Chapter 36 - Image Formation

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Francis Bell
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1 Chater 6 - Image Formation P6. The flatness of the mirror is described by R =, f = 0 f = By our general mirror euation, or = = 0 f FIG. P6. Thus, the image is as far behind the mirror as the erson is in front. The magnification is then so h = = = h h = h= 70.0 inches The reuired height of the mirror is defined by the triangle from the erson s eyes to the to bottom of his image, as shown. From the geometry of the triangle, we see that the mirror height must be: h h h = = Thus, the mirror must be at least 5.0 inches high. P6.9 (a) becomes R = 45.0 cm = 60.0 cm 90.0 cm 45.0 cm = = = cm (b) becomes R = 60.0 cm = 60.0 cm 0.0 cm ( 60.0 cm ) ( 0.0 cm ) = = =.00 (c) The image (a) is real, inverted, diminished. That of (b) is virtual, uright, enlarged. The ray diagrams are similar to Figures 6.(a) 6.(b) in the text, resectively. P6. (a) = 4 = = 4 FIG. P6.9 = 0.60 m = 4 = 0. m = 0.8 m f = + = 0. m + f = 0.8 m 60 mm

2 (b) = = + = 0.0 m = + = = 66.7 mm = mm + = R = 0. m + R = m 67 mm P6.7 (a) = ( m ), since the image must be real, = = 5 or = 5 Therefore, m = 5 or =.5 m = 6.5 m From, R ( )( ) R = = = ( ).08 m concave FIG. P6.7 (b) From art (a), =. 5 m ; the mirror should be.5 m in front of the object. P6.9 (a) The image starts from a oint whose height above the mirror vertex is given by = f R Therefore,.00 m m = m As the ball falls, decreases increases. Ball image ass when =. When this is true, + = m = or =.00 m As the ball asses the focal oint, the image switches from infinitely far above the mirror to infinitely far below the mirror. As the ball aroaches the mirror from above, the virtual image aroaches the mirror from below, reaching it together when = = 0. (b) The falling ball asses its real image when it has fallen.00 m.00 m =.00 m = gt, or when ( ).00 m t = = 9.80 m s 0.69 s The ball reaches its virtual image when it has traversed ( ).00 m.00 m 0 =.00 m = gt, or at t = = 9.80 m s 0.78 s

3 P6. From Euation 6.8 Solve for to find n n n n R nr = n n n R In this case, n =.50, n =.00, R = 5.0 cm So Therefore, the = 0.0 cm (.00)( 5.0 cm )( 0.0 cm ) ( 0.0 cm )(.00.50) (.50)( 5.0 cm ) = = 8.57 cm aarent deth is 8.57 cm *P6.5 (a) The center of curvature is on the object side, so the radius of curvature is negative. n n n n R becomes = 4.9 cm 0 cm 80 cm So the image is inside the tank, 4.9 cm behind the front wall; virtual, right side u, enlarged. (b) Now we have = 9.9 cm 90 cm 80 cm So the image is inside the tank, 9.9 cm behind the front wall; virtual, right side u, enlarged. (c) In case (a) the result of roblem 4 gives In case (b) we have. 9.9 = = (90) n. 4.9 = = = +.0 n.00(0) (d) In case (a) h' = h =.0(9.00 cm) = 9.9 cm. In case (b), the farther lobster looms larger: h' = h =.0(9.00 cm) =.5 cm (e) The lastic has uniform thickness, so the surfaces of entry exit for any articular ray are very nearly arallel. The ray is slightly dislaced, but it would not be changed in direction by going through the lastic wall with air on both sides. Only the difference between the air water is resonsible for the refraction of the light = = f R R.0 cm 8.0 cm P6.7 (a) ( n ) ( ) f = 6.4 cm

4 = f 8.0 cm (.0 cm ) (b) ( ) FIG. P6.7 f = 6.4 cm P6. We are looking at an enlarged, uright, virtual image: h = = = so h (.84 cm ) = = =+.4 cm gives f.4 cm.84 cm f f =.84 cm FIG. P6. P6. : f We may differentiate through with resect to : + = constant d = 0 d d = = d *P6.8 (a) + = becomes f a a or a = 6. cm 0.0 cm 4.0 cm a hb = ha = h = ( 0.0 cm )( 0.875) = 8.75 cm a or d = 46.7 cm 0.0 cm d 4.0 cm ( )( ) h = h = 0.0 cm. =. cm c d The suare is imaged as a traezoid. FIG. P6.8(b)

5 (b) becomes f or / = / 4 cm / 4 cm (c) h = h = h = ( 0.0 cm ) 4 cm d The uantity hd adds u the geometrical (ositive) areas of thin vertical a ribbons comrising the whole area of the image. We have hd = d = a 4 cm 8 cm d d ( 0.0 cm ) ( 0.0 cm ) a A rea = ( 0.0 cm ) cm = 8 cm cm 6. cm P6.47 f o = 0.0 m f = m e fo (a) The angular magnification roduced by this telescoe is m = = 800. f (b) Since m < 0, the image is inverted. e P6.48 (a) The mirror--lens euation gives Then, f f = = = f f f f ( ) h f = = = h f gives h = hf f (b) For >> f, f. Then, hf h = (c) Suose the telescoe observes the sace station at the zenith: ( 08.6 m)( 4.00 m) hf h = = =.07 mm m

6 *P6.54 Start with the first ass through the lens. = f = 80.0 cm 00 cm = 400 cm to right of lens For the mirror, = 00 cm = = f ( 50.0 cm) ( 00 cm) = 60.0 cm For the second ass through the lens, = 60 cm FIG. P6.54 = f = 80.0 cm 60 cm = 60 cm to the left of lens 400 cm = = = cm ( 60.0 cm) = = = ( 00 cm) 5 60 cm = = = = = cm Since < 0 the final image is inverted. P6.56 = = f 0.0 cm.5 cm so = 50.0 cm (to left of mirror) This serves as an object for the lens (a virtual object), so = = f ( 6.7 cm ) ( 5.0 cm ) = 50. cm meaning 50. cm to the right of the lens. Thus, the final image is located 5. cm to right of mirror cm = = = cm ( 50. cm ) ( 5.0 cm ) = = =.0 = = 8.05 Thus, the final image is virtual, uright right of the mirror., 8.05 times the size of object, 5. cm to P6.59 A hemishere is too thick to be described as a thin lens. The light is undeviated on entry into the flat face. We next consider the light s exit from the second surface, for which R = 6.00 cm. The incident rays are arallel, so =

7 Then, becomes n n n n R cm = 0.7 cm FIG. P6.59 P6.6 From the thin lens euation, ( 6.00 cm )(.0 cm ).0 cm ( 6.00 cm ) f = = = 4.00 cm f When we reuire that, the thin lens euation becomes = f. In this case, = d 4.00 cm Therefore, d cm = f =.0 cm d = 8.00 cm FIG. P6.6 = + f R R P6.64 (a) The lens makers euation, ( n ) = 5.00 cm 9.00 cm (.0 cm ) becomes: ( n ) giving n =.99. (b) As the light asses through the lens for the first time, the thin lens euation f becomes: or =. cm, 8.00 cm 5.00 cm This image becomes the object for the concave mirror with:. cm = = = cm m = 0.0 cm = 0.0 cm. cm = 6.67 cm R f = = cm

8 The mirror euation becomes: giving 6.67 cm m 4.00 cm = 0.0 cm m 0.0 cm = = = cm The image formed by the mirror serves as a real object for the lens on the second ass of the light through the lens with: The thin lens euation yields: m m = 0.0 cm m = cm 0.0 cm 5.00 cm or = 0.0 cm 0.0 cm = = = cm The final image is a real image located 0.0 cm to the left of the lens. The overall magnification is total = =.50 (c) Since the total magnification is negative, this final image is inverted. P6.66 The object is located at the focal oint of the uer mirror. Thus, the uer mirror creates an image at infinity (i.e., arallel rays leave this mirror). The lower mirror focuses these arallel rays at its focal oint, located at the hole in the uer mirror. Thus, the image is real, inverted, actual size. For the uer mirror: : f = 7.50 cm 7.50 cm For the lower mirror: 7.50 cm = 7.50 cm Light directed into the hole in the uer mirror reflects as shown, to behave as if it were reflecting from the hole. FIG. P6.66

9 P6.67 (a) For lens one, as shown in the first figure, 40.0 cm 0.0 cm = 0 cm 0 cm = = = cm This real image I = O is a virtual object for the second lens. That is, it is behind the lens, as shown in the second figure. The object distance is = 0 cm 0 cm = 0.0 cm : 0.0 cm 0.0 cm = 0.0 cm 0.0 cm = = =+.00 overall ( 0.0 cm ) = = 6.00 (b) overall < 0, so final image is inverted. (c) If lens two is a converging lens (third figure): 0.0 cm 0.0 cm = 6.67 cm 6.67 cm = = ( 0.0 cm ) = =.00 overall FIG. P6.67 Again, overall < 0 the final image is inverted.

( ) = + ANSWERS TO EVEN NUMBERED CONCEPTUAL QUESTIONS

( ) = + ANSWERS TO EVEN NUMBERED CONCEPTUAL QUESTIONS Mirrors and Lenses 39 7. A concave mirror forms inverted, real images of real objects located outside the focal oint ( > f ), and uright, magnified, virtual images of real objects located inside the focal