A person wears glasses of power -2.0 , D. The | vedclass

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(A) The power of the lens is $P = -2.0 \, D$. Since the power is negative,the lens used is a concave lens,which is used to correct myopia (nearsighted

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Nearsighted,$50 \, cm$

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(A) The power of the lens is $P = -2.0 \, D$. Since the power is negative,the lens used is a concave lens,which is used to correct myopia (nearsightedness).For a myopic eye,the far point is at a finite distance $x$ instead of infinity.The lens formula is $\frac{1}{f} = \frac{1}{v} - \frac{1}{u}$.For a person to see an object at infinity $(u = \infty)$,the image must be formed at their far point $(v = -x)$.Since $P = \frac{1}{f}$,we have $P = \frac{1}{v} - \frac{1}{u} = \frac{1}{-x} - \frac{1}{\infty} = -\frac{1}{x}$.Thus,$x = -\frac{1}{P} = -\frac{1}{-2.0} = 0.5 \, m = 50 \, cm$.Therefore,the person is nearsighted and their far point is $50 \, cm$.

$(I)$ Presbyopia	$(A)$ Sphero-cylindrical lens
$(II)$ Hypermetropia	$(B)$ Convex lens of proper power may be used close to the eye
$(III)$ Astigmatism	$(C)$ Concave lens of suitable focal length
$(IV)$ Myopia	$(D)$ Bifocal lens of suitable focal length

$A$ person wears glasses of power $-2.0 \, D$. The defect of the eye and the far point of the person without the glasses will be

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