A physicist's left eye is myopic (i.e., nearsighted). This eye can see clearly only out to a distance of 35 cm.Find the focal length and the power of a lens that will correct this myopia when worn 2.0 cm in front of the eye.

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Question:

A physicist's left eye is myopic (i.e., nearsighted). This eye can see clearly only out to a distance of 35 cm.Find the focal length and the power of a lens that will correct this myopia when worn 2.0 cm in front of the eye.

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Answer:Explanation:To get the focal length, we will use the lens formula;1/f = 1/u + 1/vf is the focal lengthu is the object distancev is the image distanceGiven since the physicist's left eye is myopic, it will be corrected using concave lens and the image distance is negative.u = 35cmv = -2.01/f = 1/35-1/21/f = 2-35/701/f = -33/70f = -70/33f = -2.12 cmf = -0.0212mPower of a lend is the reciprocal of its focal lengthPower of the lens = 1/fP = 1/-0.0212P = -47.17dioptresThe power of the lens is -47.17D

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