A man can see objects clearly between 15 cm | vedclass

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(B) The man has a far point of $30 \ cm$ and a near point of $15 \ cm$. To see far objects (at infinity),he needs a concave lens such that an object a

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$30$

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(B) The man has a far point of $30 \ cm$ and a near point of $15 \ cm$. To see far objects (at infinity),he needs a concave lens such that an object at infinity forms an image at his far point $(v = -30 \ cm)$.Using the lens formula: $\frac{1}{f} = \frac{1}{v} - \frac{1}{u} = \frac{1}{-30} - \frac{1}{\infty} = -\frac{1}{30}$.Thus,the focal length $f = -30 \ cm$.Now,to find the new near point,we use the same lens to see an object at distance $u$ such that the image is formed at his actual near point $(v = -15 \ cm)$.$\frac{1}{f} = \frac{1}{v} - \frac{1}{u} \Rightarrow -\frac{1}{30} = \frac{1}{-15} - \frac{1}{u}$.$\frac{1}{u} = \frac{1}{30} - \frac{1}{15} = \frac{1-2}{30} = -\frac{1}{30}$.Therefore,$u = -30 \ cm$. The new near point is at $30 \ cm$.

$A$ man can see objects clearly between $15 \ cm$ and $30 \ cm$. He uses a lens to see far objects. Due to the lens used,the new near point will be at........$cm$.

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