Calculate the focal length of a reading glass | vedclass

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

Question

(B) The person has a near point of $ 75 \ cm $. To read at a normal distance of $ 25 \ cm $,the lens must form a virtual image of an object placed at

Vedclass · Accepted Answer

$37.5$

Vedclass · Answer

(B) The person has a near point of $ 75 \ cm $. To read at a normal distance of $ 25 \ cm $,the lens must form a virtual image of an object placed at $ 25 \ cm $ at the person's near point of $ 75 \ cm $.Given: Object distance $ u = -25 \ cm $,Image distance $ v = -75 \ cm $.Using the lens formula: $ \frac{1}{f} = \frac{1}{v} - \frac{1}{u} $$ \frac{1}{f} = \frac{1}{-75} - \frac{1}{-25} $$ \frac{1}{f} = -\frac{1}{75} + \frac{3}{75} = \frac{2}{75} $$ f = \frac{75}{2} = 37.5 \ cm $Thus,the focal length of the reading glass is $ 37.5 \ cm $.

Calculate the focal length of a reading glass of a person if his distance of distinct vision is $ 75 \ cm $. (in $cm$)

Similar Questions

$A$ person uses a lens of power $+3 \text{ D}$ to normalize vision. The near point of the hypermetropic eye is ....... $m$.

$A$ far-sighted person has a near point of $60\, cm$. What power lens should be used for eyeglasses such that the person can read a book at a distance of $25\, cm$?

When objects at different distances are seen by the eye,which of the following remains constant?

If the distance of the far point for a myopia patient is doubled,the focal length of the lens required to cure it will become

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

Vedclass Test Series

Exam Paper Generator

Online Exam Module