the near and far points of an woman are 30cm and 180cm find the power of lens she should use while reading at 25cm
varsha35:
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the near point is always greater than the least distance .so here I WL first solve the problem of hypermetropia. so u=-25cm.and v=-30cm.by the lens formula , 1/-30+1/25=1/f
or,1/25-1/30=1/f,or, f=+150cm.and
now in myopia, u=-∞and v=the far point, -180cm.again,
1/-180+1/∞=1/f., f=-180cm.so in myopia a concave and in hypermetropia a convex lens is used.so if we add the two powers we will get a bifocal lens.so their sum =2/3-5/9=1/9D.
or,1/25-1/30=1/f,or, f=+150cm.and
now in myopia, u=-∞and v=the far point, -180cm.again,
1/-180+1/∞=1/f., f=-180cm.so in myopia a concave and in hypermetropia a convex lens is used.so if we add the two powers we will get a bifocal lens.so their sum =2/3-5/9=1/9D.
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