Human Eye and Defects of Vision Questions in English

(D) The condition described is $Hypermetropia$ (farsightedness).
In a normal eye,the image of a close object is formed on the retina.
In a hypermetropic eye,the eyeball is too short or the lens is too flat,causing the light rays from a nearby object to converge at a point behind the retina.
Therefore,the image is formed beyond the retina,making close objects appear blurry.

(A) The size of the image formed on the retina is directly proportional to the size of the object placed in front of the eye. As the object size increases,the angle subtended by the object at the eye increases,leading to a larger image on the retina. Therefore,the correct option is $A$.

(D) The student suffers from myopia (nearsightedness). To see objects at infinity (or a far distance),the lens must form a virtual image of the object at the student's far point.
Given: Far point $v = -15 \, cm$,Object distance $u = -300 \, cm$ $(3 \, m = 300 \, cm)$.
Using the lens formula: $\frac{1}{f} = \frac{1}{v} - \frac{1}{u}$.
Substituting the values: $\frac{1}{f} = \frac{1}{-15} - \frac{1}{-300} = \frac{-20 + 1}{300} = \frac{-19}{300}$.
Therefore,$f = -\frac{300}{19} \approx -15.8 \, cm$.
The power of the lens is given by $P = \frac{100}{f (cm)} = \frac{100}{-15.789} \approx -6.33 \, D$.
Thus,the focal length is $-15.8 \, cm$ and the power is $-6.3 \, D$.

(D) The ability of the human eye to adjust its focal length by changing the curvature of the crystalline lens,allowing it to focus on objects at varying distances,is known as accommodation.
Therefore,the correct option is $D$.

(A) In figure $1$,parallel rays from infinity focus before the retina,which is the condition for myopia (short-sightedness). Therefore,the statement '$1$ represents far-sightedness' is incorrect.
In figure $2$,a concave lens is used to correct myopia (short-sightedness).
In figure $3$,parallel rays focus behind the retina,which is the condition for hypermetropia (far-sightedness).
In figure $4$,a convex lens is used to correct hypermetropia (far-sightedness).
Thus,option $A$ is the wrong description.

(B) For a myopic eye,the far point is at a finite distance $v$ instead of infinity. The corrective lens must form an image of an object at infinity at the person's far point.
Using the lens formula: $\frac{1}{f} = \frac{1}{v} - \frac{1}{u}$.
Given power $P = -0.66 \ D$ (since it is for myopia,the lens is concave).
Focal length $f = \frac{1}{P} = \frac{1}{-0.66} \ m = -\frac{100}{66} \ cm \approx -151.5 \ cm$.
For an object at infinity $(u = -\infty)$,the image is formed at the far point $v$:
$\frac{1}{f} = \frac{1}{v} - \frac{1}{-\infty} \implies \frac{1}{f} = \frac{1}{v} \implies v = f$.
Thus,the far point is $151.5 \ cm$.

(D) The near point of a normal human eye is the minimum distance at which an object can be seen clearly without strain,which is $25 \, cm$.
The far point of a normal human eye is the maximum distance at which an object can be seen clearly,which is at infinity for a healthy eye.

(A) Myopia,also known as near-sightedness,is a defect of the human eye where distant objects cannot be seen clearly because the image is formed in front of the retina.
To correct this defect,a concave lens (diverging lens) is used to diverge the incoming light rays before they enter the eye,allowing the image to be focused correctly on the retina.
Therefore,the correct option is $A$.

(A) The eye lens is composed of a fibrous,jelly-like material. Its curvature can be modified to some extent by the ciliary muscles. The change in the curvature of the eye lens changes its focal length.
When the ciliary muscles are relaxed,the lens becomes thin,which increases its focal length. This state enables the eye to see distant objects clearly with minimal strain.
When we look at objects closer to the eye,the ciliary muscles contract to increase the curvature of the eye lens,making it thicker and decreasing its focal length. This process requires more effort from the muscles.
Therefore,the muscles of a normal eye are least strained when the eye is focused on an object far away.

(B) For $(i)$ reading,the person needs to see an object at the standard near point $(u = -25 \, cm)$ as if it were at their actual near point $(v = -50 \, cm)$.
Using the lens formula $\frac{1}{f} = \frac{1}{v} - \frac{1}{u}$:
$\frac{1}{f} = \frac{1}{-50} - \frac{1}{-25} = \frac{-1 + 2}{50} = \frac{1}{50} \, cm^{-1}$.
Power $P = \frac{100}{f(cm)} = \frac{100}{50} = +2 \, D$.
For $(ii)$ seeing distant stars,the person needs to see an object at infinity $(u = \infty)$ as if it were at their far point $(v = -3 \, m)$.
Using the lens formula $\frac{1}{f} = \frac{1}{v} - \frac{1}{u}$:
$\frac{1}{f} = \frac{1}{-3} - \frac{1}{\infty} = -\frac{1}{3} \, m^{-1}$.
Power $P = \frac{1}{f(m)} = -\frac{1}{3} \approx -0.33 \, D$.

(A) Myopia,also known as nearsightedness,is a vision condition in which people can see close objects clearly,but objects farther away appear blurred.
This occurs because the eyeball is too long or the cornea is too curved,causing light entering the eye to focus in front of the retina instead of directly on it.
Therefore,a person with myopia experiences difficulty in seeing distant objects clearly.

(B) The person is suffering from myopia (nearsightedness), as they cannot see distant objects clearly.
To correct this, the lens must form a virtual image of an object at infinity $(u = -\infty)$ at the person's far point $(v = -100 \, cm = -1 \, m)$.
Using the lens formula: $\frac{1}{f} = \frac{1}{v} - \frac{1}{u}$.
Substituting the values: $\frac{1}{f} = \frac{1}{-1} - \frac{1}{-\infty} = -1 - 0 = -1 \, m^{-1}$.
The power of the lens $P$ is given by $P = \frac{1}{f(\text{in meters})}$.
Therefore, $P = -1 \, D$.

(C) The person wants to see an object placed at $u = -25\, cm$ as if it were at their near point $v = -50\, cm$.
Using the lens formula: $\frac{1}{f} = \frac{1}{v} - \frac{1}{u}$.
Substituting the values: $\frac{1}{f} = \frac{1}{-50} - \frac{1}{-25}$.
$\frac{1}{f} = -\frac{1}{50} + \frac{1}{25} = \frac{-1 + 2}{50} = \frac{1}{50}$.
Thus,$f = +50\, cm$.
$A$ positive focal length indicates a convex lens of focal length $50\, cm$.

(B) The person is suffering from myopia (nearsightedness),as they cannot see distant objects clearly. The far point of the person is $v = -40 \, cm$.
To correct this,we need to use a concave lens that forms a virtual image of an object at infinity at the person's far point $(u = -\infty)$.
Using the lens formula: $\frac{1}{f} = \frac{1}{v} - \frac{1}{u}$.
Substituting the values: $\frac{1}{f} = \frac{1}{-40} - \frac{1}{-\infty} = \frac{1}{-40} - 0 = -\frac{1}{40} \, cm^{-1}$.
The power of the lens $P$ is given by $P = \frac{100}{f}$ (where $f$ is in $cm$).
$P = \frac{100}{-40} = -2.5 \, D$.

(D) The person has hypermetropia (farsightedness) with a near point of $60\, cm$.
To read a book at $25\, cm$, the lens must form a virtual image of the object placed at $u = -25\, cm$ at the person's near point $v = -60\, cm$.
Using the lens formula: $\frac{1}{f} = \frac{1}{v} - \frac{1}{u}$
Substituting the values: $\frac{1}{f} = \frac{1}{-60} - \frac{1}{-25} = \frac{1}{-60} + \frac{1}{25}$
$\frac{1}{f} = \frac{-5 + 12}{300} = \frac{7}{300}\, cm^{-1}$
$f = \frac{300}{7}\, cm = \frac{3}{7}\, m$
Power $P = \frac{1}{f(\text{in } m)} = \frac{1}{3/7} = \frac{7}{3} \approx +2.33\, D$.

(D) The person has a near point of $u = -120 \, cm$ due to hypermetropia.
To read at a distance of $v = -40 \, cm$,the lens must form a virtual image of an object placed at $40 \, cm$ at the person's actual near point of $120 \, cm$.
Using the lens formula: $\frac{1}{f} = \frac{1}{v} - \frac{1}{u}$.
Substituting the values: $\frac{1}{f} = \frac{1}{-40} - \frac{1}{-120}$.
$\frac{1}{f} = -\frac{1}{40} + \frac{1}{120} = \frac{-3 + 1}{120} = \frac{-2}{120} = -\frac{1}{60}$.
Therefore,$f = -60 \, cm$.

(A) The human eye is in a relaxed state when it is focused on objects at infinity.
In this state,the ciliary muscles are least strained,and the focal length of the eye lens is at its maximum.
As the object moves closer to the eye,the ciliary muscles must contract to increase the curvature of the lens,thereby decreasing the focal length to maintain focus on the retina.
Therefore,the strain on the ciliary muscles is minimum when the object is at infinity.

(C) The person is suffering from myopia (nearsightedness). The far point of the person is $x = 3 \, m$. The person wants to see objects at $y = 12 \, m$.
To correct this,the lens must form a virtual image of an object placed at $y = 12 \, m$ at the person's far point $x = 3 \, m$.
Using the lens formula: $\frac{1}{f} = \frac{1}{v} - \frac{1}{u}$.
Here,$v = -3 \, m$ and $u = -12 \, m$.
$\frac{1}{f} = \frac{1}{-3} - \frac{1}{-12} = -\frac{1}{3} + \frac{1}{12} = \frac{-4 + 1}{12} = -\frac{3}{12} = -\frac{1}{4} \, m^{-1}$.
Therefore,the power of the lens is $P = \frac{1}{f} = -0.25 \, D$.

(C) The total converging power of the eye is the sum of the power of the cornea and the eye lens.
Total power $P = P_{c} + P_{e} = 40\, D + 20\, D = 60\, D$.
For a normal eye,the image of an object at infinity is formed on the retina.
The focal length $f$ of the eye system is given by $f = \frac{1}{P}$.
$f = \frac{1}{60}\, m = \frac{100}{60}\, cm = \frac{5}{3}\, cm$.
$f \approx 1.67\, cm$.
Since the image is formed on the retina,the distance between the retina and the cornea-eye lens is equal to the focal length of the eye system,which is $1.67\, cm$.

(A) The person suffers from myopia (nearsightedness) because they cannot see objects beyond $400\, cm$.
To correct this, we need to place a lens such that an object at infinity $(\infty)$ forms a virtual image at the person's far point $(400\, cm)$.
Given: Object distance $u = -\infty$, Image distance $v = -400\, cm = -4\, m$.
Using the lens formula: $\frac{1}{f} = \frac{1}{v} - \frac{1}{u}$.
Substituting the values: $\frac{1}{f} = \frac{1}{-4} - \frac{1}{-\infty} = -0.25\, m^{-1}$.
Since the focal length $f$ is negative, the lens must be concave.
The power of the lens $P = \frac{1}{f} = -0.25\, D$.

(C) The person wants to read at a distance of $d = 22\; cm$ from the eyes.
Since the spectacles are placed $2\; cm$ from the eyes,the object distance $u$ for the lens is $u = -(22 - 2) = -20\; cm$.
The lens must form a virtual image at the person's near point,which is $60\; cm$ from the eyes.
Since the lens is $2\; cm$ from the eyes,the image distance $v$ is $v = -(60 - 2) = -58\; cm$.
Using the lens formula $\frac{1}{f} = \frac{1}{v} - \frac{1}{u}$:
$\frac{1}{f} = \frac{1}{-58} - \frac{1}{-20} = \frac{1}{20} - \frac{1}{58} = \frac{58 - 20}{20 \times 58} = \frac{38}{1160}$.
$f = \frac{1160}{38} \approx 30.5\; cm$.
Given the options provided,the closest integer value is $30\; cm$.

(D) For a normal eye,the image distance $v$ is equal to the diameter of the eyeball,$v = 2.5\, cm = 0.025\, m$.
When viewing distant objects (at infinity),the eye is relaxed. Using the lens formula $\frac{1}{f} = \frac{1}{v} - \frac{1}{u}$:
$P = \frac{1}{f} = \frac{1}{0.025} - \frac{1}{-\infty} = 40\, D$.
When viewing objects at the near point $(u = -25\, cm = -0.25\, m)$,the eye is under maximum strain:
$P = \frac{1}{f} = \frac{1}{0.025} - \frac{1}{-0.25} = 40 + 4 = 44\, D$.
Thus,the power of the eye lens varies from $44\, D$ to $40\, D$.

(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 $P = +3 \ D$.

(C) For a person with myopia (nearsightedness),the far point is limited. To see distant objects (at infinity) clearly,the corrective lens must form a virtual image of the object at the patient's actual far point.
Here,the object distance $u = \infty$ and the image distance $v = -40 \ cm$ (since the image must be formed on the same side as the object).
Using the lens formula: $\frac{1}{f} = \frac{1}{v} - \frac{1}{u}$
$\frac{1}{f} = \frac{1}{-40} - \frac{1}{\infty} = \frac{1}{-40} - 0 = -\frac{1}{40} \ cm^{-1}$
Power $P$ in Diopters is given by $P = \frac{100}{f \text{ (in cm)}}$
$P = \frac{100}{-40} = -2.5 \ D$

(A) The human eye acts like a convex lens system.
When light rays from an object enter the eye,they are refracted by the cornea and the crystalline lens.
These rays converge to form an image on the light-sensitive screen called the retina.
Since the light rays actually meet at the retina,the image formed is real.
Furthermore,due to the nature of the convex lens,the image formed on the retina is inverted.
Therefore,the image formed on the retina is real and inverted.

(D) person suffering from short-sightedness (myopia) cannot see distant objects clearly. To correct this,a concave lens is used to bring the image of distant objects (at infinity) to the person's far point $(2\,m)$.
Using the lens formula: $\frac{1}{f} = \frac{1}{v} - \frac{1}{u}$.
Here,$v = -2\,m$ (the far point) and $u = -\infty$.
$\frac{1}{f} = \frac{1}{-2} - \frac{1}{-\infty} = -0.5\,m^{-1}$.
Power $P = \frac{1}{f(m)} = -0.5\,D$.

(B) For a long-sighted person,the near point is shifted to $x \ m$. To read a newspaper at $u = -x/2 \ m$,the lens must form a virtual image at the person's near point,so $v = -x \ m$.
Using the lens formula: $\frac{1}{f} = \frac{1}{v} - \frac{1}{u}$.
Substituting the values: $P = \frac{1}{f} = \frac{1}{-x} - \frac{1}{-x/2} = -\frac{1}{x} + \frac{2}{x} = \frac{1}{x}$.
Given that the power $P = x \ D$,we have $x = \frac{1}{x}$.
This implies $x^2 = 1$,so $x = 1$ (since distance must be positive).

(B) Near-sightedness,or myopia,is a common refractive error of the eye.
Myopia occurs when the eyeball is too long relative to the focusing power of the cornea and the lens of the eye.
This causes light rays to focus at a point in front of the retina,rather than directly on its surface.
To correct this,a diverging lens is required to shift the focal point back onto the retina.
$A$ concave lens is a diverging lens,which is why it is used to treat myopia.

(C) For a far-sighted person (hypermetropia), the near point is shifted further away. The lens must form a virtual image of an object placed at the normal near point $(u = -25\, cm)$ at the person's actual near point $(v = -60\, cm)$.
Using the lens formula: $\frac{1}{f} = \frac{1}{v} - \frac{1}{u}$
Substituting the values: $\frac{1}{f} = \frac{1}{-60} - \frac{1}{-25}$
$\frac{1}{f} = -\frac{1}{60} + \frac{1}{25} = \frac{-5 + 12}{300} = \frac{7}{300}$
Power $P = \frac{100}{f(cm)} = \frac{100 \times 7}{300} = +2.33\, D$.

(A) For a hypermetropic person,the near point is $d' = 75\, cm$. The person wants to read a book at the normal near point $d = 25\, cm$.
Using the lens formula: $\frac{1}{f} = \frac{1}{v} - \frac{1}{u}$.
Here,$v = -75\, cm$ (the image must be formed at the person's actual near point) and $u = -25\, cm$ (the object distance).
$\frac{1}{f} = \frac{1}{-75} - \frac{1}{-25} = \frac{-1 + 3}{75} = \frac{2}{75}\, cm^{-1}$.
The power $P$ in Diopters is given by $P = \frac{100}{f(cm)}$.
$P = 100 \times \frac{2}{75} = \frac{200}{75} = +2.67\, D$.

(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 $40 \, cm$.

(C) For a person with myopia (far point limited),the corrective lens must form an image of an object at infinity at the patient's far point.
Using the lens formula: $\frac{1}{f} = \frac{1}{v} - \frac{1}{u}$.
Here,the object distance $u = \infty$ and the image distance $v = -40 \, cm = -0.4 \, m$.
Power $P = \frac{1}{f} = \frac{1}{v} - \frac{1}{u} = \frac{1}{-0.4} - \frac{1}{\infty}$.
$P = -2.5 \, D$.

(A) For a myopic person,the far point is at a finite distance $v = -80\, cm$.
To see objects at infinity $(u = \infty)$ clearly,the lens must form a virtual image at the person's far point.
Using the lens formula: $\frac{1}{f} = \frac{1}{v} - \frac{1}{u}$.
Substituting the values: $\frac{1}{f} = \frac{1}{-80} - \frac{1}{\infty} = \frac{1}{-80} - 0 = -\frac{1}{80}\, cm^{-1}$.
Thus,the focal length $f = -80\, cm = -0.8\, m$.
The power of the lens is given by $P = \frac{1}{f(m)}$.
$P = \frac{1}{-0.8} = -1.25\, D$.

(C) The phenomenon where an image persists on the retina after the stimulus is removed is known as persistence of vision.
This persistence of vision lasts for approximately $\frac{1}{16} \, s$.
To perceive motion in a movie, the frame rate must be higher than the reciprocal of the persistence time, which is $16 \, \text{frames per second}$.
Since $24 \, \text{frames per second}$ is greater than $16 \, \text{frames per second}$, the motion appears smooth.
The Reason provided states the persistence time is $1/10 \, s$, which is factually incorrect as it is approximately $1/16 \, s$.
Therefore, the Assertion is correct, but the Reason is incorrect.

(A) The human eye is considered a natural optical instrument. It functions similarly to a camera,where light enters through the pupil,is refracted by the cornea and lens,and forms an image on the retina.

(A) $(i)$ For a normal relaxed eye,the power required to focus at the far point $(0.1\, m)$ is given by the lens formula $\frac{1}{f} = \frac{1}{v} - \frac{1}{u}$. With $v = 0.02\, m$ and $u = -0.1\, m$,the power $P_f = \frac{1}{0.02} - \frac{1}{0.1} = 50 - (-10) = 60\, D$.
To see distant objects $(u = \infty)$,the required power is $P_f' = \frac{1}{0.02} - \frac{1}{\infty} = 50\, D$.
The power of the corrective lens $P_g$ is $P_f' - P_f = 50 - 60 = -10\, D$.
$(ii)$ The power of accommodation is $4\, D$. The power at the near point $P_n = P_f + 4 = 60 + 4 = 64\, D$.
Using $\frac{1}{f} = P_n = \frac{1}{v} - \frac{1}{u_n}$,where $v = 0.02\, m$:
$64 = \frac{1}{0.02} - \frac{1}{u_n} \implies 64 = 50 - \frac{1}{u_n} \implies \frac{1}{u_n} = 50 - 64 = -14$.
$u_n = -\frac{1}{14} \approx -0.0714\, m$ or $7.14\, cm$.
$(iii)$ With glasses of $P_g = -10\, D$,the total power at the near point is $P_{total} = P_n + P_g = 64 - 10 = 54\, D$.
Using $\frac{1}{v} - \frac{1}{u_n'} = 54$,with $v = 0.02\, m$:
$50 - \frac{1}{u_n'} = 54 \implies \frac{1}{u_n'} = 50 - 54 = -4$.
$u_n' = -\frac{1}{4} = -0.25\, m$ or $25\, cm$.

(B) For the spectacles:
Let the far point of the eye be at a distance $x$ from the spectacles.
The lens equation is $\frac{1}{v} - \frac{1}{u} = \frac{1}{f}$.
For a distant object,$u = \infty$ and $v = -x \, cm$.
$\frac{1}{-x} - \frac{1}{\infty} = \frac{1}{f} \implies f = -x \, cm = -\frac{x}{100} \, m$.
Power $P = \frac{1}{f} = -\frac{100}{x} \, D$.
Given $P = -5 \, D$,so $-5 = -\frac{100}{x} \implies x = 20 \, cm$.
For contact lenses:
The distance between the lens and the eye is $0$. The far point is at a distance $x + 2 \, cm$ from the eye (or the lens).
So,the new object distance $u' = -(20 + 2) = -22 \, cm = -0.22 \, m$.
To see distant objects $(u = \infty)$,the image must be formed at the far point $v' = -22 \, cm = -0.22 \, m$.
Using the lens formula: $P' = \frac{1}{f'} = \frac{1}{v'} - \frac{1}{u'} = \frac{1}{-0.22} - \frac{1}{\infty} = -\frac{1}{0.22} \approx -4.54 \, D$.

(B) The inability to see distant objects clearly is a symptom of $Myopia$ (nearsightedness).
However,the specific observation that a uniform mesh appears distorted or non-uniform is the characteristic symptom of $Astigmatism$.
$Astigmatism$ occurs due to the irregular curvature of the cornea or lens,causing light to focus at different points on the retina rather than a single point.
Since the question describes both the inability to see distant objects clearly and the distortion of the image,the diagnosis is $Myopia$ with $Astigmatism$.

(C) The correct answer is $A-C$.
$I$. Both the human eye and the camera use convex lenses to converge light rays,which is correct.
$II$. The eye focuses by changing the focal length of the lens (accommodation),while a camera focuses by changing the distance between the lens and the film/sensor,which is correct.
$III$. Both the eye and the camera form real and inverted (upside down) images on the retina and film/sensor respectively. Therefore,statement $III$ is incorrect.
$IV$. The retina acts as the light-sensitive screen in the eye,similar to the film or digital sensor in a camera,which is correct.
$V$. $A$ camera uses an aperture to control light,and the iris (not the cornea) controls the pupil size in the eye to regulate light. However,in the context of standard physics curriculum comparisons,statement $V$ is often accepted as correct regarding the mechanism of light control. Given the options,$I, II, IV,$ and $V$ are the intended correct statements.

(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 $P = \frac{1}{f(m)} = \frac{1}{0.5} = +2 \, D$.

(C) The person is suffering from hypermetropia (farsightedness) because the near point has shifted from the normal $25\,cm$ to $40\,cm$.
To read a book at $u = -25\,cm$,the lens must form a virtual image at the person's near point,$v = -40\,cm$.
Using the lens formula: $\frac{1}{f} = \frac{1}{v} - \frac{1}{u}$.
Substituting the values: $P = \frac{1}{-0.4} - \frac{1}{-0.25}$.
$P = -2.5 + 4.0 = 1.5\,D$.
Thus,the required power of the lens is $1.5\,D$.

(D) The reading glass is used to correct the near point of the eye to the standard near point of $25 \, cm$.
For a reading glass,the object is placed at the standard near point $u = -25 \, cm$,and the image is formed at the person's actual near point $v$.
The power of the reading glass is $P = +2.0 \, D$.
The focal length $f$ is given by $f = \frac{1}{P} = \frac{1}{2} \, m = 50 \, cm$.
Using the lens formula $\frac{1}{v} - \frac{1}{u} = \frac{1}{f}$:
$\frac{1}{v} - \frac{1}{-25} = \frac{1}{50}$
$\frac{1}{v} + \frac{1}{25} = \frac{1}{50}$
$\frac{1}{v} = \frac{1}{50} - \frac{1}{25} = \frac{1-2}{50} = -\frac{1}{50}$
$v = -50 \, cm$.
The least distance of distinct vision for this person is $50 \, cm$.

(B) The person has a near point of $ 75 \ cm $. To read at a normal distance of $ 25 \ cm $,the lens must form a virtual image of an object placed at $ 25 \ cm $ at the person's near point of $ 75 \ cm $.
Given: Object distance $ u = -25 \ cm $,Image distance $ v = -75 \ cm $.
Using the lens formula: $ \frac{1}{f} = \frac{1}{v} - \frac{1}{u} $
$ \frac{1}{f} = \frac{1}{-75} - \frac{1}{-25} $
$ \frac{1}{f} = -\frac{1}{75} + \frac{3}{75} = \frac{2}{75} $
$ f = \frac{75}{2} = 37.5 \ cm $
Thus,the focal length of the reading glass is $ 37.5 \ cm $.

(C) The normal least distance of distinct vision $(D)$ is $25 \ cm$. Since the boy's least distance is $35 \ cm$,he is suffering from hypermetropia (farsightedness).
To correct this,we need to use a convex lens that forms a virtual image of an object placed at $25 \ cm$ at the boy's actual near point of $35 \ cm$.
Here,object distance $u = -25 \ cm$ and image distance $v = -35 \ cm$.
Using the lens formula: $\frac{1}{f} = \frac{1}{v} - \frac{1}{u}$.
$\frac{1}{f} = \frac{1}{-35} - \frac{1}{-25} = \frac{1}{25} - \frac{1}{35}$.
$\frac{1}{f} = \frac{7 - 5}{175} = \frac{2}{175}$.
$f = \frac{175}{2} = 87.5 \ cm$.
Since the focal length is positive,it is a convex lens.

(B) The person suffers from myopia (nearsightedness) because they cannot see objects beyond $400 \ cm$. To correct this, we need a lens that forms an image at the far point $(v = -400 \ cm)$ for an object placed at infinity $(u = -\infty)$.
Using the lens formula: $\frac{1}{f} = \frac{1}{v} - \frac{1}{u}$.
Substituting the values: $\frac{1}{f} = \frac{1}{-400} - \frac{1}{-\infty} = \frac{1}{-400} - 0$.
Therefore, $f = -400 \ cm = -4 \ m$.
The power of the lens is $P = \frac{1}{f(\text{in } m)} = \frac{1}{-4} = -0.25 \ D$.
$A$ negative power indicates a concave lens.

(D) person with short-sightedness (myopia) cannot see distant objects clearly because the image forms in front of the retina. To correct this, a concave lens is used.
For a person to see objects at infinity $(u = \infty)$, the lens must form a virtual image at the person's far point $(v = -400 \ cm = -4 \ m)$.
Using the lens formula: $\frac{1}{f} = \frac{1}{v} - \frac{1}{u}$.
Substituting the values: $\frac{1}{f} = \frac{1}{-4} - \frac{1}{\infty} = -0.25 \ m^{-1}$.
Since power $P = \frac{1}{f(\text{in meters})}$, we get $P = -0.25 \ D$.

(B) Given,power of the lens $P = 2 \ D$.
The focal length of the lens is $f = \frac{1}{P} = \frac{100}{2} = 50 \ cm$.
For a person with a near point defect (hypermetropia),the lens forms an image of an object placed at the standard near point $(u = -25 \ cm)$ at the person's actual near point $(v)$.
Using the lens formula: $\frac{1}{f} = \frac{1}{v} - \frac{1}{u}$.
Substituting the values: $\frac{1}{50} = \frac{1}{v} - \frac{1}{-25}$.
$\frac{1}{50} = \frac{1}{v} + \frac{1}{25}$.
$\frac{1}{v} = \frac{1}{50} - \frac{1}{25} = \frac{1 - 2}{50} = -\frac{1}{50}$.
Therefore,$v = -50 \ cm$.
The magnitude of the near point is $50 \ cm$.

(B) For a long-sighted person (hypermetropia),the person cannot see objects clearly at the normal near point $(u = -25 \ cm)$. However,the problem states the person's current near point is $v = -60 \ cm$. We want to bring the near point to $u = -12 \ cm$ using a lens.
Using the lens formula: $\frac{1}{f} = \frac{1}{v} - \frac{1}{u}$
Here,$v = -60 \ cm$ (image formed at the person's actual near point) and $u = -12 \ cm$ (object placed at the desired near point).
$\frac{1}{f} = \frac{1}{-60} - \frac{1}{-12} = \frac{-1 + 5}{60} = \frac{4}{60} = \frac{1}{15} \ cm^{-1}$.
Since $f$ is in $cm$,$f = 15 \ cm = 0.15 \ m$.
The power of the lens $P = \frac{1}{f(m)} = \frac{1}{0.15} = \frac{100}{15} = +\frac{20}{3} \ D$.