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Where Is the Final Image? Two thin converging lenses of focal lengths f, - 10.0 cm and f, - 18.0 cm are separated by 20.0 cm as illustrated in the figure. An object is placed 31.0 cm to the left of Lens 1. Find the position and the magnification of the final image. Figure Description Lens 1 Lens 2 IN Pi 20.0 cm- SOLUTION Conceptualize Imagine light rays passing through the first lens and forming a real image (because p> v n in the absence of a second lens. The figure shows these light rays forming the inverted image I,. Once the light rays converge to the image point, they do not stop. They continue through the image point and interact with the second lens. The rays leaving the image point behave in the same way as the rays leaving an object. Therefore, the image of the first lens serves as the object of the second lens. Categorize We categorize this problem as one in which the thin lens equation v is applied in a stepwise fashion to the two lenses. Analyze Find the location of the image formed by Lens 1 from the thin lens equation: (Enter all distances in cm. Enter values of q, to at least one decimal place.) Find the location of the image formed by Lens 1 from the thin lens equation: 1 1 1 cm Find the magnification of the image: M,-- P1 The image formed by this lens acts as the object for the second lens. Therefore, the object distance for the second lens is 20 cm - cm= cm. Find the location of the image formed by Lens 2 from the thin lens equation: 1 1 18.0 cm 92 cm 92 - cm Find the magnification of the image: 92 M2 -- P2 Find the overall magnification of the system from this equation: M= M,M2 = ( Finalize The negative sign on the overall magnification indicates that the final image is inverted with respect to the initial object. Because the absolute value of the magnification is less than 1, the final image is smaller than the object. Because q, is negative, the final image is on the front, or --?- diagram in the figure. side of Lens 2. These conclusions are consistent with the ray

Where Is the Final Image? Two thin converging lenses of focal lengths f, - 10.0 cm and f, - 18.0 cm are separated by 20.0 cm as illustrated in the figure. An object is placed 31.0 cm to the left of Lens 1. Find the position and the magnification of the final image. Figure Description Lens 1 Lens 2 IN Pi 20.0 cm- SOLUTION Conceptualize Imagine light rays passing through the first lens and forming a real image (because p> v n in the absence of a second lens. The figure shows these light rays forming the inverted image I,. Once the light rays converge to the image point, they do not stop. They continue through the image point and interact with the second lens. The rays leaving the image point behave in the same way as the rays leaving an object. Therefore, the image of the first lens serves as the object of the second lens. Categorize We categorize this problem as one in which the thin lens equation v is applied in a stepwise fashion to the two lenses. Analyze Find the location of the image formed by Lens 1 from the thin lens equation: (Enter all distances in cm. Enter values of q, to at least one decimal place.) Find the location of the image formed by Lens 1 from the thin lens equation: 1 1 1 cm Find the magnification of the image: M,-- P1 The image formed by this lens acts as the object for the second lens. Therefore, the object distance for the second lens is 20 cm - cm= cm. Find the location of the image formed by Lens 2 from the thin lens equation: 1 1 18.0 cm 92 cm 92 - cm Find the magnification of the image: 92 M2 -- P2 Find the overall magnification of the system from this equation: M= M,M2 = ( Finalize The negative sign on the overall magnification indicates that the final image is inverted with respect to the initial object. Because the absolute value of the magnification is less than 1, the final image is smaller than the object. Because q, is negative, the final image is on the front, or --?- diagram in the figure. side of Lens 2. These conclusions are consistent with the ray

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Where Is the Final Image? Two thin converging lenses of focal lengths f, = 10.0 cm and f, = 18.0 cm are separated by 20.0 cm as illustrated in the figure. An object is placed 31.0 cm to the left of Lens 1. Find the position and the magnification of the final image. Figure Description Lens 1 Lens 2 fi Pi 20.0 cm- SOLUTION Conceptualize Imagine light rays passing through the first lens and forming a real image (because p n in the absence of a second lens. The figure shows these light rays forming the inverted image I . Once the light rays converge to the image point, they do not stop. They continue through the image point and interact with the second lens. The rays leaving the image point behave in the same way as the rays leaving an object. Therefore, the image of the first lens serves as the object of the second lens. Categorize We categorize this problem as one in which the thin lens equation v is applied in a stepwise fashion to the two lenses. Analyze Find the location of the image formed by Lens 1 from the thin lens equation: (Enter all distances in cm. Enter values of q, to at least one decimal place.) Find the location of the image formed by Lens 1 from the thin lens equation: 1 1 = 91 1 P1 91 = cm Find the magnification of the image: 91 M, = - P1 The image formed by this lens acts as the object for the second lens. Therefore, the object distance for the second lens is 20 cm - cm = ст. cm. Find the location of the image formed by Lens 2 from the thin lens equation: 1 1 92 18.0 cm cm 92 = cm Find the magnification of the image: 92 M2 = P2 Find the overall magnification of the system from this equation: M = M,M2 =| Finalize The negative sign on the overall magnification indicates that the final image is inverted with respect to the initial object. Because the absolute value of the magnification is less than 1, the final image is smaller than the object. Because q, is negative, the final image is on the front, or --?-- , side of Lens 2. These conclusions are consistent with the ray diagram in the figure. EXERCISE A system consists of two lenses mounted on an optical bench. Lens 1 has focal length 5.20 cm. An object is placed 7.5 cm to the left of this lens. A second lens, Lens 2, with focal length 8.90 cm, is placed to the right of the first lens. A screen is to the right of the second lens. The image on the screen has a magnification of +1.00. What is the distance (in cm) between the object and the screen? Hint cm Need Help? Read It
Transcribed Image Text:Where Is the Final Image? Two thin converging lenses of focal lengths f, = 10.0 cm and f, = 18.0 cm are separated by 20.0 cm as illustrated in the figure. An object is placed 31.0 cm to the left of Lens 1. Find the position and the magnification of the final image. Figure Description Lens 1 Lens 2 fi Pi 20.0 cm- SOLUTION Conceptualize Imagine light rays passing through the first lens and forming a real image (because p n in the absence of a second lens. The figure shows these light rays forming the inverted image I . Once the light rays converge to the image point, they do not stop. They continue through the image point and interact with the second lens. The rays leaving the image point behave in the same way as the rays leaving an object. Therefore, the image of the first lens serves as the object of the second lens. Categorize We categorize this problem as one in which the thin lens equation v is applied in a stepwise fashion to the two lenses. Analyze Find the location of the image formed by Lens 1 from the thin lens equation: (Enter all distances in cm. Enter values of q, to at least one decimal place.) Find the location of the image formed by Lens 1 from the thin lens equation: 1 1 = 91 1 P1 91 = cm Find the magnification of the image: 91 M, = - P1 The image formed by this lens acts as the object for the second lens. Therefore, the object distance for the second lens is 20 cm - cm = ст. cm. Find the location of the image formed by Lens 2 from the thin lens equation: 1 1 92 18.0 cm cm 92 = cm Find the magnification of the image: 92 M2 = P2 Find the overall magnification of the system from this equation: M = M,M2 =| Finalize The negative sign on the overall magnification indicates that the final image is inverted with respect to the initial object. Because the absolute value of the magnification is less than 1, the final image is smaller than the object. Because q, is negative, the final image is on the front, or --?-- , side of Lens 2. These conclusions are consistent with the ray diagram in the figure. EXERCISE A system consists of two lenses mounted on an optical bench. Lens 1 has focal length 5.20 cm. An object is placed 7.5 cm to the left of this lens. A second lens, Lens 2, with focal length 8.90 cm, is placed to the right of the first lens. A screen is to the right of the second lens. The image on the screen has a magnification of +1.00. What is the distance (in cm) between the object and the screen? Hint cm Need Help? Read It
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