Question

if a certain person has minimum distance of distinct vision of 150 cm he wishes to read at a distance of 25cm what focal length should be used what is the nature of eyes defect​

Answer 1

The defect is hypermetropia

Gn,

v=-150

u=-25

1/f=1/-150 - 1/-25

=1/-150+1/25

=1/30

f=30cm

Answer 2

Determine the focal length of lens

Explanation:

Given data:

Distance of distinct vision is v = - 150 cm

Read distance u = -25 cm

Focal length can be calculated by using the following formula

$\frac{1}{f} =\frac{1}{v} + \frac{1}{u}$

$\frac{1}{f} = \frac{1}{-150} +\frac{1}{-25}$

$\frac{1}{f} = \frac{1}{30}$

F = 30 cm

The eye defect is Hypermetropia

learn more:

https://brainly.in/question/1148940

https://brainly.in/question/8838219