the image by nting that this stance is the hyptenuse f a right triangle. The base f the triangle has a length f 3.7 m +. m and the height f the triangle is 4.3 m. Bird 4.3 m Hedge. m 4.3 m Image SOLUTION The stance d frm the camera t the image f the bird can be btained by using the Pythagrean therem: 3.7 m. m d = ( 3.7 m +. m) + ( 4.3 m) = 7. m

2 Prblem 5-5 REASONING The bject stance (d = cm) is shrter than the fcal length (f = 8 cm) f the mirrr, s we expect the image t be virtual, appearing behind the mirrr. Taking Figure 5.8a as ur mdel, we will trace ut: three rays frm the tip f the bject t the surface f the mirrr, then three reflected rays, and finally three virtual rays extenng behind the mirrr and meeting at the tip f the image. The scale f the ray tracing will determine the lcatin and height f the image. The three sets f rays are:. An incident ray frm the bject t the mirrr, parallel t the principal axis and then reflected thrugh the fcal pint F.. An incident ray frm the bject t the mirrr, rectly away frm the fcal pint F and then reflected parallel t the principal axis. (The incident ray cannt pass frm the bject thrugh the fcal pint, as this wuld take it away frm the mirrr, and it wuld nt be reflected.) 3. An incident ray frm the bject t the mirrr, rectly away frm the center f curvature C, then reflected back thrugh C. SOLUTION C Image (virtual) 7.6 cm tall F cm 8 cm Object 3.0 cm tall Scale: 6.0 cm a. The ray agram incates that the image is 8 cm behind the mirrr. b. We see frm the ray agram that the image is 7.6 cm tall. Prblem 5- REASONING a. We are dealing with a cncave mirrr whse raus f curvature is 56.0 cm. Thus, the fcal length f the mirrr is f = R= 8.0 cm (Equatin 5.) The bject stance is d = 3.0 cm. With knwn values fr f and d, we can use the mirrr equatin t (Equatin 5.3) find the image stance. b. T determine the image height h i, recall that it is related t the bject height h by the magnificatin m; h i = mh. The magnificatin is related t the image and bject stances by

3 the magnificatin equatin, m= / d (Equatin 5.4). Since we knw values fr h, d i, and d, we can find the image height h i. SOLUTION a. The image stance is given by the mirrr equatin as fllws: = = s = 90 cm (5.3) d f d 8.0 cm 3.0 cm i b. Using the magnificatin equatin, we find that the image height is 90 cm hi = mh = h = ( 0.95 cm ) = 8.9 cm d 3.0 cm (5.4) c. Since h i is negative, the image is inverted relative t the bject. Thus, t make the picture n the wall appear nrmal, the slide must be riented upside dwn in the prjectr. Prblem 5-9 REASONING We need t knw the fcal length f the mirrr and can btain it frm the mirrr equatin, Equatin 5.3, as applied t the first bject: + = + = r f = 4.0 cm d d 4.0 cm 7.00 cm f i Accrng t the magnificatin equatin, Equatin 5.4, the image height h i is related t the bject height h as fllws:. SOLUTION Applying this result t each bject, we find that h i = h, r i Therefre, Using the fact that h = h, we have Using this result in the mirrr equatin, as applied t the secnd bject, we find that

4 r Therefre, + = d d f i + = d 0.50 d d = cm 4.0 cm Prblem 5-33 REASONING The raus f curvature R f a cncave mirrr is related t the fcal length f f the mirrr by f = R (Equatin 5.), s we have that R= f () The mirrr s fcal length is given by the mirrr equatin = + (Equatin 5.3), f d where d = 4 cm is the bject stance and d i is the image stance. The image stance is related t the bject stance via the magnificatin equatin m = (Equatin 5.4). d Because the image is virtual, the image stance d i is negative. The bject stance is psitive, s we cnclude frm Equatin 5.4 that the magnificatin is psitive: m > 0. The image is twice the size f the bject, s we have that the magnificatin is m = +.0. SOLUTION Taking the reciprcal f bth sides f = + (Equatin 5.3) yields f d f =. Substituting this result int Equatin (), we find that + d R= f = () + d Slving m = (Equatin 5.4) fr d d i and substituting m = +.0 yields d = md = + (.0) d = d (3) i Substituting Equatin (3) int Equatin (), we btain R= = = = 4d ( ) = 4 4 cm = 56 cm + + d d ( d) d

5 Prblem 5-37 REASONING The mirrr equatin relates the bject and image stances t the fcal length. Thus, we can apply the mirrr equatin nce with the given bject and image stances t determine the fcal length. Then, we can use the mirrr equatin again with the fcal length and the secnd bject stance t determine the unknwn image stance. SOLUTION Accrng t the mirrr equatin, we have + = and + = d d f d d f i i First psitin f bject Secnd psitin f bject Since the fcal length is the same in bth cases, it fllws that The negative value fr d i incates that the image is lcated 4 cm behind the mirrr. Prblem 5-4 REASONING This prblem can be slved by using the mirrr equatin, Equatin 5.3, and the magnificatin equatin, Equatin 5.4. SOLUTION a. Using the mirrr equatin with d = d and f = R /, we have Therefre, we find that d = R. i d f d R / = = r = d d R i b. Accrng t the magnificatin equatin, the magnificatin is m d d = i = = d d c. Since the magnificatin m is negative, the image is inverted.

6 Prblem 5-44 REASONING The fcal length f f the cnvex mirrr determines the first image stance d i via = + (Equatin 5.3), where d f d d = 5.0 cm is the stance between the candle i and the cnvex mirrr. The secnd image, frmed by the plane mirrr, is lcated as far behind the mirrr as the bject is in frnt f the mirrr. Therefre, when the plane mirrr replaces the cnvex mirrr, the secnd image stance d i is given by = d = 5.0 cm () The negative sign in Equatin () arises because the image lies behind the plane mirrr and, therefre, is a virtual image. When the plane mirrr replaces the cnvex mirrr, the image mves a stance x farther away frm the mirrr, s the initial and final image stances are related by d = d x () i The negative sign in Equatin () ccurs because the image mves farther frm the mirrr, thus making the secnd image stance d i a negative number f greater magnitude than the first image stance d i. SOLUTION Substituting Equatin () int Equatin () and slving fr d i yields i d x= d r d = x d (3) i i Substituting Equatin (3) int = + (Equatin 5.3), we btain f d d i = + = + (4) f d d d x d i Taking the reciprcal f bth sides f Equatin (4), we find that the fcal length f the cnvex mirrr is f = = = 7 cm + + d x d 5.0 cm 7.0 cm 5.0 cm