You have a converging lens with a focal length of 1.2 centimeters. At a distance

Question

You have a converging lens with a focal length of 1.2 centimeters. At a distance of 1.29 centimeters in front of it, you place a small item that you are trying to view. Then, 20 centimeters behind the first lens, you place a second converging lens with a focal length of 2.0 centimeters. Use the think lens equation to quantify the image that the first lens produces. Then use the first image as an object for the second lens, and locate the image from the second lens. Describe the final image in words, and also quantify its location and magnification (which can simple be the product of the two individual magnifications). What is the general name given to this type of instrument? And, what is the name given to the first and second lens?

Explanation / Answer

The focal length of the first lens f1 = 1.2 cm The object disatnce p = 1.29 cm then from lens formula        1/f1 = 1/p + 1/q then the image distance             q = fp/p -f                = (1.29)(1.2) / 1.29 - 1.2 = 17.2 cm the magnification m = - q/p = - 17.2 / 1.29 = -13.33 now the object distance for second lens is p2 = 20-17.2 = 2.8 cm gain from lens formula             q2 = f2 p2/p2 - f2                  = (2)(2.8) / 2.8 - 2                  = 7 cm magnification m = - 7/2.8 = -2.5 overal magnification M = (-13.33)(-2.5) = 33.32 Therefore the final image is real, upright and enlarge

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