Human Eye and Defects of Vision Questions in English

(C) far-sighted person suffers from hypermetropia,meaning they cannot see nearby objects clearly because the image forms behind the retina.
By looking through a small hole (a pinhole),the person effectively increases the depth of focus of their eye.
This reduction in the aperture of the eye allows the light rays to enter more parallel to the optical axis,which reduces the blur circle on the retina.
Effectively,this acts like a convex lens,allowing the eye to focus on closer objects by decreasing the required accommodation power,which is equivalent to an effective decrease in the focal length of the eye's optical system to bring the image onto the retina.

(A) In optometry,the least distance of distinct vision $(LDDV)$ is the closest distance at which an object can be placed so that it can be seen clearly by a normal human eye without any strain.
For a normal human eye,this distance is standardly defined as $25 \ cm$.
To convert this into meters,we divide by $100$:
$25 \ cm = 25 / 100 \ m = 0.25 \ m$.
Therefore,the correct option is $A$.

(B) In myopia,the far point of the eye is at a finite distance $d$ instead of infinity. To correct this,we need a lens that forms an image of an object at infinity at the far point $d$ of the myopic eye.
Using the lens formula: $\frac{1}{f} = \frac{1}{v} - \frac{1}{u}$
Here,$u = -\infty$ and $v = -d$ (since the image must be formed at the far point).
Substituting these values: $\frac{1}{f} = \frac{1}{-d} - \frac{1}{-\infty} = -\frac{1}{d} + 0$
Therefore,$f = -d$.
Since the focal length $f$ is negative,a concave lens is used to correct myopia.

(D) Hypermetropia,also known as long-sightedness,is a vision defect where a person can see distant objects clearly but cannot see nearby objects distinctly. This happens because the light rays from a nearby object focus behind the retina. To correct this,a convex lens is used to converge the incoming light rays before they enter the eye,allowing the image to be formed exactly on the retina. Therefore,the correct option is $D$.

(B) The choroid is the vascular layer of the eye,lying between the retina and the sclera. It contains specialized cells known as retinal pigmented epithelial cells. These cells contain a dark-colored pigment called Nigrim (or melanin),which helps in absorbing excess light and preventing internal reflection within the eye. This layer also provides oxygen and nourishment to the outer layers of the retina.

(C) Astigmatism is a defect of the human eye where the cornea or lens has an irregular curvature,causing blurred vision at all distances.
To correct this defect,a cylindrical lens is used.
The axis of the cylindrical lens is oriented to compensate for the irregular curvature of the eye's cornea,thereby focusing light correctly onto the retina.

(B) The retina contains a small,yellowish,circular area located at the centre of the posterior part of the eye,directly opposite the lens. This area is known as the $Yellow$ $spot$ or $macula$ $lutea$. It is the region of highest visual acuity,containing a high density of cone cells. Therefore,the correct option is $B$.

(A) The human eye acts like a convex lens. Light rays from an object enter the eye and are refracted by the cornea and the crystalline lens. These rays converge to form an image on 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) With two eyes,the field of view is wider (about $180^{\circ}$),which allows for a larger visible region. If a person has only one eye,the field of view is reduced. Furthermore,binocular vision (using two eyes) is essential for perceiving depth and three-dimensional images. With only one eye,the brain cannot compare two slightly different perspectives,so the ability to judge depth is lost. Therefore,both $(b)$ and $(c)$ are correct.

(A) The person suffers from hypermetropia (farsightedness) as they cannot see objects closer than $1 \, m$ $(100 \, cm)$.
To read a book at $u = -25 \, cm$, the lens must form a virtual image at the person's near point, $v = -100 \, cm$.
Using the lens formula: $\frac{1}{f} = \frac{1}{v} - \frac{1}{u}$.
Substituting the values: $\frac{1}{f} = \frac{1}{-100} - \frac{1}{-25} = -\frac{1}{100} + \frac{1}{25} = \frac{-1 + 4}{100} = \frac{3}{100} \, cm^{-1}$.
Since power $P = \frac{1}{f(\text{in } meters)} = \frac{100}{f(\text{in } cm)}$, we have $P = \frac{100}{100/3} = +3.0 \, D$.

(C) The man suffers from myopia (nearsightedness),where the far point is limited to $5 \, m$.
To see distant objects (stars) clearly,the lens must form a virtual image of an object at infinity at the man's far point.
Given: Object distance $u = \infty$,Image distance $v = -5 \, m$ (using sign convention).
Using the lens formula: $\frac{1}{f} = \frac{1}{v} - \frac{1}{u}$.
Substituting the values: $\frac{1}{f} = \frac{1}{-5} - \frac{1}{\infty}$.
Since $\frac{1}{\infty} = 0$,we get $\frac{1}{f} = -\frac{1}{5}$.
Therefore,$f = -5 \, m$.

(A) To correct the near point, we need to place an object at the standard near point $(u = -25 \, cm)$ such that its virtual image is formed at the man's actual near point $(v = -75 \, cm)$.
Using the lens formula: $\frac{1}{f} = \frac{1}{v} - \frac{1}{u}$.
Substituting the values: $\frac{1}{f} = \frac{1}{-75} - \frac{1}{-25} = \frac{-1 + 3}{75} = \frac{2}{75} \, cm^{-1}$.
Thus, $f = \frac{75}{2} \, cm = 0.375 \, m$.
The power of the lens is given by $P = \frac{1}{f(\text{in meters})} = \frac{1}{0.375} = \frac{100}{37.5} = +\frac{8}{3} \, D$.

(B) In short-sightedness (myopia),the focal length of the eye lens decreases or the eyeball becomes too long. As a result,the light rays from a distant object converge at a point before reaching the retina. Therefore,the image is formed before the retina.

(D) The man is suffering from myopia (nearsightedness), where the far point is limited to $100 \, cm$. To see objects at infinity, the lens must form a virtual image of an object at infinity at the person's far point of $100 \, cm$.
Using the lens formula: $\frac{1}{f} = \frac{1}{v} - \frac{1}{u}$
Here, $v = -100 \, cm = -1 \, m$ (image formed at the far point) and $u = -\infty$ (object at infinity).
$\frac{1}{f} = \frac{1}{-1} - \frac{1}{-\infty} = -1 \, m^{-1}$
Power $P = \frac{1}{f (\text{in meters})} = -1 \, D$.

(B) The man has a far point of $30 \ cm$ and a near point of $15 \ cm$. To see far objects (at infinity),he needs a concave lens such that an object at infinity forms an image at his far point $(v = -30 \ cm)$.
Using the lens formula: $\frac{1}{f} = \frac{1}{v} - \frac{1}{u} = \frac{1}{-30} - \frac{1}{\infty} = -\frac{1}{30}$.
Thus,the focal length $f = -30 \ cm$.
Now,to find the new near point,we use the same lens to see an object at distance $u$ such that the image is formed at his actual near point $(v = -15 \ cm)$.
$\frac{1}{f} = \frac{1}{v} - \frac{1}{u} \Rightarrow -\frac{1}{30} = \frac{1}{-15} - \frac{1}{u}$.
$\frac{1}{u} = \frac{1}{30} - \frac{1}{15} = \frac{1-2}{30} = -\frac{1}{30}$.
Therefore,$u = -30 \ cm$. The new near point is at $30 \ cm$.

(C) For a myopic eye,the far point is at a finite distance instead of infinity. To correct this,a concave lens is used to form an image of an object at infinity at the person's far point.
Given,the far point $d = 40 \, cm = 0.4 \, m$.
The focal length $f$ of the required concave lens is equal to the negative of the far point distance: $f = -40 \, cm = -0.4 \, m$.
The power of the lens $P$ is given by the formula $P = \frac{1}{f \text{ (in meters)}}$.
$P = \frac{1}{-0.4} = -2.5 \, D$.
Therefore,the power of the lens required is $-2.5 \, D$.

(C) For a myopic eye,the lens must form a virtual image of an object placed at the desired distance $(u)$ at the person's far point $(v)$.
Given: Desired object distance $u = -60 \ cm$.
Given: Far point of the person $v = -10 \ cm$.
Using the lens formula: $\frac{1}{f} = \frac{1}{v} - \frac{1}{u}$.
Substituting the values: $\frac{1}{f} = \frac{1}{-10} - \frac{1}{-60}$.
$\frac{1}{f} = -\frac{1}{10} + \frac{1}{60} = \frac{-6 + 1}{60} = -\frac{5}{60}$.
$\frac{1}{f} = -\frac{1}{12}$.
Therefore,$f = -12 \ cm$.

(B) For a myopia patient,the far point is at a finite distance $x$. To correct this,a concave lens is used such that the image of an object at infinity is formed at the far point $x$.
Using the lens formula: $\frac{1}{f} = \frac{1}{v} - \frac{1}{u}$.
Here,$v = -x$ and $u = \infty$.
So,$\frac{1}{f} = \frac{1}{-x} - \frac{1}{\infty} = -\frac{1}{x}$,which gives $f = -x$.
If the far point distance is doubled,the new far point is $x' = 2x$.
The new focal length $f'$ will be $-x' = -2x$.
Since $f' = 2f$,the magnitude of the focal length becomes double.

(B) An imaginary line joining the optical centre of the eye lens and the yellow point (fovea centralis) is known as the $Vision \text{ axis}$.
The $Vision \text{ axis}$ represents the path along which light rays travel to reach the fovea, which is the area of sharpest vision on the retina.

(A) The Cornea is the outer membrane of the eyeball. It has a refractive index,$\mu = 1.38$,which is quite large compared to the refractive index of the surrounding air medium,$\mu_{air} \approx 1.00$.
As light falls on the Cornea first before hitting the retina,and due to the significant difference in refractive indices between air and the Cornea,light undergoes most of its refraction (moving from an optically rarer to a denser medium) while passing through the Cornea.
Therefore,when light enters the human eye,it experiences the majority of its refraction while passing through the Cornea.

(A) The phenomenon where the image of an object persists on the retina for a short duration after the object is removed from the field of view is known as persistence of vision.
This duration is approximately $1/16$ of a second,which is equal to $0.0625$ seconds,commonly rounded to $0.1$ seconds in physics textbooks.
Therefore,the correct option is $A$.

(A) For a myopic eye,a concave lens is used to correct the vision.
Given:
Object distance $u = -60 \,cm$ (the distance at which the person wants to see).
Image distance $v = -15 \,cm$ (the far point of the myopic eye).
Using the lens formula: $\frac{1}{f} = \frac{1}{v} - \frac{1}{u}$.
Substituting the values: $\frac{1}{f} = \frac{1}{-15} - \frac{1}{-60}$.
$\frac{1}{f} = -\frac{1}{15} + \frac{1}{60} = \frac{-4 + 1}{60} = -\frac{3}{60} = -\frac{1}{20}$.
Therefore,$f = -20 \,cm$.
The negative sign indicates a concave lens of focal length $20 \,cm$.

(D) The sensation of vision in the retina is carried to the brain by the $Optic \text{ } nerve$.
Once light reflected from the surrounding object enters the retina through the eye lens, chemical and electrical impulses (sensory information) are generated.
The $Optic \text{ } nerve$ then transmits this visual information to the brain via electrical impulses for processing.

(A) The power of a lens $(P)$ is inversely proportional to its focal length $(f)$,given by $P = 1/f$.
When the power of the eye lens increases,the focal length of the eye lens decreases.
This causes the image of distant objects to be formed in front of the retina instead of on the retina.
This condition is known as short-sightedness or myopia.

(D) Colour blindness is a genetic disorder caused by the absence or malfunction of specific cone cells in the retina.
Since it is a hereditary condition related to the genetic makeup of the individual,it cannot be corrected or cured by wearing any type of corrective lenses or goggles.
Therefore,none of the options provided can remove this defect.

(C) The human eye focuses on objects at varying distances through a process called accommodation.
This is achieved by the ciliary muscles,which adjust the shape (convexity) of the crystalline lens.
When the ciliary muscles contract,the lens becomes more convex (thicker),increasing its power to focus on nearby objects.
When the ciliary muscles relax,the lens becomes less convex (thinner),decreasing its power to focus on distant objects.
Therefore,the correct mechanism is the change in the convexity of the lens surface.

(B) short-sighted person (myopic) cannot see objects beyond their far point clearly.
Here,the far point of the person is $d = 100\, cm = 1\, m$.
To see a distant object (at infinity),the lens must form a virtual image of the object at the person's far point.
Using the lens formula: $\frac{1}{f} = \frac{1}{v} - \frac{1}{u}$.
Here,$u = \infty$ and $v = -100\, cm = -1\, m$ (since the image must be formed on the same side as the object).
$\frac{1}{f} = \frac{1}{-1} - \frac{1}{\infty} = -1 - 0 = -1\, m^{-1}$.
Power $P = \frac{1}{f} = -1\, D$.

(C) The person is suffering from myopia (nearsightedness) as he cannot see objects beyond $25 \, cm$. To correct this,a concave lens is required.
Given:
Object distance $u = -50 \, cm$ (the position where he wants to read).
Image distance $v = -25 \, cm$ (his far point,where the virtual image must be formed).
Using the lens formula: $\frac{1}{f} = \frac{1}{v} - \frac{1}{u}$
$\frac{1}{f} = \frac{1}{-25} - \frac{1}{-50} = -\frac{2}{50} + \frac{1}{50} = -\frac{1}{50}$
So,$f = -50 \, cm = -0.5 \, m$.
The power of the lens is $P = \frac{1}{f(m)} = \frac{1}{-0.5} = -2.0 \, D$.
Thus,a concave lens of power $-2.0 \, D$ is required.

(C) The human eye lens is a crystalline,transparent,and biconvex structure. Unlike a glass lens which has a uniform material and constant refractive index,the human eye lens is composed of layers of protein fibers. These fibers are more densely packed towards the center of the lens compared to the periphery. Consequently,the refractive index of the human eye lens is not constant; it gradually increases from the outer surface towards the center. This gradient refractive index helps in reducing spherical aberration and focusing light effectively on the retina. Therefore,the correct option is $C$.

(C) The man is suffering from myopia (nearsightedness) because he cannot see distant objects clearly.
To correct this defect, we use a concave lens that forms a virtual image of an object at infinity at the man's far point $(60 \, cm)$.
Here, the object distance $u = \infty$ and the image distance $v = -60 \, cm = -0.6 \, m$.
Using the lens formula: $\frac{1}{f} = \frac{1}{v} - \frac{1}{u}$.
$\frac{1}{f} = \frac{1}{-0.6} - \frac{1}{\infty} = -\frac{1}{0.6}$.
Power $P = \frac{1}{f(\text{in meters})} = -\frac{1}{0.6} = -1.66 \, D$.

(B) For reading,the person needs to see an object at the standard near point $(u = -25 \, cm)$ as if it were at his own near point $(v = -50 \, cm)$.
Using the lens formula $\frac{1}{f} = \frac{1}{v} - \frac{1}{u}$:
$\frac{1}{f} = \frac{1}{-50} - \frac{1}{-25} = \frac{-1 + 2}{50} = \frac{1}{50} \, cm^{-1}$.
$f = 50 \, cm = 0.5 \, m$.
Power $P = \frac{1}{f(m)} = \frac{1}{0.5} = +2 \, D$.
For seeing distant stars,the person needs to see an object at infinity $(u = \infty)$ as if it were at his 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 $P = \frac{1}{f(m)} = -\frac{1}{3} \approx -0.33 \, D$.

(B) The power of the lens is given as $P = -2.5 \, D$.
Since the power is negative,the lens used is a concave lens,which indicates that the person is suffering from myopia or nearsightedness.
The focal length $f$ of the lens is given by $f = \frac{1}{P} = \frac{1}{-2.5} \, m = -0.4 \, m = -40 \, cm$.
For a person with nearsightedness,the far point is the focal length of the corrective lens used.
Therefore,the far point of the person without glasses is $40 \, cm$.

(A) Myopia,also known as nearsightedness,occurs when the eye ball becomes too long or the cornea is too curved.
As a result,light rays entering the eye are focused at a point in front of the retina rather than directly on it.
Therefore,the correct cause among the given options is the elongation of the eye ball.

(C) Astigmatism is a common vision defect caused by an irregular curvature of the cornea or the lens of the eye.
When the cornea is not perfectly spherical,it causes light rays to focus at different points on the retina rather than a single point.
This results in blurred or distorted vision at all distances.
Therefore,the correct reason is that the cornea is not spherical.

(D) The person is suffering from myopia (nearsightedness) because they cannot see objects beyond a certain distance $(2.0 \, m)$.
To correct this defect, a concave lens is used.
The far point of the person is $v = - 2.0 \, m$.
We want the person to be able to see objects at infinity, so the object distance is $u = \infty$.
Using the lens formula: $\frac{1}{f} = \frac{1}{v} - \frac{1}{u}$.
$\frac{1}{f} = \frac{1}{- 2.0} - \frac{1}{\infty} = - 0.5 \, m^{-1}$.
Power $P = \frac{1}{f (in \, meters)} = - 0.5 \, D$.

(D) The human eye adjusts its focal length to focus objects at varying distances onto the retina.
Since the distance between the eye lens and the retina is fixed in a healthy eye,the image distance remains constant.
Therefore,the image distance from the eye lens is the correct answer.

(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 $50 \, cm$.

(C) The correct matches are as follows:
$1$. Presbyopia: This is a defect where the eye loses its power of accommodation with age. It is corrected using a bifocal lens $(I-D)$.
$2$. Hypermetropia (Farsightedness): This is corrected by using a convex lens of appropriate power to converge light rays onto the retina $(II-B)$.
$3$. Astigmatism: This defect occurs due to the irregular curvature of the cornea. It is corrected using a sphero-cylindrical lens $(III-A)$.
$4$. Myopia (Nearsightedness): This is corrected by using a concave lens of suitable focal length to diverge light rays before they enter the eye $(IV-C)$.
Therefore, the correct combination is $I-D, II-B, III-A, IV-C$.

(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 $(\infty)$.
Therefore,the near point and far point are $25\, cm$ and $\infty$ respectively.

(B) The back of the eye is lined by a layer called the retina,which acts like the film inside a camera.
The retina is a thin layer of nerve tissue that contains photoreceptors.
Photoreceptors convert light rays into electrical impulses,which are then sent through the optic nerve to the brain,where an image is perceived.
As with a camera,if the film is bad in the eye (i.e.,the retina is damaged or diseased),no matter how well the rest of the eye is functioning,a good picture is not possible.

(B) Hypermetropia,also known as long-sightedness or farsightedness,is a vision defect where a person can see distant objects clearly but cannot see nearby objects distinctly.
This occurs because the light rays from a nearby object focus behind the retina instead of on it.
Therefore,the correct option is $B$.

(D) The man is suffering from myopia (nearsightedness),as he cannot see objects beyond $20 \, cm$.
To correct this defect,the lens must form a virtual image of an object placed at infinity $(u = \infty)$ at the man's far point $(v = -20 \, cm)$.
Using the lens formula: $\frac{1}{f} = \frac{1}{v} - \frac{1}{u}$.
Substituting the values: $\frac{1}{f} = \frac{1}{-20} - \frac{1}{\infty}$.
Since $\frac{1}{\infty} = 0$,we get $\frac{1}{f} = -\frac{1}{20}$.
Therefore,$f = -20 \, cm$.
$A$ negative focal length indicates a concave lens. Thus,a concave lens of focal length $20 \, cm$ is required.

(B) The power of the lens is given as $P = +2D$.
Since the power of the lens is positive,the lens used is a convex lens.
$A$ convex lens is used to correct hypermetropia (long-sightedness),where the person cannot see nearby objects clearly because the image forms behind the retina.
Therefore,the person is suffering from hypermetropia.

(B) For a myopic eye,the corrective lens must form an image of an object at infinity at the person's far point $(x)$.
Using the lens formula: $\frac{1}{f} = \frac{1}{v} - \frac{1}{u}$.
Here,$u = -\infty$ and $v = -x$ (where $x$ is the distance of the far point).
Since $P = \frac{1}{f}$ (in meters),we have $P = \frac{1}{-x}$.
Thus,$x = -\frac{1}{P}$.
Given $P = -0.66 \, D$ (a concave lens is required for myopia).
$x = \frac{1}{0.66} \, m = 1.515 \, m$.
Converting to centimeters: $x \approx 151 \, cm \approx 150 \, cm$.

(C) Presbyopia is a vision defect where the eye loses its ability to focus on near objects due to the hardening of the lens with age. It often occurs alongside myopia (nearsightedness) and hypermetropia (farsightedness).
To correct this,a bifocal lens is used.
The upper portion of the bifocal lens is a concave lens,which is used to correct myopia (distance vision).
The lower portion of the bifocal lens is a convex lens,which is used to correct hypermetropia (near vision).
Therefore,the correct option is $C$.

(A) The person suffers from myopia (nearsightedness) and needs to see an object placed at $u = -30 \, cm$ as if it were at their near point of $v = -10 \, cm$.
Using the lens formula: $\frac{1}{f} = \frac{1}{v} - \frac{1}{u}$.
Substituting the values: $\frac{1}{f} = \frac{1}{-10} - \frac{1}{-30}$.
$\frac{1}{f} = -\frac{1}{10} + \frac{1}{30} = \frac{-3 + 1}{30} = \frac{-2}{30} = -\frac{1}{15}$.
Therefore,$f = -15 \, cm$.
The negative sign indicates a concave lens with a focal length of $15 \, cm$.

(A) For a myopic eye,the far point is at a finite distance $d = 250 \, cm$. To correct this defect,the lens must form an image of an object at infinity at the far point of the eye.
Using the lens formula: $\frac{1}{f} = \frac{1}{v} - \frac{1}{u}$.
Here,$u = \infty$ (object at infinity) and $v = -250 \, cm$ (image formed at the far point).
Substituting these values: $\frac{1}{f} = \frac{1}{-250} - \frac{1}{\infty}$.
Since $\frac{1}{\infty} = 0$,we get $\frac{1}{f} = -\frac{1}{250}$.
Therefore,$f = -250 \, cm$.

(C) The man suffers from myopia (nearsightedness). To see objects at infinity or a distant point $u = -12 \, m$,the lens must form a virtual image at his far point $v = -3 \, m$.
Using the lens formula: $\frac{1}{f} = \frac{1}{v} - \frac{1}{u}$.
Substituting the values: $P = \frac{1}{f} = \frac{1}{-3} - \frac{1}{-12}$.
$P = -\frac{1}{3} + \frac{1}{12} = \frac{-4 + 1}{12} = -\frac{3}{12} = -0.25 \, D$.
Thus,the required power is $-1/4 \, D$.

(A) The person is suffering from myopia (nearsightedness) because he cannot see distant objects clearly.
For a myopic eye, the far point is limited to $40 \, cm$.
To correct this defect, we need to use a concave lens that forms a virtual image of an object at infinity at the person's far point $(40 \, cm)$.
Here, the object distance $u = \infty$ and the image distance $v = -40 \, cm$.
Using the lens formula: $\frac{1}{f} = \frac{1}{v} - \frac{1}{u}$.
$\frac{1}{f} = \frac{1}{-40} - \frac{1}{\infty} = -\frac{1}{40} \, cm^{-1}$.
Thus, the focal length $f = -40 \, cm = -0.4 \, m$.
The power of the lens $P$ is given by $P = \frac{1}{f(\text{in } meters)}$.
$P = \frac{1}{-0.4} = -2.5 \, D$.

(A) The power of the lens is $P = +3 \text{ D}$.
Focal length $f = \frac{1}{P} = \frac{1}{3} \text{ m} = \frac{100}{3} \text{ cm}$.
For a hypermetropic eye,the near point is shifted beyond the normal near point $(u = -25 \text{ cm})$. The lens forms a virtual image of an object placed at the normal near point $(u = -25 \text{ cm})$ at the person's actual near point $(v = -x)$.
Using the lens formula $\frac{1}{f} = \frac{1}{v} - \frac{1}{u}$:
$\frac{1}{100/3} = \frac{1}{-x} - \frac{1}{-25}$
$\frac{3}{100} = -\frac{1}{x} + \frac{1}{25}$
$\frac{1}{x} = \frac{1}{25} - \frac{3}{100} = \frac{4-3}{100} = \frac{1}{100}$
$x = 100 \text{ cm} = 1 \text{ m}$.