Science Part I (solutions) Solutions for Class 10 Science Chapter 7 - Lenses

Answer:

Farsightedness: This defect is also known as Hypermetropia. It is an eye defect in which a person is unable to see nearby objects clearly but can see the far away objects clearly.

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, as you already have seen in the animation. The ray diagram for the corrective measure for a hypermetropic eye is shown in the given figure.

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.

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. A presbyopic eye has its near point greater than 25 cm and it gradually increases as the eye becomes older.

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.
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Answer:

Terms Associated with Lenses:
Optical centre
Optical centre is a point at the centre of the lens. It always lies inside the lens and not on the surface. It is denoted by ‘O’.

Centre of curvature
It is the centre point of arcs of the two spheres from which the given spherical lens (concave or convex) is made. Since a lens constitutes two spherical surfaces, it has two centers of curvature.

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

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.

Focus
The focus (F) is the point on the principal axis of a lens where all incident parallel rays, after refraction from the lens meet or appear to diverge from. For lenses there are two foci (F₁ and F₂) depending on the direction of incident rays.         

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

Answer:

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

Answer:

a. Simple microscope has convex lens which has the ability to produce 20 times larger as well as erect image of an object. This means the magnifying power of the microscope is very high. Thus, simple microscopes are used by watch makers to see the small parts and screws of the watch while repairing it.

b. The cells present on the retina and responsible for colour vision are known as cone cells. These cells become active only under bright light and remain inactive under dark. Thus, we are able sense only in bright light.

c. We cannot clearly see an object kept at a distance less than 25 cm from the eye. This is because ciliary muscles of our eyes are unable to contract beyond certain limit. If the object is placed at a distance less than 25 cm from the eye, then the object appears blurred because light rays coming from the object meet behind the retina.

Answer:

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. 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. The same has been shown in the figure below.

Answer:

a.

b.

Answer:

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.

Answer:

i. Given:
Power of lens, P = +1.5 D
Now, focal length of lens, $f=\frac{1}{P}=+\frac{1}{1.5}=+0.67 \mathrm{m}$. 
Since, the focal length is positive, the lens prescribed for correction is convex lens. Thus, the defect of vision is farsightedness or hypermetropia.

ii. Given:
Height of object, h_o = 5 cm
Object distance, u = $-$25 cm
Since the lens is converging, thus it is a convex lens.
Focal length of the lens, f = 10 cm
Using lens formula,
$$\begin{gathered} \frac{1}{v}-\frac{1}{u}=\frac{1}{f} \\ \Rightarrow \frac{{\displaystyle 1}}{{\displaystyle v}}=\frac{1}{10}+\frac{1}{-25}=\frac{3}{50} \\ \Rightarrow v=\frac{50}{3} = 16.7 \mathrm{cm} \end{gathered}$$
Thus, the image is formed $16.7 \mathrm{cm}$ right of the lens. 
Now, we know
$$\begin{gathered} \frac{v}{u}=\frac{h_{\mathrm{i}}}{h_{\mathrm{o}}} \\ \Rightarrow h_{\mathrm{i}}=\frac{50}{3\times -25}\times 5=\frac{10}{3}=-3.3 \mathrm{cm} \end{gathered}$$
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 diminished.

iii. Given:
P₁ = 2 D, P₂ = 2.5 D, P₃ = 1.7 D
Let the total power of the lens combination be P. Thus,
$P=P_1+P_2+P_3=2+2.5+1.7=6.6 \mathrm{D}$

iv. Given:
Object distance, u = $-$60 cm
Image distance, v = $-$20 cm
Using lens formula,
$$\begin{gathered} \frac{1}{v}-\frac{1}{u}=\frac{1}{f} \\ \Rightarrow \frac{1}{f}=\frac{{\displaystyle 1}}{{\displaystyle -20}}-\frac{{\displaystyle 1}}{{\displaystyle -60}}=-\frac{1}{30} \\ \Rightarrow f=-30 \mathrm{cm} \end{gathered}$$
Since, the focal length is negative, the lens is a diverging lens or a concave lens.