Physics Solution Manual for 1100 and 2101

# 615 and y 240 10 m 2 the distance between the fringes

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Unformatted text preview: positive indicates that the image is real . c. The magnification equation indicates that the magnification m is m= hi d =− i ho do where ho and hi are the object height and image height, respectively. Solving for the image height gives d 204 cm hi = ho − i = (13.0 cm ) − = −17.1 cm d 155.0 cm o The negative value indicates that the image is inverted with respect to the object. 113. SSM REASONING We begin by using Snell's law (Equation 26.2: n1 sin θ1 = n2 sin θ 2 ) to find the index of refraction of the material. Then we will use Equation 26.1, the definition of the index of refraction ( n = c / v ) to find the speed of light in the material. SOLUTION From Snell's law, the index of refraction of the material is Chapter 26 Problems n2 = 1419 n1 sin θ1 (1.000) sin 63.0° = = 1.22 sin θ 2 sin 47.0° Then, from Equation 26.1, we find that the speed of light v in the material is c 3.00 × 10 8 m/s = = 2.46 × 10 8 m/s n2 1.22 ______________________________________________________________________________ v= 114. REASONING The angular size θ of the image is the angular magnification M times the reference angular size θ of the object: θ ′ = Mθ (Equation 26.9). The reference angular size is that seen by the naked eye when the object is located at the near point and is given in the problem statement as θ = 0.060 rad. To determine the angular magnification, we can utilize Equation 26.10, assuming that the angles involved are small: 1 1 M ≈ − N f d i where f is the focal length of the magnifying glass, di is the image distance, and N is the distance between the eye and the near point. In this expression we note that the image distance is negative since the image in a magnifying glass is virtual (di = −64 cm). SOLUTION Substituting the expression for the angular magnification into Equation 26.9 gives 1 1 1 1 − θ ′ = M θ ≈ − Nθ = ( 32 cm )( 0.060 rad ) = 0.15 rad f d 16 cm ( −64 cm ) i 115. REASONING Nearsightedness is corrected using diverging lenses to form a virtual image at the far point of the eye, as Section 26.10 discusses. The far point is given as 5.2 m, so we know that the image distance for the contact lenses is di = –5.2 m. The minus sign indicates that the image is virtual. The thin-lens equation can be used to determine the focal length. SOLUTION According to the thin-lens equation, we have 11 1 1 1 += + = or f = –9.2 m d o di 12.0 m –5.2 m f ______________________________________________________________________________ 1420 THE REFRACTION OF LIGHT: LENSES AND OPTICAL INSTRUMENTS 116. REASONING When light goes from air into the plastic, the light is refracted. Snell’s law relates the incident and refracted angles (θ1 and θ2) to the indices of refraction (n1 and n2) of the incident and refracting media by: Violet light n1 sin θ1 = n2, Violet sin θ 2, Violet (26.2) Red light n1 sin θ1 = n2, Red sin θ2, Red (26.2) By using these relations, and the fact that n2, Violet − n2, Red = 0.0400, we will be able to determine n2, Violet. SOLUTION Since the angle of incidence θ1 is the same for both colors and since n1 = nair for both colors, the left-hand sides of the two equations above are equal. Thus, the righthand sides of these equations must also be equal: n2, Violet sin θ 2, Violet = n2, Red sin θ 2, Red (1) We are given that n2, Violet − n2, Red = 0.0400, or n2, Red = n2, Violet − 0.0400 . Substituting this expression for n2, Red into Equation (1), we have that n2, Violet sin θ 2, Violet = ( n2, Violet − 0.0400 ) sin θ 2, Red Solving this equation for n2, Violet gives n2, Violet = − ( 0.0400 ) sin θ 2, Red sin θ 2, Violet − sin θ 2, Red = − ( 0.0400 ) sin 31.200° = 1.73 sin 30.400° − sin 31.200° ______________________________________________________________________________ 117. REASONING We can use the magnification equation (Equation 26.7) to determine the image height hi. This equation is hi d =− i ho do or d hi = ho − i d o (26.7) We are given the object height ho and the object distance do. Thus, we need to begin by finding the image distance di, for which we use the thin-lens equation (Equation 26.6): 111 += d o di f or 1 1 1 do − f =− = di f d o fdo or di = fdo do − f (26.6) Chapter 26 Problems 1421 Substituting this result into Equation 26.7 gives d hi = ho − i d o 1 = ho − do fd o d o − f f = ho f − do (1) SOLUTION a. Using Equation (1), we find that the image height for the 35.0-mm lens is f 35.0 × 10−3 m = −0.00625 m hi = ho = (1.60 m ) f −d 35.0 × 10−3 m − 9.00 m o ( ) b. Using Equation (1), we find that the image height for the 150.0-mm lens is f 150.0 × 10−3 m = −0.0271 m hi = ho = (1.60 m ) f −d 150.0 × 10−3 m − 9.00 m o ( ) Both heights are negative because the images are inverted with respect to the object. 118. REASONING AND SOLUTION We note that the object is placed 20.0 cm from the lens. Since the focal point of the lens is f = –20.0 cm, the object is situated at the focal point. In the scale drawin...
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## 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.

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