Question

A patient can't see objects closer than 32.5 cm and wishes to clearly see objects that are 20.0 cm from his eye. (a) is the patient nearsighted or farsighted? O nearsighted O farsighted (b) If the eye-lens distance is 2.05 cm, what is the minimum object distance p from the lens? (Give your answer to at least three significant digits.) cm (c) What image position with respect to the lens will allow the patient to see the object? (Give your answer to at least three significant digits.) cm (a) is the image real or virtual? Is the image distance a positive or negative? (Select all that apply.) real image virtual image distance is positive distance is negative (e) Calculate the required focal length cm () Find the power of the lens in diopters dopters

Answer 1

Ques.soln (a) since the patients can't see objects closer than 32.5 im then • The patient is farsighted Ans (b) The minimum distance of object from lens p=x-Y Here x = 20.0um y = = 2.05 um p= 20.0-2.05 17.95 cm And,, (c) The image positim is given by q=2-y Here 2= 32.5 y = 2.05 q = 32.5-2.05 = 30.45 cm Ans, (d) since image and object are on same side then image will be uirtual and distance will be negative le) using the lens formula ㅗ I + 1 f q Р 1 ->) f:(17.95) (-30.45) f -30.45 17.95 17.95- 30.45 43.7262 Ans - +

If) The power of the lens will be 2.285 D p=100 f 100 43.762 (g) for prescription the distance of the eye lens is ignored then p=x 120 un Ary, Distance of the image from lens. q = -x = -32.5 cm ans, focal length will be 1 f -32.5cm 200m + I - P -lt 20cm 32.5 ( 20cm) (-32.5cm) 52 cm Ans, - 1.923 D Ans, P = 100 = 100 52 f