If you are using eyeglasses of power 2 D, wha | vedclass

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

Question

(B) Given, power of the lens $P = 2 \ D$. The focal length of the lens is $f = \frac{1}{P} = \frac{100}{2} = 50 \ cm$. For a person with a near point

Vedclass · Accepted Answer

$50$

Vedclass · Answer

(B) Given, power of the lens $P = 2 \ D$. The focal length of the lens is $f = \frac{1}{P} = \frac{100}{2} = 50 \ cm$. For a person with a near point defect (hypermetropia), the lens forms an image of an object placed at the standard near point $(u = -25 \ cm)$ at the person's actual near point $(v)$. Using the lens formula: $\frac{1}{f} = \frac{1}{v} - \frac{1}{u}$. Substituting the values: $\frac{1}{50} = \frac{1}{v} - \frac{1}{-25}$. $\frac{1}{50} = \frac{1}{v} + \frac{1}{25}$. $\frac{1}{v} = \frac{1}{50} - \frac{1}{25} = \frac{1 - 2}{50} = -\frac{1}{50}$. Therefore, $v = -50 \ cm$. The magnitude of the near point is $50 \ cm$.

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

Similar Questions

$A$ person cannot see objects beyond $3 \, m$. What should be the focal length of the lens required to see objects at $12 \, m$?

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

$A$ person can see objects clearly only when they lie between $50\, cm$ and $400\, cm$ from his eyes. In order to increase the maximum distance of distinct vision to infinity, the type and power of the correcting lens the person has to use will be

$A$ person cannot see objects clearly placed at a distance of more than $40 \, cm$. He is advised to use a lens of power:

The impact of an image on the retina remains for......$sec$.

Vedclass Test Series

Exam Paper Generator

Online Exam Module