Solution since the angle of incidence θ 1 is the

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SOLUTION Since the angle of incidence θ 1 is the same for both colors and since n 1 = n air for both colors, the left-hand sides of the two equations above are equal. Thus, the right- hand sides of these equations must also be equal:
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Chapter 26 Problems 142 2, Violet 2, Violet 2, Red 2, Red sin sin n n θ θ = (1) We are given that 2, Violet 2, Red 0.0400, n n - = or 2, Red 2, Violet 0.0400 n n = - . Substituting this expression for n 2, Red into Equation (1), we have that ( 29 2, Violet 2, Violet 2, Violet 2, Red sin 0.0400 sin n n θ θ = - Solving this equation for n 2, Violet gives ( 29 ( 29 2, Red 2, Violet 2, Violet 2, Red 0.0400 sin 0.0400 sin31.200 1.73 sin sin sin30.400 sin31.200 n θ θ θ - - ° = = = - °- ° _____________________________________________________________________________ _ 117. REASONING We can use the magnification equation (Equation 26.7) to determine the image height h i . This equation is i i i i o o o o or h d d h h h d d = - = - (26.7) We are given the object height h o and the object distance d o . Thus, we need to begin by finding the image distance d i , for which we use the thin-lens equation (Equation 26.6): o o i o i i o o o 1 1 1 1 1 1 or or d f fd d d d f d f d fd d f - + = = - = = - (26.6) Substituting this result into Equation 26.7 gives o i i o o o o o o o 1 fd d f h h h h d d d f f d  = - = - =   - -  (1) SOLUTION a. Using Equation (1), we find that the image height for the 35.0-mm lens is ( 29 ( 29 3 i o 3 o 35.0 10 m 1.60 m 0.00625 m 35.0 10 m 9.00 m f h h f d - - × = = = - - × - b. Using Equation (1), we find that the image height for the 150.0-mm lens is
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143 THE REFRACTION OF LIGHT: LENSES AND OPTICAL INSTRUMENTS ( 29 ( 29 3 i o 3 o 150.0 10 m 1.60 m 0.0271 m 150.0 10 m 9.00 m f h h f d - - × = = = - - × - 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 drawing that follows, we locate the image using the two rays labeled a and b, which originate at the top of the object. 20.0 cm F Image Object a b 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 d i = . b. Measuring the heights of the image and the object in the drawing, we find that the magnification is m = . _____________________________________________________________________________ _ 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/ d o ) + (1/ d i ) = (1/ f ) ]. Since the distance between the moon and the camera is so large, the object distance d o is essentially infinite, and 1/ d o = 1/ ∞ = 0 . Therefore the thin-lens equation becomes 1/ d i = 1/ f or d i = f . The diameter of the moon's imagine on the slide film is equal to the image height h i , as given by the magnification equation (Equation 26.7: h i / h o = d i / d o ).
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Chapter 26 Problems 144 When the slide is projected onto a screen, the situation is similar to that in Figure 26.27.
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  • Spring '11
  • rollino
  • Physics, Light, Snell's Law, NC, Total internal reflection, Geometrical optics

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