a) Calculate the distance between image and diverging lens. b) Calculate the ima

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Explanation / Answer

A converging lens with a focal length of 50cm and a diverging lens with a focal length of -52 cm are 227cm apart. A 3.9-cm-tall object is 70cm in front of the converging lens. Calculate the distance between image and diverging lens. Calculate the image height for final image distance: 1/f1 = 1/p1 + 1/q1 1/50 - 1/70 = 1/q1 q1 = 175 cm 227-175=52cm 1/f2 = 1/p2 + 1/q2 -1/52 - 1/52 = 1/q2 q2 = -26 -----> 26 cm final image distance for image height: M1 = -q1/p1 = 2.5 l-2.5l = h1(image)/h1(ogject) 9.75cm = h1(image) M2 = -q2/p2 = .5 .5 = h2(image)/h1(image) 4.9 = h2(image) Consider a diverging lens of focal length 6.8 cm. e) Find the image distance if the object distance is 6.2 cm. Answer in units of cm. he focal length of a converging lens is positive, but for a diverging lens, it is negative, therefore f = -6.8 cm We now use the object image relation for a thin lens (concave or diverging) which is defined by: 1/s + 1/s' = 1/f where s is the object distance, s' is the image distance and f is the focal length, therefore 1/6.2cm + 1/s' = 1/-6.8cm solve for s' and get approx. -3.243cm, which means the image formed is located 3.243cm to the left of the diverging lens if the object is located 6.2cm to the left of the diverging lens. The lateral magnification for a thin lens is defined by m=-s'/s therefore m = -(-3.243cm) / 6.2cm = +0.5231 The magnification is positive, which means the image formed is erect (pointing upwards) if the object is pointing upwards. You can check this by drawing a raw diagram as well. Hope this helps and is correct.

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