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

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(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

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$50$

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(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 $cm$)?

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