Question

please answer for this problem​

Answer 1

Answer:

Given:

Focal length of lens = 10 cm

Assumptions:

Considering the eye of the watch repairer to be normal , we can say that his near point will be 25 cm .

Concept:

Whenever , he is using the convex lens , actually he produces a virtual image at his near point. The virtual image acts as an object and forms image at retina.

Calculation:

As per Lens formula :

$\star \: \: \dfrac{1}{f} = \dfrac{1}{v} - \dfrac{1}{u}$

$\implies \: \: \dfrac{1}{10} = \dfrac{1}{ - 25} - \dfrac{1}{ u}$

$\implies \: \: \dfrac{1}{u} = - \dfrac{1}{25} - \dfrac{1}{10}$

$\implies \: \: \dfrac{1}{u} = - \dfrac{7}{50}$

$\implies \: \: u = - \dfrac{50}{7} cm$

$\implies \: \: u = - 7.142 \: cm$

So final answer :

$\boxed{ \red{ \sf{ \bold{ \huge{u = - 7.142 \: cm}}}}}$