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

The far point of a myopic person is $80\, cm$ in front of the eye. What is the power of the lens required to enable him to see very distant objects clearly?

To remove myopia (short-sightedness),a lens of power $0.66 \, D$ is required. The far point of the eye is approximately ....... $cm$.

When the power of the eye lens increases,a defect of vision is produced. This defect is known as:

$A$ man can see clearly up to $3 \, m$. Prescribe a lens for his spectacles so that he can see clearly up to $12 \, m$.

The far point of a myopic eye is at $40 \, cm$. To correct this defect,the power of the lens required will be: (in $, D$)

Vedclass Test Series

Exam Paper Generator

Online Exam Module