A patient has a near point of 450 cm and far point of 850


A patient has a near point of 45.0 cm and far point of 85.0 cm.

(a) Can a single lens correct the patient's vision? Explain the patient's options.

(b) Calculate the power lens needed to correct the near point so that the patient can see objects 25.0 cm away. Neglect the eye-lens distance.

(c) Calculate the power lens needed to correct the patient's far point, again neglecting the eye-lens distance.

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Physics: A patient has a near point of 450 cm and far point of 850
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