The near point of a person is 75 , cm. In ord | vedclass

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(C) The person is suffering from hypermetropia (farsightedness),as the near point has shifted to $75 \, cm$.To read a book at $u = -30 \, cm$,the lens

Vedclass · Accepted Answer

$+2 \, D$

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(C) The person is suffering from hypermetropia (farsightedness),as the near point has shifted to $75 \, cm$.To read a book at $u = -30 \, cm$,the lens must form a virtual image at the person's actual near point,$v = -75 \, cm$.Using the lens formula: $\frac{1}{f} = \frac{1}{v} - \frac{1}{u}$.Substituting the values: $\frac{1}{f} = \frac{1}{-75} - \frac{1}{-30}$.$\frac{1}{f} = -\frac{1}{75} + \frac{1}{30} = \frac{-2 + 5}{150} = \frac{3}{150} = \frac{1}{50} \, cm^{-1}$.Since $f$ is in $cm$,$f = 50 \, cm = 0.5 \, m$.The power $P$ is given by $P = \frac{1}{f(m)} = \frac{1}{0.5} = +2 \, D$.

$(I)$ 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 near point of a person is $75 \, cm$. In order for him to be able to read a book at a distance of $30 \, cm$,what should be the power of the spectacle lenses?

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