Solutions-Class 10-Science & Technology-1-Chapter-7-Lenses-Maharashtra Board

Farsightedness: This defect is also known as Hypermetropia. In this defect the human eye can see distant objects clearly but cannot see nearby objects distinctly.

It is caused due to

1. reduction in the curvature of the lens
2. decrease in the size of the eyeball

Since a convex lens has the ability to converge incoming rays, it can be used to correct this defect of vision by using a convex lens with proper focal length. This lens converges the incident rays before they reach the lens. The lens then converges them to form the image on the retina.

Presbyopia: This is a common defect of vision, which generally occurs at old age. A person suffering from this type of defect of vision cannot see nearby objects clearly and distinctively without spectacles.

Presbyopia is caused by the

1. weakening of the ciliary muscles
2. reduction in the flexibility of the eye lens

It can be corrected using bifocal lens. In such lenses, the upper part is concave lens and corrects nearsightedness while the lower part is a convex lens which corrects the farsightedness.

Nearsightedness: This defect is also known as Myopia. It is a defect of vision in which a person clearly sees all the nearby objects, but is unable to see the distant objects comfortably and his eye is known as a myopic eye. A myopic eye has its far point nearer than infinity. It forms the image of a distant object in front of its retina as shown in the figure.

It is caused by

1. increase in curvature of the lens
2. increase in length of the eyeball

Since a concave lens has an ability to diverge incoming rays, it is used to correct this defect of vision. The image is allowed to form at the retina by using a concave lens of suitable power as shown in the given figure.

Various Terms related with Lenses:

1) Optical centre : The point inside a lens on the principal axis, through which light rays pass without changing their path is called the optical centre (O) of the lens.

2) Centre of curvature : The centres of the spheres whose parts form the surfaces of a lens are called the centres of curvature of the lens. A lens has two centres of curvature C₁ and C₂ for its two spherical surfaces.

3) Radius of curvature : The distance of the optical centre from either of the centre of curvatures is termed as the radius of curvature.

4) Principal axis : The imaginary straight line joining the two centers of curvature and the optical centre (O) is called the principal axis of the lens.

5) Focus (Principal focus) (F) : When light rays parallel to the principal axis are incident on a convex lens, they converge at a point on the principal axis. This point is called the principal focus (F) of the convex lens.

Light rays travelling parallel to the principal axis of a concave lens diverge after refraction in such a way that they appear to be coming out of a point on the principal axis. This point is called the principal focus of the concave lens. A lens has two principal foci F₁ and F₂.

6) Focal length : The distance between the focus (F₁ or F₂) and the optical centre (O) is known as the focal length of the lens.

When an object is placed at the centre of curvature 2F₁ of a convex lens, we will get a real image 2F₂ of the same size as the object.

• When an object is placed within the focal length of a magnifying glass or simple microscope (convex lens), its larger and erect image is obtained on the same side of the lens as that of the object.
• By adjusting the distance between the object and the lens, the image can be obtained at the minimum distance of distinct vision.
• Thus, a watch repairer can see the minute parts of a watch more clearly with the aid of a magnifying glass (a simple microscope) than with the naked eye, without any stress on the eye.

Hence, simple microscopes are used by watch makers to see the small parts and screws of the watch while repairing it.

The cells present on the retina and responsible for colour vision are known as cone cells. These cells do not respond to faint light and function only under bright light. Thus, we are able sense only in bright light.

We cannot clearly see an object kept at a distance less than 25 cm from the eye.

• This is because when we try to see a nearby object, the eye lenses becomes more rounded, and its focal length decreases.
• Then clear image of object form on retina of eye.
• The focal length of eye are unable to contract beyond certain limit.
• Therefore We cannot clearly see an object kept at a distance less than 25 cm from the eye

The astronomical telescope consists of two lenses: objective and eyepiece. Objective has larger focal length and diameter to accommodate maximum amount of light coming from the far away (astronomical) objects.

Working :

• A parallel beam of rays from an astronomical object is made to fall on the objective lens of the telescope.
• It forms a real, inverted and diminished image A'B' of the object.
• The eyepiece is so adjusted that A'B' lies just at the focus of the eye piece.
• Therefore, a highly magnified image of the object is formed at infinity.

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Function of Iris: The iris is a muscular diaphragm that controls the size of the pupil, which, in turn, controls the amount of light entering the eye. It also gives colour to the eye.

Function of ciliary muscles: The eye lens is held in position by the ciliary muscles. The focal length of the eye lens is adjusted by the expansion and contraction of the ciliary muscles.

Given: Power of lens, P = +1.5 D,  f=?
Now, focal length of lens, f=1/P =1/(+1.5D) =10/15 m =+0.67 m
P is positive, This shows the lens is convex. Thus, the defect of vision is farsightedness or hypermetropia.

Given: Height of object (h₁) = 5 cm, Object distance (u) = −25 cm
Focal length of the lens (f) = 10 cm
Using lens formula,

\(\frac{1}{f} = \frac{1}{v} - \frac{1}{u} ∴ \frac{1}{v} = \frac{1}{f} + \frac{1}{u}\)

\(∴\frac{1}{v} = \frac{1}{10} + \frac{1}{-25} = \frac{1}{10} - \frac{1}{25}\)

\(=\frac{3}{50} cm\)

\(∴ \text{Image distance v} = \frac{50}{3}cm = 16.67 cm\)

Thus, the image is formed 16.7 cm cm right of the lens. 
Now, we know

\(\frac{h_2}{h_1} = \frac{v}{u}  ∴ h_2 = \frac{v}{u} h_1\)

∴ \(h_2 = \frac{v}{u} h_1\)

∴ \(h_2 = \frac{50/3}{-25} ×5 cm\)

= \(-3.33 cm\)

Thus, the size of the image is 3.3 cm. Negative sign shows that the image formed is real and inverted. Hence, the image formed is real and inverted and smaller than the object.

Given: P₁ = 2 D, P₂ = 2.5 D, P₃ = 1.7 D

Let the total power of the lens combination be P. Thus,

P=P₁+P₂+P₃ = 2+2.5+1.7

=6.2 D

Given:
Object distance, u = −60 cm
Image distance, v = −20 cm

f= ?
Using lens formula,

\(\frac{1}{f} = \frac{1}{v} - \frac{1}{u} = \frac{1}{-20} - \frac{1}{-60}\)

\(= -\frac{3-1}{60cm}\)

\(= -\frac{1}{30cm}\)

\(f=-30cm\)

Since, the focal length is negative, the lens is a diverging lens.