Question

A hypermetropic person having near point at a distance of $0.75m$ puts on spectacles of power $2.5D$. The near point now is at

• A. 0.75 m
• B. 0.83 m
• C. 0.36 cm
• D. 0.26 m

Answer 1

Answer: (D)

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 $u=-\frac{6}{23}m=-0.26m$