A diverging lens located in the y-z plane at x = 0 forms an image of an arrow at x = x2 = -14 cm. The image of the tip of the arrow is located at y = y2 = 6.2 cm. The magnitude of the focal length of the diverging lens is 28.5 cm. 1) What is x1, the x co-ordinate of the object arrow?. 2) What is y1, the y co-ordinate of the tip of the object arrow?
Added by Iker G.
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5}\] Solving for \(x_1\): \[\frac{1}{-x_1} = \frac{1}{14} - \frac{1}{28.5}\] \[\frac{1}{-x_1} = \frac{2.5 - 1}{28.5}\] \[\frac{1}{-x_1} = \frac{1.5}{28.5}\] \[-x_1 = \frac{28.5}{1.5}\] \[-x_1 = 19\] \[x_1 = -19 \, \text{cm}\] ** Show more…
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