Question

Near point moved 1.5 cm. What is focal length? I ho

Answer 1

$\large\underline{\bigstar \: \: {\sf Given-}}$

• Near point (v) = -1.5cm

$\large\underline{\bigstar \: \: {\sf To \: Find-}}$

• Focal Lenght (f)

$\large\underline{\bigstar \: \: {\sf Formula \: Used -}}$

$\color{violet}\bullet\underline{\boxed{\sf \dfrac{1}{f}=\dfrac{1}{v}-\dfrac{1}{u}}}$

$\large\underline{\bigstar \: \: {\sf Solution -}}$

➩ Near Point :-

The minimum distance which a person can see without help of spectacles.

• In this case image of the object is formed at the near point.
• If the reference of object is not given it is taken as 25cm

Distance of near point = Negative (-ve)

Distance of Object = Negative (-ve)

Power of lens = positive (+)

_________________________________

$\implies{\sf \dfrac{1}{f}=-\dfrac{1}{1.5}-\left[-\dfrac{1}{25}\right]}$

$\implies{\sf \dfrac{1}{f}=-\dfrac{1}{1.5}+\dfrac{1}{25}}$

$\implies{\sf \dfrac{1}{f}=\dfrac{-25+1.5}{-1.5×25} }$

$\implies{\sf \dfrac{1}{f}=\dfrac{-23.5}{-37.5}}$

$\implies{\sf \dfrac{1}{f}=0.62}$

$\implies{\sf f=\dfrac{1}{0.62}}$

$\color{red}\implies{\sf f=1.61\:cm}$

$\large\underline{\bigstar \: \: {\sf Answer-}}$

Focal Lenght of lens is $\color{red}{\sf 1.61\:cm}$