A person has a near point at 60; cm. The foca | vedclass

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(C) The person wants to read at a distance of $d = 22\; cm$ from the eyes.Since the spectacles are placed $2\; cm$ from the eyes,the object distance $

Vedclass · Accepted Answer

$30$

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(C) The person wants to read at a distance of $d = 22\; cm$ from the eyes.Since the spectacles are placed $2\; cm$ from the eyes,the object distance $u$ for the lens is $u = -(22 - 2) = -20\; cm$.The lens must form a virtual image at the person's near point,which is $60\; cm$ from the eyes.Since the lens is $2\; cm$ from the eyes,the image distance $v$ is $v = -(60 - 2) = -58\; cm$.Using the lens formula $\frac{1}{f} = \frac{1}{v} - \frac{1}{u}$:$\frac{1}{f} = \frac{1}{-58} - \frac{1}{-20} = \frac{1}{20} - \frac{1}{58} = \frac{58 - 20}{20 	imes 58} = \frac{38}{1160}$.$f = \frac{1160}{38} \approx 30.5\; cm$.Given the options provided,the closest integer value is $30\; cm$.

$A$ person has a near point at $60\; cm$. The focal length of the spectacle lenses required to read at $22\; cm$,with the glasses separated by $2\; cm$ from the eyes,is: (in $; cm$)

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