The least distance of distinct vision for a f | vedclass

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(A) The person is farsighted (hypermetropic),meaning they cannot see objects closer than $1 \ m$ clearly. We want to use a lens to form a virtual imag

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$+3 \ D$

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(A) The person is farsighted (hypermetropic),meaning they cannot see objects closer than $1 \ m$ clearly. We want to use a lens to form a virtual image of an object placed at $u = -25 \ cm = -0.25 \ m$ at the person's near point $v = -1 \ m$.Using the lens formula: $\frac{1}{f} = \frac{1}{v} - \frac{1}{u}$.Substituting the values: $\frac{1}{f} = \frac{1}{-1} - \frac{1}{-0.25}$.$\frac{1}{f} = -1 + 4 = +3 \ m^{-1}$.Since power $P = \frac{1}{f}$ (in meters),the optical power is $P = +3 \ D$.

$(I)$ 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

The least distance of distinct vision for a farsighted person is $1 \ m$. The optical power of the lens of his spectacles,which effectively reduces his least distance of distinct vision $(LDDV)$ to $25 \ cm$,is:

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