If the least distance of distinct vision for | vedclass

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

Question

(C) The normal least distance of distinct vision $(D)$ is $25 \ cm$. Since the boy's least distance is $35 \ cm$,he is suffering from hypermetropia (f

Vedclass · Accepted Answer

convex lens of focal length $87.5 \ cm$

Vedclass · Answer

(C) The normal least distance of distinct vision $(D)$ is $25 \ cm$. Since the boy's least distance is $35 \ cm$,he is suffering from hypermetropia (farsightedness).To correct this,we need to use a convex lens that forms a virtual image of an object placed at $25 \ cm$ at the boy's actual near point of $35 \ cm$.Here,object distance $u = -25 \ cm$ and image distance $v = -35 \ cm$.Using the lens formula: $\frac{1}{f} = \frac{1}{v} - \frac{1}{u}$.$\frac{1}{f} = \frac{1}{-35} - \frac{1}{-25} = \frac{1}{25} - \frac{1}{35}$.$\frac{1}{f} = \frac{7 - 5}{175} = \frac{2}{175}$.$f = \frac{175}{2} = 87.5 \ cm$.Since the focal length is positive,it is a convex lens.

If the least distance of distinct vision for a boy is $35 \ cm$,then the lens to be used by the boy for correcting the defect of his eye is

Similar Questions

$A$ person is suffering from the defect astigmatism. Its main reason is

For a normal eye,the least distance of distinct vision is......$m$.

If you are using eyeglasses of power $2 D$,what is your near point (in $cm$)?

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

The near point and far point of a normal human eye are respectively:

Vedclass Test Series

Exam Paper Generator

Online Exam Module