A person wears glasses of power -2.5 , D. Wha | vedclass

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Question

(C) The power of the lens is given as $P = -2.5 \, D$. Since the person wears glasses to correct their vision,the lens must form an image of an object

Vedclass · Accepted Answer

$40$

Vedclass · Answer

(C) The power of the lens is given as $P = -2.5 \, D$. Since the person wears glasses to correct their vision,the lens must form an image of an object at infinity at the person's far point. The lens formula is given by $\frac{1}{f} = \frac{1}{v} - \frac{1}{u}$. Since $P = \frac{1}{f}$ (in meters),we have $P = \frac{100}{f}$ (in cm). Thus,$P = \frac{100}{v} - \frac{100}{u}$. Here,$u = -\infty$ (for a normal far point) and $v$ is the far point of the person. $-2.5 = \frac{100}{v} - \frac{100}{-\infty}$. Since $\frac{100}{\infty} = 0$,we get $-2.5 = \frac{100}{v}$. $v = \frac{100}{-2.5} = -40 \, cm$. The negative sign indicates that the far point is in front of the eye. Therefore,the far point of the person without glasses is $40 \, cm$.

$A$ person wears glasses of power $-2.5 \, D$. What is the far point of the person without glasses in $cm$?

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