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Q.
A hypermetropic person having near point at a distance of $0.75 \,m$ puts on spectacles of power $2.5 \,D$. The near point now is at
Ray Optics and Optical Instruments
Solution:
As, $\frac{1}{f}=\frac{1}{v}-\frac{1}{u}$ (lens formula)
$\Rightarrow 2.5=\frac{1}{-0.75}-\frac{1}{u}$
Or $\frac{1}{u}=-\frac{100}{75}-\frac{25}{10}$
Or $\frac{1}{u}=-\frac{4}{3}-\frac{5}{2}$
Or $\frac{1}{u}=\frac{-8-15}{6}=-\frac{23}{6}$
or $u=-\frac{6}{23} m =-0.26\, m$