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? (Take the image distance from the lens of the eye to the retina to be 2 cm.)
(i) power of lens required to see clearly the object placed at infinity. U=−∞ f1=v1−u1 ⇒f1=−101−∞1;P=f1m ⇒f1=101;P=−0.11m (ii) when no corrective lens used: Let powers of eye when object is at far point, near point are Pf and Pn respectively and power ∴ Pn=Pf+Pa When object is at far point its clear image is formed at retina 2cm from eye lens ∴ u=1−cm=−0.1m,V=2cm=0.02m If f is focal length of eye lens focused at far point then f1=v1−u1=0.021−(−0.1)1 f1=50+10=60 Pf=60D ∴ Pn=Pf+Pa=60+4=64D Let the near point be xn u=xn (v=2cm=0.02m) v1−u1=f1 0.021+xn1=Power (Pn) 50+xn1=64 xn1=64−50=14D Near point without glass xn=141m=14100cm=7cm (approx.) (iii) When used corrective lens: When corrective lens is used then eye can see the object at infinity. Power of eye lens in thissituation is P∞ u=∞ and v=2cm=0.02m f1=v1−u1 P∞=0.021−∞1=50 P∞=50+0 P∞=50D If Pn= power of eye at near point when corrective lens is used Pn=P∞+Pa=50+4=54D Let near point in this situation is xn u=−xnm v=+2cm=0.02m f1=54 v1−u1=f1 0.021+xn1=54 (all distance are in m) 50+xn1=54(xn1=4) xn=41m=0.25m
