Three-lens systems. In the figure, stick figure O (the object) stands on the com

Question

Three-lens systems. In the figure, stick figure O (the object) stands on the common central axis of three thin, symmetric lenses, which are mounted in the boxed regions. Lens 1, a diverging lens with focal length -9.6 cm, is mounted within the boxed region closest to O, which is at object distance p1 = 5.9 cm. Lens 2, a converging lens with focal length 11 cm, is mounted within the middle boxed region, at distance d12 = 15 cm from lens 1. Lens 3, a converging lens with focal length 7.4 cm, is mounted in the farther boxed region, at distance d23 = 31 cm from lens 2. Find (a) the image distance i3 for the (final) image produced by lens 3 (the final image produced by the system) and (b) the overall lateral magnification M for the system, including signs.

Explanation / Answer

The object distance P1 = 4.3 cm The lens L1 is converging The focal length of the lens L1 is f1 = + 5.6 cm Apply the thin lens equation 1/f1 = 1/P1 + 1/q1 here q1 is the image distance from the lens L1 1/5.6 = 1/4.3 + 1/q1 q1 = -18.52 cm The image is virtual and upright The magnification is m1 = -q1/p1 m1 = -(-18.52)/4.3 m1 = 4.31 The image is formed at distance 18.52 cm from the lens L1 on the left side The image acts like the object for the lens L2 placed at d12 - q1 distance The object distance is p2 = d12-q1 p2 = 8 cm - (-18.52 cm) p2 = 26.52 cm The lens L2 is diverging The focal length of the lens L2 is f2 = -3.8 cm Apply thin lens equation 1/f2 = 1/p2 + 1/q2 Here q2 is the image distance from the lens L2 1/-3.8 = 1/26.52 + 1/q2 q2 = -3.32 cm The image is virtual and upright The magification is m2 = -q2/p2 m2 = -(-3.32 cm)/26.52 cm m2 = 0.125 The image is formed at 3.32 cm from the lens L2 on the left side The image acts like the object for the lens L3 placed at d23 - q2 distance The object distance is p3 = d23-q2 p3 = 5.7 cm - (-3.32 cm) p3 = 9.02 cm The lens L3 is diverging The focal length of the lens L3 is f3 = -12 cm Apply thin lens equation 1/f3 = 1/p3 + 1/q3 Here q3 is the image distance from the lens L3 1/-12 = 1/9.02 + 1/q3 q3 = -5.15 cm The image is virtual and upright The magification is m3 = -q2/p2 m3 = -(-5.15 cm)/9.02 cm m3 = 0.57 The final image of the system is formed at 5.15 cm from the lens L3 on the left side The final image is Virtual, upright and formed on the same side of the object of L3 The overall magnification M = m1m2m3 M = (4.31)(0.125)(0.57) M = 0.307

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