Question

A particular patient's eyes are unable to focus on objects closer than 34.5 cm and corrected lenses are to be prescribed so that the patient can focus on objects 22.0 cm from their eyes HINT (a) is the patient nearsighted or farsighted? O nearsighted O farsighted (b) If contact lenses are to be prescribed, determine the required lens power (in diopters). diopters (c) veglasses are to be prescribed instead and the distance between the eyes and the lenses is 1.80 cm, determine the power (in diopters) of the required corrective diopters Verses (4) Are the required lenses converging or diverging? converging diverging Need Help?

Answer 1

(a) near Patient's is far sighted, since Patient's Point (34,5cm) is greater than the normal Point 2 2 cm Ans: far sighted (b) is Power 1 is, I f focal length f, the Pal P = Giren, Distance of the object, u=22cm Distance of the image, v=-34.5cm Lens eruation, t ttu 334.5 + 1 => 1 From the को V tit 34.5 22 0.0164 6 cm-l tt 14 V 21.65 mal a ) Pa 1.65 mny Pa 1.65 diapter (C) Distance of the object from the eyeglass is, u= (22 -1.80) cm 2012 em Distance of the image from me eye glass is, ve - (34,5 -1.80) 32.7 cam

from, Lens equation, t + t = te »- 32.7 + bor2 = > th 0.01892 cmrl tt = 1.89 my 2) Pa 1.89 Diopter (d) since Power of lenses from both part are positive. so it will be converging lenses. Aus :- converging