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Q.
A person wears glasses of power $-2.5\,D$ . What is the far point of the person without the glasses ?
Solution:
$P =-2.5 D$ i.e. concave lens, the person is nearsighted,
$f =\frac{100}{-2.5}=-40$
$\frac{1}{-( F . P .)}-\frac{1}{\text{-( distance )}}=\frac{1}{ f } $
$\frac{1}{-( F . P .)}-\frac{1}{\infty}=\frac{1}{-40} $
$\Rightarrow F . P .=40\, cm$