Physics Solution Manual for 1100 and 2101

Thus in this case the minimum condition for

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

View Full Document Right Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: g that follows, we locate the image using the two rays labeled a and b, which originate at the top of the object. a b F Object Image 2 0.0 cm 1422 THE REFRACTION OF LIGHT: LENSES AND OPTICAL INSTRUMENTS a. Measuring according to the scale used in the drawing, we find that the image is located 10.0 cm to the left of the lens. The lens is a diverging lens and forms a virtual image, so the image distance is di = –10.0 cm . b. Measuring the heights of the image and the object in the drawing, we find that the magnification is m = +0.500 . ______________________________________________________________________________ 119. SSM WWW REASONING The optical arrangement is similar to that in Figure 26.26. We begin with the thin-lens equation, [Equation 26.6: (1 / do ) + (1 / di ) = (1 / f ) ]. Since the distance between the moon and the camera is so large, the object distance do is essentially infinite, and 1 / do = 1 / ∞ = 0 . Therefore the thin-lens equation becomes 1 / di = 1 / f or di = f . The diameter of the moon's imagine on the slide film is equal to the image height hi, as given by the magnification equation (Equation 26.7: hi / ho = – di / do ). When the slide is projected onto a screen, the situation is similar to that in Figure 26.27. In this case, the thin-lens and magnification equations can be used in their usual forms. SOLUTION a. Solving the magnification equation for hi gives h i = – ho 50.0 × 10 –3 m –4 = (–3.48 × 10 6 m) m = –4.52 × 10 do 3.85 × 10 8 m di The diameter of the moon's image on the slide film is, therefore, 4.52 × 10 –4 m . b. From the magnification equation, hi = – ho ( d i / d o ) . We need to find the ratio di / do . Beginning with the thin-lens equation, we have 1 1 1 + = do di f or 1 11 =– do f di which leads to di do = di f – di di = di f –1 Therefore, ( ) d 15.0 m hi = – ho i –1 = – 4.52 ×10 –4 m –1 = –6.12 ×10 –2 m 110.0 ×10 –3 m f The diameter of the image on the screen is 6.12 × 10 –2 m . ______________________________________________________________________________ Chapter 26 Problems 1423 120. REASONING The angular magnification of a magnifying glass is given by Equation 26.10: M ≈ (1/ f ) – (1/ di ) N , where N is the distance from the eye to the near-point. For maximum magnification, the closest to the eye that the image can be is at the near point, with di = – N (where the minus sign indicates that the image lies to the left of the lens and is virtual). In this case, Equation 26.10 becomes Mmax ≈ N / f + 1 . At minimum magnification, the image is as far from the eye as it can be ( di = – ∞ ); this occurs when the object is placed at the focal point of the lens. In this case, Equation 26.10 simplifies to Mmin ≈ N / f . Since the woman observes that for clear vision, the maximum angular magnification is 1.25 times larger than the minimum angular magnification, we have Mmax = 1.25Mmin . This equation can be written in terms of N and f using the above expressions, and then solved for f. SOLUTION We have N N + 1 = 1.25 f f { 123 M M min max Solving for f, we find that f = 0.25 N = (0.25)(25 cm) = 6.3 cm ______________________________________________________________________________ 121. SSM REASONING A contact lens is placed directly on the eye. Therefore, the object distance, which is the distance from the book to the lens, is 25.0 cm. The near point can be determined from the thin-lens equation [Equation 26.6: (1 / do ) + (1 / di ) = (1 / f ) ]. SOLUTION a. Using the thin-lens equation, we have 111 1 1 =− = − di f do 65.0 cm 25.0 cm or di = −40.6 cm In other words, at age 40, the man's near point is 40.6 cm. Similarly, when the man is 45, we have 111 1 1 =− = − or di = −52.4 cm di f do 65.0 cm 29.0 cm and his near point is 52.4 cm. 52.4 cm – 40.6 cm = 11.8 cm . Thus, the man’s near point has changed by 1424 THE REFRACTION OF LIGHT: LENSES AND OPTICAL INSTRUMENTS b. With do = 25.0 cm and di = –52.4 cm , the focal length of the lens is found as follows: 1 1 1 1 1 = + = + or f = 47.8 cm f d o d i 25.0 cm (–52.4 cm) ______________________________________________________________________________ 122. REASONING The image distance di2 produced by the 2nd lens is related to the object distance do2 and the focal length f2 by the thin-lens equation (Equation 26.6). The focal length is known, but the object distance is not. However, the problem states that the object distance is equal to that (do1) of the 1st lens, so do2 = do1. Since the final image distance di1 and the focal length f1 of the 1st lens are known, we can determine the object distance for this lens by employing the thin-lens equation. SOLUTION The image distance di2 produced by the 2nd lens is related to the object distance do2 and the focal length f2 by the thin-lens equation: 1 1 1 = − di2 f 2 do2 (26.6) Since do2 = do1 (the image distance for the 1st lens), Equation 26.6 can be written as 1 1 1 = − di2 f 2 do1 (1) The object distance for the...
View Full Document

This note was uploaded on 04/30/2013 for the course PHYS 1100 and 2 taught by Professor Chastain during the Spring '13 term at LSU.

Ask a homework question - tutors are online