The near point of a hypermetropic person is 7 | vedclass

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

Question

(A) For a hypermetropic person,the near point is $d' = 75\, cm$. The person wants to read a book at the normal near point $d = 25\, cm$.Using the lens

Vedclass · Accepted Answer

$+ 2.67\, D$

Vedclass · Answer

(A) For a hypermetropic person,the near point is $d' = 75\, cm$. The person wants to read a book at the normal near point $d = 25\, cm$.Using the lens formula: $\frac{1}{f} = \frac{1}{v} - \frac{1}{u}$.Here,$v = -75\, cm$ (the image must be formed at the person's actual near point) and $u = -25\, cm$ (the object distance).$\frac{1}{f} = \frac{1}{-75} - \frac{1}{-25} = \frac{-1 + 3}{75} = \frac{2}{75}\, cm^{-1}$.The power $P$ in Diopters is given by $P = \frac{100}{f(cm)}$.$P = 100 	imes \frac{2}{75} = \frac{200}{75} = +2.67\, D$.

The near point of a hypermetropic person is $75\, cm$ from the eye. What is the power of the lens required to enable the person to read clearly a book held at $25\, cm$ from the eye?

Similar Questions

$A$ person uses spectacles of power $+2D$. He is suffering from

$A$ 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 $cm$?

Astigmatism (for a human eye) can be removed by using

The substance present in the choroid is:

$A$ 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 $25\, cm$?

Vedclass Test Series

Exam Paper Generator

Online Exam Module