The far point of a myopic eye is 250 , cm. Wh | vedclass

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

Question

(A) For a myopic eye,the far point is at a finite distance $d = 250 \, cm$. To correct this defect,the lens must form an image of an object at infinit

Vedclass · Accepted Answer

$-250 \, cm$

Vedclass · Answer

(A) For a myopic eye,the far point is at a finite distance $d = 250 \, cm$. To correct this defect,the lens must form an image of an object at infinity at the far point of the eye.Using the lens formula: $\frac{1}{f} = \frac{1}{v} - \frac{1}{u}$.Here,$u = \infty$ (object at infinity) and $v = -250 \, cm$ (image formed at the far point).Substituting these values: $\frac{1}{f} = \frac{1}{-250} - \frac{1}{\infty}$.Since $\frac{1}{\infty} = 0$,we get $\frac{1}{f} = -\frac{1}{250}$.Therefore,$f = -250 \, cm$.

The far point of a myopic eye is $250 \, cm$. What is the focal length of the lens required to correct this defect?

Similar Questions

The image formed on the retina is:

The human eye has a lens which has a

$A$ person with hypermetropia has a near point of $60\, cm$. What power of lens should be used for the eye to read a book at $25\, cm$?

$A$ person wears normal spectacles in which the distance between the glasses and the eyes is approximately $2 \, cm$. The power required is $-5 \, D$. If he wears contact lenses,then the required power is ......... $D$.

$A$ person has a near point of $50\, cm$. What lens should they use to reduce it to $25\, cm$?

Vedclass Test Series

Exam Paper Generator

Online Exam Module