The Human Eye and Defects of Vision Questions in English

(N/A) The power of accommodation is the ability of the human eye to adjust the focal length of its lens to focus on objects at varying distances.
When the ciliary muscles are relaxed,the eye lens becomes thin,its focal length increases,and distant objects are clearly visible.
To see nearby objects clearly,the ciliary muscles contract,making the eye lens thicker.
This contraction decreases the focal length of the eye lens,allowing the image of nearby objects to be formed clearly on the retina.
Thus,the ability of the eye to change its focal length to view both distant and nearby objects is known as the power of accommodation.

(B) The person is suffering from myopia (near-sightedness),which means they can see nearby objects clearly but cannot see distant objects beyond a certain point (the far point).
In this case,the far point of the person is $1.2\, m$.
This happens because the image of an object placed beyond $1.2\, m$ is formed in front of the retina instead of on the retina.
To correct this defect,a concave lens (diverging lens) is used.
The concave lens diverges the incoming light rays from distant objects,effectively shifting the image back onto the retina,as shown in the figure.

(A) The near point of the eye is the minimum distance of an object from the eye,which can be seen distinctly without strain. For a normal human eye,this distance is $25 \ cm$.
The far point of the eye is the maximum distance to which the eye can see objects clearly. The far point of the normal human eye is infinity.

A student who has difficulty reading the blackboard while sitting in the last row is unable to see distant objects clearly. This condition is known as $Myopia$ (near-sightedness). In this defect, the image of a distant object is formed in front of the retina rather than on the retina itself. This defect can be corrected by using a concave lens of suitable focal length, which diverges the light rays before they enter the eye, allowing the image to be focused correctly on the retina.

(A) The human eye can change the focal length of the eye lens to see objects situated at various distances from the eye.
This ability of the eye lens to adjust its focal length with the help of ciliary muscles is known as the power of accommodation.

(B) The human eye functions like a camera. Light enters the eye through the cornea and passes through the pupil. The eye lens focuses this light onto the light-sensitive screen located at the back of the eye,which is known as the retina. Therefore,the image of an object is formed on the retina.

(C) The least distance of distinct vision is the minimum distance at which an object can be placed so that a clear and distinct image is formed on the retina without any strain on the eye.
For a young adult with normal vision,this distance is approximately $25\, cm$.

(D) The relaxation or contraction of ciliary muscles changes the curvature of the eye lens.
The change in curvature of the eye lens changes the focal length of the eyes.
Hence,the change in focal length of an eye lens is caused by the action of ciliary muscles.

(A) The power $P$ of a lens is related to its focal length $f$ (in meters) by the formula $P = \frac{1}{f}$.
To find the focal length in centimeters,we use the formula $f (cm) = \frac{100}{P}$.
For correcting distant vision,the power required is $P = -5.5 \, D$.
Substituting the value of $P$ into the formula:
$f = \frac{100}{-5.5} \, cm$
$f \approx -18.18 \, cm$.
Rounding to one decimal place,the focal length is $-18.2 \, cm$.

(B) The power $P$ of a lens is related to its focal length $f$ (in meters) by the formula $P = \frac{1}{f}$.
For correcting near vision,the required power $P = +1.5 \, D$.
Using the formula,$f = \frac{1}{P} = \frac{1}{1.5} \, m$.
$f = 0.6666... \, m \approx 0.667 \, m$.
To convert the focal length into centimeters $(cm)$,we multiply by $100$:
$f = 0.667 \times 100 \, cm = 66.7 \, cm$.
Therefore,the focal length of the lens required for correcting near vision is $66.7 \, cm$.

(C) The person is suffering from an eye defect called myopia. In this defect,the image is formed in front of the retina. Hence,a concave lens is used to correct this defect of vision.
Object distance,$u = \infty$
Image distance,$v = -80 \,cm = -0.8 \,m$
Focal length = $f$
According to the lens formula,$\frac{1}{v} - \frac{1}{u} = \frac{1}{f}$
$\frac{1}{-0.8} - \frac{1}{\infty} = \frac{1}{f}$
$\frac{1}{f} = -1.25 \,m^{-1}$
$f = -0.8 \,m$
We know,Power,$P = \frac{1}{f(\text{in metres})}$
$P = \frac{1}{-0.8} = -1.25 \,D$
$A$ concave lens of power $-1.25 \,D$ is required by the person to correct his defect.

(D) person suffering from hypermetropia can see distant objects clearly but faces difficulty in seeing nearby objects clearly. It happens because the eye lens focuses the incoming divergent rays beyond the retina. This defect of vision is corrected by using a convex lens. $A$ convex lens of suitable power converges the incoming light in such a way that the image is formed on the retina,as shown in the following figure.
The convex lens actually creates a virtual image of a nearby object ( $N'$ in the figure) at the near point of vision $(N)$ of the person suffering from hypermetropia.
The given person will be able to clearly see the object kept at $25 \, cm$ (near point of the normal eye),if the image of the object is formed at his near point,which is given as $1 \, m$.
Object distance,$u = -25 \, cm = -0.25 \, m$
Image distance,$v = -1 \, m$
Focal length,$f$
Using the lens formula,
$\frac{1}{v} - \frac{1}{u} = \frac{1}{f}$
$\frac{1}{-1} - \frac{1}{-0.25} = \frac{1}{f}$
$\frac{1}{f} = \frac{1}{0.25} - 1 = 4 - 1 = 3 \, D$
Power,$P = +3.0 \, D$
$A$ convex lens of power $+3.0 \, D$ is required to correct the defect.

(D) The human eye has a limited power of accommodation.
To see objects closer than $25 \,cm$,the ciliary muscles need to contract significantly to increase the curvature of the eye lens,thereby decreasing its focal length.
However,the ciliary muscles have a limit to their contraction,meaning the focal length of the eye lens cannot be decreased below a certain minimum value.
Consequently,the image of an object placed closer than $25 \,cm$ is formed behind the retina,resulting in a blurred image.
This distance of $25 \,cm$ is known as the least distance of distinct vision.

(C) Since the size of the eyeball cannot change,the distance between the eye lens and the retina remains constant.
Therefore,the image distance in the eye remains constant.
When we increase the distance of an object from the eye,the image distance does not change.
The increase in the object distance is compensated by the change in the focal length of the eye lens through the action of ciliary muscles.
The focal length of the eye lens adjusts such that the image is always formed on the retina.

(A) The person is suffering from myopia (nearsightedness) because they cannot see distant objects beyond $2\, m$.
To correct this defect,a concave lens is used.
The focal length $(f)$ of the required lens should be equal to the far point of the person,which is $-2\, m$ (negative because it is a concave lens).
The power $(P)$ of a lens is given by the formula $P = 1/f$ (in meters).
Substituting the value: $P = 1 / (-2) = -0.5\, D$.
Therefore,the correct power of the lens is $-0.5\, D$.

(B) The student is suffering from hypermetropia (farsightedness).
In hypermetropia,a person can see distant objects clearly but cannot see nearby objects clearly.
This happens because the near point of the eye,which is normally $25 \text{ cm}$,has receded away from the eye.
Therefore,the student is unable to read the textbook held at a normal distance.

(D) When light rays enter the eye,they first encounter the cornea. The cornea is the transparent,curved,outermost layer of the eye. Because of the significant difference in the refractive index between air and the cornea,the majority of the light refraction occurs at the outer surface of the cornea. The crystalline lens then provides the fine-tuning of the focus to form a sharp image on the retina.

(A) The focal length of the eye lens is controlled by the ciliary muscles.
When the ciliary muscles are relaxed,the suspensory ligaments become taut,which pulls the lens and makes it thinner.
$A$ thinner lens has a larger radius of curvature,which results in an increase in the focal length of the eye lens.
This adjustment allows the eye to focus on distant objects.

(B) Myopia,also known as near-sightedness,is a defect of vision in which a person can see nearby objects clearly but cannot see distant objects distinctly.
Hypermetropia,also known as far-sightedness,is a defect of vision in which a person can see distant objects clearly but cannot see nearby objects distinctly.
Therefore,the statement '$A$ person with myopia can see nearby objects clearly' is correct.

(N/A) $(i)$ Myopic eye (Near-sightedness): In a myopic eye,the image of a distant object is formed in front of the retina instead of on the retina. This occurs because the eye lens becomes too curved or the eyeball becomes too long.
$(ii)$ Hypermetropic eye (Far-sightedness): In a hypermetropic eye,the image of a nearby object is formed behind the retina. This occurs because the eye lens becomes too flat or the eyeball becomes too short.

The student is suffering from myopia,also known as near-sightedness. In this defect,a person can see nearby objects clearly but cannot see distant objects distinctly because the image of a distant object is formed in front of the retina.
To correct this defect,a doctor will advise her to use a concave lens of appropriate power. The concave lens diverges the incoming light rays from a distant object,effectively shifting the image back onto the retina.

(B) The human eye is able to see both nearby and distant objects clearly by adjusting the focal length of the eye lens.
This process is known as the power of accommodation.
The ciliary muscles change the curvature of the lens,thereby altering its focal length to focus images of objects at varying distances onto the retina.

(N/A) The person is suffering from $Myopia$ (near-sightedness) because the power of the lens is negative.
$(b)$ The focal length $f$ is given by the formula $f = \frac{1}{P}$,where $P$ is the power in diopters $(D)$.
$f = \frac{1}{-4.5\, D} = -0.222\, m$ or $-22.2\, cm$.
$(c)$ Since the power and focal length are negative,the nature of the corrective lens is a $Concave$ lens.

(N/A) The human eye is a complex organ that functions like a camera. Its main parts are:
$1$. Cornea: The thin membrane covering the front of the eyeball,which acts as the primary refractive surface.
$2$. Iris: $A$ dark muscular diaphragm that controls the size of the pupil.
$3$. Pupil: The aperture that regulates the amount of light entering the eye.
$4$. Crystalline Lens: $A$ transparent,flexible lens that focuses light onto the retina.
$5$. Ciliary Muscles: Muscles that change the curvature and focal length of the lens.
$6$. Retina: The light-sensitive screen where the image is formed.
$7$. Optic Nerve: Transmits visual information from the retina to the brain.
Functioning and Power of Accommodation:
The ability of the eye to adjust its focal length to see objects at varying distances is called the power of accommodation.
- To see distant objects: The ciliary muscles relax,making the lens thin,which increases its focal length.
- To see nearby objects: The ciliary muscles contract,making the lens thicker,which decreases its focal length.

(B) Step $1$: $A$ person is considered myopic (near-sighted) when they can see nearby objects clearly but cannot see distant objects distinctly. This occurs because the image of a distant object is formed in front of the retina. It is corrected using a concave lens of suitable focal length.
Step $2$: $A$ person is considered hypermetropic (far-sighted) when they can see distant objects clearly but cannot see nearby objects distinctly. This occurs because the image of a nearby object is formed behind the retina. It is corrected using a convex lens of suitable focal length.

(N/A) The essential parts of the human eye are as follows:
$(i)$ Cornea: The transparent outer layer that protects the eye and helps focus light.
$(ii)$ Iris: The colored part of the eye that controls the size of the pupil.
$(iii)$ Pupil: The opening in the center of the iris that allows light to enter the eye.
$(iv)$ Eye lens: $A$ transparent,biconvex structure that focuses light onto the retina.
$(v)$ Ciliary muscles: Muscles that change the shape of the lens to adjust focus.
$(vi)$ Retina: The light-sensitive layer at the back of the eye where images are formed.

(B) The cornea is the thin,transparent,spherical membrane that covers the front surface of the eyeball.
It acts as the primary refractive element of the eye,allowing light to enter and focusing it onto the lens.

(B) The space between the cornea and the lens is filled with a thin,watery fluid called the $Aqueous$ $humour$. This fluid helps in maintaining the intraocular pressure and provides nourishment to the cornea and the lens.

(B) The iris is a dark,muscular diaphragm located behind the cornea and in front of the lens. It is responsible for controlling the size of the pupil,thereby regulating the amount of light entering the eye.

(B) The $iris$ is a dark muscular diaphragm that controls the size of the $pupil$. The $pupil$ regulates and controls the amount of light entering the eye by adjusting its size based on the intensity of light present in the environment.

(A) The eye lens is a transparent,biconvex structure in the eye.
It is composed of a jelly-like,proteinaceous material.
This material is organized into layers,which allows the lens to change its shape to focus light on the retina,a process known as accommodation.

(B) The eye lens is a transparent,biconvex structure made of a jelly-like proteinaceous material.
It is held in its position by the ciliary muscles.
These muscles not only hold the lens in place but also change its focal length by adjusting its curvature,which allows the eye to focus on objects at varying distances.

(B) The sense of vision is carried from the eye to the brain through the optic nerve. When light enters the eye,it forms an image on the retina. The photoreceptor cells in the retina generate electrical signals,which are then transmitted to the brain via the optic nerve for visual processing.

(B) The space between the eye lens and the retina is filled with a transparent,jelly-like substance known as the vitreous humour. It helps in maintaining the shape of the eyeball and supports the retina.

(B) blind spot is a small area on the retina where the optic nerve exits the eye to travel to the brain.
This region lacks photoreceptor cells (rods and cones).
Therefore,no image can be formed at this specific point,making it 'blind'.

(C) The blind spot is a small area on the retina where the optic nerve exits the eye to connect to the brain.
This region lacks both rod and cone cells,which are the photoreceptor cells responsible for detecting light.
Because there are no photoreceptor cells present at this specific point,no visual information is captured or transmitted to the brain.
Therefore,any image formed at this point cannot be perceived,which is why it is called the blind spot.

(B) The retina is the light-sensitive layer of the eye that contains photoreceptor cells known as rods and cones.
These cells are responsible for converting light energy into electrical signals.
The human retina contains more than $125$ million of these photoreceptor cells.

(B) The ability of the eye lens to adjust its focal length to focus objects at varying distances on the retina is known as accommodation. This is achieved by the action of ciliary muscles,which change the curvature and,consequently,the focal length of the eye lens.

(A) The ability of the eye lens to adjust its focal length to focus objects at varying distances is known as the $Power$ of $Accommodation$.

(N/A) The farthest point up to which an eye can see objects clearly is called the far point of the eye. For a normal human eye, this point is at infinity.

(D) The far point of a normal human eye is the maximum distance up to which the eye can see objects clearly.
For a normal human eye,this distance is considered to be infinity.

(B) The nearest point up to which an eye can see objects clearly without any strain is called the near point of the eye.
For a normal human eye,this distance is approximately $25 \ cm$.

(B) For a normal human eye of an adult,the near point is the minimum distance at which an object can be seen clearly without strain.
This distance is approximately $25 \ cm$ from the eye.

(N/A) The minimum distance at which an object can be placed so that the human eye can see it clearly without any strain is known as the least distance of distinct vision. For a normal adult eye,this distance is approximately $25 \ cm$.

(C) The least distance of distinct vision is the minimum distance at which an object can be seen clearly without any strain on the eye.
For a normal human eye,this distance is standardly accepted as $25 \, cm$.

(N/A) No,this is not a contradiction. The lens of the eye converges the divergent beam of light (which appears to come from the virtual image) onto the retina. This process forms a real image of the virtual object on the retina,allowing us to perceive it.

(D) $(i)$ Myopia: Also known as near-sightedness,where a person can see nearby objects clearly but distant objects appear blurred.
$(ii)$ Hypermetropia: Also known as far-sightedness,where a person can see distant objects clearly but nearby objects appear blurred.
$(iii)$ Astigmatism: $A$ defect where the cornea or lens has an irregular curvature,causing blurred or distorted vision at all distances.
$(iv)$ Presbyopia: $A$ gradual loss of the eye's ability to focus on nearby objects,typically occurring with aging.

(B) Near-sightedness,also known as $Myopia$,is a common vision defect of the eye.
In this condition,a person can see nearby objects clearly because the image is formed in front of the retina.
However,the person cannot see far-away (distant) objects clearly because the light rays from distant objects converge before reaching the retina.

(A) The other name for near-sightedness is $Myopia$. In this condition,a person can see nearby objects clearly but cannot see distant objects distinctly.

(N/A) One of the primary causes of near-sightedness (myopia) is the excessive curvature of the eye lens,which leads to a decrease in its focal length. This causes the image of a distant object to be formed in front of the retina rather than on it.