A presbyopic patient has a near point of 30 , | vedclass

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(C) For a person with myopia or presbyopia, to see distant objects clearly, the corrective lens must form a virtual image of an object at infinity at

Vedclass · Accepted Answer

$-2.5$

Vedclass · Answer

(C) For a person with myopia or presbyopia, to see distant objects clearly, the corrective lens must form a virtual image of an object at infinity at the patient's far point.Given the far point $d = 40 \, cm = 0.4 \, m$.The lens formula is $\frac{1}{f} = \frac{1}{v} - \frac{1}{u}$.For distant objects, $u = \infty$. The image must be formed at the far point, so $v = -40 \, cm = -0.4 \, m$.Thus, $\frac{1}{f} = \frac{1}{-0.4} - \frac{1}{\infty} = -2.5 \, m^{-1}$.Power $P = \frac{1}{f(	ext{in } meters)} = -2.5 \, D$.

$A$ presbyopic patient has a near point of $30 \, cm$ and a far point of $40 \, cm$. The dioptric power of the corrective lens required for seeing distant objects is: (in $D$)

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