Physics Solution Manual for 1100 and 2101

# Without magnifying glasses the smallest object

This preview shows page 1. Sign up to view the full content.

This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: n (1), we find that D = di, f − di, 0 = do, f f do, f − f − do, 0 f do, 0 − f =f do, f do, f − f − do, 0 do, 0 − f (4) Chapter 26 Problems 1389 Substituting Equation (4) into Equation 2.1, we obtain f s= D = ∆t d o, f d o, f − f − d o, 0 d o, 0 − f (5) ∆t During the first two seconds after the camera begins rolling, the horse is initially a distance do, 0 = 40.0 m from the lens, and ends up (7.0 m/s)(2.0 s) = 14.0 m closer to the lens. The final object distance, then, is do, f = 40.0 m − 14.0 m = 26.0 m. From Equation (5), the average speed of the image during this interval is f s= d o, f − d o, f − f d o, 0 d o, 0 − f ∆t = ( 50.0 m ) 26.0 m 40.0 m − 26.0 m − 50.0 m 40.0 m − 50.0 m = 73 m/s 2.0 s b. During the next 2.0 s of filming, the horse’s initial position is do, 0 = 26.0 m from the lens, and its final position is do, f = 26.0 m − 14.0 m = 12.0 m. Once again employing Equation (5), we find that f s= d o, f d o, f − f − ∆t d o, 0 d o, 0 − f = ( 50.0 m ) 12.0 m 26.0 m − 12.0 m − 50.0 m 26.0 m − 50.0 m = 19 m/s 2.0 s 65. REASONING We will consider one lens at a time, using the thin-lens equation for each. The key to the solution is the fact that the image formed by the first lens serves as the object for the second lens. SOLUTION Using the thin-lens equation, we find the image distance for the first lens: 111 1 1 =– = – di f d o –8.0 cm 4.0 cm or d i = –2.7 cm The negative value for di indicates that the image is virtual and located 2.7 cm to the left of the first lens. The lenses are 16 cm apart, so this image is located 2.7 cm + 16 cm = 18.7 cm from the second lens. Since this image serves as the object for the second lens, we can locate the image formed by the second lens with the aid of the thin-lens equation, with do = 18.7 cm: 111 1 1 =– = – di f d o –8.0 cm 18.7 cm or d i = –5.6 cm 1390 THE REFRACTION OF LIGHT: LENSES AND OPTICAL INSTRUMENTS ______________________________________________________________________________ 66. REASONING In a two-lens situation, the image produced by the first (converging) lens serves as the object for the second (diverging) lens. We will use the thin-lens equation 111 += (Equation 26.6) to determine the object distance do2 for the diverging lens d o di f from the focal length f2 = −28.0 cm and the image distance di2. This object is the image produced by the first (converging) lens. A second application of Equation 26.6 will then permit us to find the distance do1 between the converging lens and the object. SOLUTION Because the final image appears to the left of the second (diverging) lens, the 1 1 1 + = image distance di2 is negative: di2 = −20.7 cm. Solving (Equation 26.6) for do2 di2 f 2 do2 yields 1 1 1 1 1 = − = − = 0.0126 cm −1 do2 f 2 di2 −28.0 cm −20.7 cm or do2 = 79.4 cm This is a positive object distance, so our sign convention indicates that the object for the diverging lens is located 79.4 cm to the left of the diverging lens. Because this distance is greater than the distance L = 56.0 cm between the lenses, the object for the diverging lens is located do2 − L = 79.4 cm − 56.0 cm = 23.4 cm left of the converging lens (see the drawing). First image di2 do1 Object Final image di1 L do2 This object is the image produced by the first (converging) lens. However, as the image appears left of the converging lens, the image distance di1 is negative: di1 = −23.4 cm. 1 1 1 Solving + = (Equation 26.6) for do1, we find that do1 di1 f1 1 1 1 1 1 =− = − = 0.0844 cm −1 do1 f1 di1 24.0 cm −23.4 cm or do1 = 11.8 cm Chapter 26 Problems 1391 67. REASONING The thin-lens equation can be used to find the image distance of the first image (the image produced by the first lens). This image, in turn, acts as the object for the second lens. The thin-lens equation can be used again to determine the image distance for the final image (the image produced by the second lens). SOLUTION For the first lens, the object and image distances, do,1 and di,1, are related to the focal length f of the lens by the thin-lens equation 1 1 1 + = do1 di1 f (26.6) Solving this expression for the image distance produced by the first lens, we find that 1 1 1 1 1 =− = − di1 f do1 12.00 cm 36.00 cm or di1 = 18.0 cm This image distance indicates that the first image lies between the lenses. Since the lenses are separated by 24.00 cm, the distance between the image produced by the first lens and the second lens is 24.00 cm − 18.0 cm = 6.0 cm. Since the image produced by the first lens acts as the object for the second lens, we have that do2 = 6.0 cm. Applying the thin-lens equation to the second lens gives 1 1 1 1 1 =− = − di2 f do2 12.00 cm 6.0 cm or di 2 = −12 cm The fact that this image distance is negative means that the final image is virtual and lies to the left of the second lens. ______________________________________________________________________________ 68. REASONING In part a of the drawing, the object...
View Full Document

{[ snackBarMessage ]}

Ask a homework question - tutors are online