Question

The Focal length of a lens suggested to a person with
Hypermetuopia is 100cm. Find the distance of near point and
the power of the lens

Answer 1

$\Large\underline{\underline{\sf Given:}}$

• Focal lenght of lens (f) = 100cm or 1m

$\Large\underline{\underline{\sf To\:Find:}}$

• Distance of near point (d)

• Power of lens (P)

$\Large\underline{\underline{\sf Formula\:Used:}}$

$\large{\boxed{\sf \pink{Power\:(P)=\dfrac{1}{Focal\: Lenght (f)}} }}$

$\large{\boxed{\sf \pink{f=\dfrac{25d}{d-25}}}}$

$\Large\underline{\underline{\mathfrak \orange{Solution:}}}$

✮ Power of Lens (P) :

$\implies{\sf P=\dfrac{1}{f} }$

f should be in meter

$\implies{\sf P=\dfrac{1}{1} }$

$\implies{\sf \red{P=1\:diopter}}$

✮ Distance of Near Point :

$\Large\implies{\sf f=\dfrac{25}{d-25}}$

$\implies{\sf 100=\dfrac{25d}{d-25} }$

$\implies{\sf 100d-2500=25d }$

$\implies{\sf 100d-25d=2500 }$

$\implies{\sf 75d=2500 }$

$\implies{\sf d=\dfrac{2500}{75} }$

$\implies{\sf \red{d=33.33\:cm} }$

$\Large\underline{\underline{\mathfrak \orange{Answer:}}}$

☆Power of lens (P) = 1 diopter

☆Distance of near point (d) = 33.33cm