A presbyopic patient has a near point of 30 | vedclass

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(C) For a person with myopia (nearsightedness),the far point is limited. To see distant objects (at infinity) clearly,the corrective lens must form a

Vedclass · Accepted Answer

$-2.5$

Vedclass · Answer

(C) For a person with myopia (nearsightedness),the far point is limited. To see distant objects (at infinity) clearly,the corrective lens must form a virtual image of the object at the patient's actual far point.Here,the object distance $u = \infty$ and the image distance $v = -40 \ cm$ (since the image must be formed on the same side as the object).Using the lens formula: $\frac{1}{f} = \frac{1}{v} - \frac{1}{u}$$\frac{1}{f} = \frac{1}{-40} - \frac{1}{\infty} = \frac{1}{-40} - 0 = -\frac{1}{40} \ cm^{-1}$Power $P$ in Diopters is given by $P = \frac{100}{f 	ext{ (in cm)}}$$P = \frac{100}{-40} = -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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