A person suffering from hypermetropia cannot | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(D) The person has a near point of $u = -120 \, cm$ due to hypermetropia.To read at a distance of $v = -40 \, cm$,the lens must form a virtual image o

Vedclass · Accepted Answer

$-60$

Vedclass · Answer

(D) The person has a near point of $u = -120 \, cm$ due to hypermetropia.To read at a distance of $v = -40 \, cm$,the lens must form a virtual image of an object placed at $40 \, cm$ at the person's actual near point of $120 \, cm$.Using the lens formula: $\frac{1}{f} = \frac{1}{v} - \frac{1}{u}$.Substituting the values: $\frac{1}{f} = \frac{1}{-40} - \frac{1}{-120}$.$\frac{1}{f} = -\frac{1}{40} + \frac{1}{120} = \frac{-3 + 1}{120} = \frac{-2}{120} = -\frac{1}{60}$.Therefore,$f = -60 \, cm$.

$A$ person suffering from hypermetropia cannot focus on objects closer than $120 \, cm$. To read at a distance of $40 \, cm$,what should be the focal length of the lens required?

Similar Questions

$A$ man can see only between $75 \, cm$ and $100 \, cm$. The power of the lens required to correct the near point will be:

The sensation of vision in the retina is carried to the brain by

If the far point of a short-sighted person is $400 \ cm$, then the power of the lens required to enable him to see very distant objects clearly is

$A$ defective eye cannot see close objects clearly because their image is formed

$A$ person has been using spectacles of power $-1.0$ diopter for distant vision and a separate reading glass of power $+2.0$ diopters. What is the least distance of distinct vision for this person in $cm$?

Vedclass Test Series

Exam Paper Generator

Online Exam Module