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A myopic adult has a far point at 0.1 m. His power of accomodation is 4 Diopters.(i) What power lenses are required to see distant objects ? (ii) What is his near point without glasses ? (iii) What is his near point with glasses ? (iii) What is his near point with glasses ? (Take the image distance from the lens of the eye to the retina to be 2 cm.) |
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Answer» Solution :`rArr` (i) Required power for normal relaxed eye at far PIONT, `P_f = 1/f= 1/f_1 + 1/f_2` `thereforeP_f = (1)/(0.1) + (1)/(0.02) = (0.02 + 0.1)/(0.002) = (0.12)/(0.002)` `thereforeP_f = 60 D` `rArr` Now required power for shifting far point to INFINITY (using corrective lens) `P._(f) = (1)/(f.) = (1)/(oo) + (1)/(0.02) = 0 +(1)/(0.02)` `therefore P.f = 50 D` For the combined system of relaxed eye and spectacles, `P._(f) = P_(f) +P_(g)` `therefore 50 = 60 + P_g` `therefore P_g = -10 D` (ii) Power of normal eye P = 4D Now, if power eye P = 4D Now, if power of normal eye of given person for near vision is `P_n` then , `P = P_n - P_g` `therefore4 = P_n -60` `therefore P_n = 64 D` `rArr` SUPPOSE DISTANCE of near point without spectacles is `x_n` `therefore 1/(x_n) + (1)/(0.02) = 64` `therefore (1)/(x_n) + 50 = 64` `therefore 1/(x_n)= 14` `therefore x_n= (1)/(14)` `therefore x_n = 0.07 m ` (iii) With spectacles, `P._n = P._f +P` `54 = (1)/(x._n) +(1)/(0.02) = (1)/(x._n) + 50` `therefore (1)/(x._n) = 4` `therefore x._(n) = 1/4 = 0.25 m ` |
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