The far point of a nearsighted person is 6.0 | vedclass

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(D) The far point of the person is $6.0 \, m$,so the contact lens must form a virtual image of a distant object at $v = -6.0 \, m$ when $u = \infty$.

Vedclass · Accepted Answer

$0.50$

Vedclass · Answer

(D) The far point of the person is $6.0 \, m$,so the contact lens must form a virtual image of a distant object at $v = -6.0 \, m$ when $u = \infty$. Thus,the focal length $f$ of the lens is $-6.0 \, m$.Now,for a tree at $u = -18.0 \, m$ with height $h = 2.0 \, m$,we use the lens formula $\frac{1}{f} = \frac{1}{v} - \frac{1}{u}$:$\frac{1}{-6} = \frac{1}{v} - \frac{1}{-18}$$\frac{1}{v} = -\frac{1}{6} - \frac{1}{18} = \frac{-3-1}{18} = -\frac{4}{18}$$v = -\frac{18}{4} = -4.5 \, m$.The magnification $m$ is given by $m = \frac{v}{u} = \frac{-4.5}{-18} = 0.25$.The height of the image $h'$ is $h' = m 	imes h = 0.25 	imes 2.0 = 0.50 \, m$.

The far point of a nearsighted person is $6.0 \, m$ from her eyes,and she wears contact lenses that enable her to see distant objects clearly. $A$ tree is $18.0 \, m$ away and $2.0 \, m$ high. How high is the image formed by the contact lenses?

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