Given,

Minimum distance for clear vision of healthy eye is  $25 \ cm$ and maximum distance is infinite.

Answer is A.

A person who is short sighted can focus clearly on near objects but cannot focus on distant objects. This is because the eyeball is too long. Light from distant objects is focused at a point in front of the retina resulting in a blurred image.
This defect can be corrected by wearing a concave (diverging) spectacle lens. The rays of light from a near object are diverged before entering the eye so that the cornea and eye lens can direct the focal point onto the retina.
Hence, the statement is true.

Here, $u = \infty$ and $ v = -200\; cm$

\begin{array}{l} According\, \, to\, question................... \ Near\, point\, of\, the\, child\, (u)=10cm \ Far\, point\, of\, the\, child\, (u)=100cm \ and,the\, \, retina\, is\, \, \, 2\, cm\, behind\, the\, eye\, lens. \ \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, V=2\, cm=0.02\, m \ Now, \ \frac { 1 }{ f } =\frac { 1 }{ { 0.02 } } -\frac { 1 }{ { (-0.1) } } \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \left| { from\, lens\, formula:\frac { 1 }{ v } -\frac { 1 }{ u } =\frac { 1 }{ f }  } \right.  \ \, \, \, \, \, \, \, \Rightarrow 50+10=60\, m \ \therefore \, \, Power\, \, of\, the\, lens:\, \, P=\frac { 1 }{ f } =60\, D \ Now,\, \, \, consider\, the\, far\, point\, is\, \, 100\, cm. \ where, \ \, u=-100cm=-1m,\, \, \, and\, \, \, \, V=2\, cm=0.02\, m \ then, \ \frac { 1 }{ f } =\frac { 1 }{ { 0.02 } } -\frac { 1 }{ { (-1) } } \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \left[ { we\, have\, the\, lens\, formla: } \right. \frac { 1 }{ v } -\frac { 1 }{ u } =\frac { 1 }{ f }  \ \frac { 1 }{ f } =50+1=51\, m \ \therefore \, \, \, Power\, \, of\, the\, lens:\, \, P=\frac { 1 }{ f } =51\, D \ so,\, that\, the\, rangeof\, the\, \, power\, of\, eye\, lens\, is\, from\, \, \, \underline { 60\, D\, \, to\, \, 51\, D }  \ and\, the\, correct\, option\, is\, B. \ \, \,  \ so\, the\, correct\, optuon\, is\, A. \end{array}

Hyperopia or Far-sightedness is the condition where person can see the far objects clearly but cannot focus on near objects. This occurs since the image of nearby objects is formed behind retina. This corrected by using a convex lens which increases the power of the eye, and image is obtained on the retina.

Near sightedness, or myopia, is the most common refractive error of the eye. Myopia occurs when the eyeball is too long, relative to the focusing power of the cornea and lens of the eye. This causes light rays to focus at a point in front of the retina, rather than directly on its surface. So, to treat myopia the glasses will be preceded by a minus sign ().
Hence, to treat myopia concave lens should be used.

A person cannot see beyond 50 cm. Hence, he should use a concave lens of 50 cm focal length.Hence $Power=\dfrac{-100}{50} =-2 D$

Long-sighted people who have lost there spectacles can still read a book by looking through a small (3-4 mm) hole in a sheet of paper because by doing so due to the diffraction at the hole the focal length of the eye lens is effectively decreased. Hence correct option is C.

$u= +20cm$

$v= \infty $

$\dfrac{1}{v}-\dfrac{1}{u}= \dfrac{1}{f}$

$\dfrac{1}{\infty }-\dfrac{1}{+20}= \dfrac{1}{f}$

$f=-20\ cm$

As $p= \dfrac{1}{f}$

$u=+2m$

$v= \infty $

$\dfrac{1}{v}-\dfrac{1}{u}=\dfrac{1}{f}$

$\dfrac{1}{\infty }-\dfrac{1}{2}=\dfrac{1}{f}$

$f= -2m$

$P= \dfrac{1}{f}$

$= \dfrac{1}{-2}$

$= -0.5 D$

Given, $u= \infty $

Here, $u= 2m$
$v= \infty $

$f= \dfrac{1}{p}$

$= \dfrac{100}{2}cm$

$= 50\ cm$

So he can see upto a distance of 50 cm.

$f= \dfrac{1}{p}$

  $=\dfrac{100}{2}cm$

  $=50\ cm$

The defect of eye will be farsighted.

$u = -\infty $ 

$ v = -250 cm$

$\dfrac{1}{v}-\dfrac{1}{u}= \dfrac{1}{f}$

$(-\dfrac{1}{\infty}-\dfrac{1}{250})= \dfrac{1}{f}$

$f= -250cm$

The correcting lens of focal length $250 cm$.

Our normal vision length = 25 cm which is eye lens focal length focal length  $f = 0.25 m$

The focal length of the concave lens should be -1 m as object at infinity forms image at focus.

Astigmatism occurs when the cornea does not have a truly spherical shape. This defect can be cured by the use of a convex lens.

Hypermetropia is a condition of the eye where the person is not able to see things clearly when nearer to the eye.

The normal near point of the eye is \[u=25\text{ }cm\]

Focal length, $f$

Image formed at distance, $v=100\,cm$

$ \dfrac{1}{f}=\dfrac{1}{v}-\dfrac{1}{u} $

$ \dfrac{1}{f}=\dfrac{1}{25}-\dfrac{1}{100}=\dfrac{3}{100} $

$ f=33.33\,cm=0.33\,m $

Power, $P=\dfrac{1}{f}=\,\dfrac{1}{0.333}=3D$

Hence, Power of lens is $3D$

Given,

Near point, $v=-120\,cm$

Reading Distance, $u=-40\,cm$

From lens formula,

  $ \dfrac{1}{f}=\dfrac{1}{v}-\dfrac{1}{u}=\dfrac{1}{-120}-\dfrac{1}{-40}=\dfrac{1}{60} $

 $ f=+60\,cm $

Focal length of lens $f=+60\,cm$ 

Maximum distance of distinct vision $=400$cm. So image of object at infinity is to be formed at $400$cm

$ \dfrac { 1 }{ { f }^{ ' } } =\dfrac { 1 }{ 25 } -\dfrac { 1 }{ 50 } =\dfrac { 1 }{ 50 } \quad or$

A convex lens of appropriate focal length is used to correct farsightedness.
Farsightedness is the inability to see nearby objects clearly due to the formation of image by convergence of rays at a point behind the retina. A convex or converging lens with the correct focal length allows the rays to converge at a point on the retina. The convex lens first converge the rays before it reaches the eye and thus, reducing the image distance. The eye-lens will now be able to converge the refracted rays ton the retina.

Short sighted is also known as myopia, in which person cannot see distant objects clearly but able to see the near object clearly,  this also leads to blurred distance vision and is a very common problem which can be corrected by glasses or contact lenses.

Nearsightedness results in blurred vision when the visual image is focused in front of the retina, rather than directly on it.

Answer is B.

The lens equation is given as $\displaystyle \frac{1}{u}+ \frac{1}{v} = \frac{1}{f}$
$\displaystyle \frac{1}{1200} + \frac{1}{-300} - \frac{1}{f}$ (u and v are in cm) 

$Answer:-$ C option

In the problem, it is given that the near point of defective eye is 1 m and that of a normal eye is 25 cm. Hence $u = -25$ cm. The lens used forms its virtual image at near point of hypermetropic eye i.e., $v = - 1 m = -100$ cm. 

Using lens formula, we have :-

$\dfrac { 1 }{ v } -\dfrac { 1 }{ u } =\dfrac { 1 }{ f } \\ \dfrac { 1 }{ -100 } -\dfrac { 1 }{ -25 } =\dfrac { 1 }{ f } \\ f=\dfrac { 100 }{ 3 } =0.33\quad m\\ power=\dfrac { 1 }{ f(in\quad metres) } =+3.0\quad D$

To see object at infinity, eye uses its least converging power.

That lens should be given to the person which forms the image of an object at infinity at distance 40 cm in front of the eye.

When a person cannot see far situated object we assume object distance to infinity.

Then image distance be v = -50cm

$u=\infty \\ v=-50cm$

then,

$\dfrac { 1 }{ f } =\dfrac { 1 }{ v } -\dfrac { 1 }{ u } \\ \dfrac { 1 }{ f } =-\dfrac { 1 }{ 50 } \\ f=-50cm=-0.5m$    (concave lens)

$power=\dfrac { 1 }{ f } =\dfrac { 1 }{ 0.5 } =-2D$

Myopia - Diverging lens (Concave)

Given, $v= \infty $

Focal length, $f= \dfrac{1}{p}$

$= \dfrac{100}{0.8}$

$= 125 cm$

To see the objects before 15 cm 

Image distance, $v=-4\ m$

Given: A far sighted person has his near point 50cm, 

$u=-2m;v=-0.4m$
$\cfrac { 1 }{ f } =\cfrac { 1 }{ v } -\cfrac { 1 }{ u } $
$P=\cfrac { 1 }{ f } =-\cfrac { 1 }{ 0.4 } -\cfrac { 1 }{ (-2) } =-2$
$P=-2D$
that means concave lens of power $2D$

Here, angular resolution of human eye,
$\phi \, = \, 5.8 \, \times \, 10^{-4} \, red$
The linear distance between two successive dots in a typical photo printer is $l \, = \, \dfrac{2.54}{300} \, cm \, = \, 0.84 \, \times \, 10^{-2} \, cm.$
At a distance of z cm, the gap distance l will subtend an angle
$\phi \, = \, \dfrac{l}{z} \, \therefore \, z \, = \, \dfrac{l}{\phi} \, = \, \dfrac{0.84 \, \times \, 10^{-2} \, cm}{5.8 \, \times \, 10^{-4}} \, = \, 14.5 \, cm$

The inability among the elderly to see nearby objects clearly because of the weakening of the ciliary muscles is called presbyopia.

Given that,

Power ${{P} _{1}}=40\,D$

Power ${{P} _{2}}=20\,D$

Total power of the combination

  $ P={{P} _{1}}+{{P} _{2}} $

 $ P=40+20 $

 $ P=60\,D $

Now, focal length of the combination

  $ f=\dfrac{1}{60}\times 100 $

 $ f=\dfrac{5}{3}\,cm $

Now, for minimum converging state of the eye lens object is at infinity

  $ u=-\infty  $

 $ f=\dfrac{5}{3} $

Now, from lens equation

  $ \dfrac{1}{f}=\dfrac{1}{v}-\dfrac{1}{u} $

 $ \dfrac{3}{5}=\dfrac{1}{v}-\dfrac{1}{\infty } $

 $ \dfrac{1}{v}=\dfrac{3}{5} $

 $ v=\dfrac{5}{3}\,cm $