7 Lenses Exercise, Maharashtra Board Class 10 Science Solutions Part 1 Chapter 7 Lenses Exercise, 7 Lenses Class 10 question and answers, 7 Lenses class 10 important questions and answers PDF

7 Lenses Exercise Solutions Science 1 – Maharashtra board

Question No. 1]

Match the columns in the following table and explain them:

Answer:

1. Farsightedness:
Hypermetropia or farsightedness is the defect of vision in which a human eye can see distant objects clearly but is unable to see nearby objects clearly.
In this case the image of a nearby object would fall behind the retina instead of on the retina.

Possible reasons for hypermetropia:
(1) Curvature of the cornea and the eye lens decreases. Hence, the converging power of the eye lens becomes less.
(2) The distance between the eye lens and retina decreases (relative to the normal eye) and the focal length of the eye lens becomes very large due to the flattening of the eyeball.

2. Presbyopia:
Presbyopia is the defect of vision in which aged people find it difficult to see nearby objects
comfortably and clearly without spectacles.

Reason of presbyopia: The power of accommodation of eye usually decreases with ageing. The muscles near the eye lens lose their ability to change the focal length of the lens.

Therefore, the near point of the eye lens shifts rarther from the eye, This defect is corrected using a. convex lens of appropriate power. The lens converges light rays before they fall on the eye lens such that the action of the eye lens forms the image on the retina.

3. Nearsightedness:
Myopia or nearsightedness is the defect of vision in which a human eye can see nearby objects distinctly but is unable to see distant objects clearly as they appear indistinct.
In this case the image of a distant object is formed in front of the retina instead of on the retina.

Possible reasons for myopia:
(1) The curvature of the cornea and the eye lens increases. The muscles near the lens cannot relax so that the converging power of the lens remains large. (2) The distance between the eye lens and the retina increases as the eyeball elongates.

Myopia is corrected using a suitable concave lens. Light rays are diverged by the concave lens before they strike the eye lens. A concave lens of proper focal length is chosen to produce the required divergence. Hence, after the converging action of the eye lens, the image is formed on the retina.

7 Lenses Exercise

Question No.2]

Draw a figure explaining various terms related to a lens.

Answer:

(1) Centre of curvature (C): 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.

(2) Radii of curvature (R₁, R₂): The radii of the spheres whose parts form surfaces of a lens are called the radii of curvature of the lens.

(3) Principal axis: The imaginary straight line passing through the two centres of curvature of a lens is called the principal axis of the lens.

(4) Optical centre (O): 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.

(5) 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.
[Note: In this chapter, the terms focus and the principal focus are used in the same sense.]

(6) Focal length (f): The distance between the optical centre and the principal focus of a lens is called the focal length (f) of the lens.

C₁, C₂: Centres of curvature, R₁, R₂: Radii of curvature, O: Optical centre.
The cross sections of convex and concave lenses are shown in parts (a) and (b) of Fig. 7.4. The surface marked as 1 is part of sphere S₁ while the surface marked as 2 is part of sphere S₂.

P₁, P₂, P₃: Incident rays of light,
Q₁, Q₂, Q₃: Refracted rays of light, O: Optical centre

F₁, F₂: Principal foci of the lens, f: Focal length of the lens

7 Lenses Exercise

Question No.3]

At which position will you keep an object in front of a convex lens so as to get a real image of the same size as the object? Draw a figure.

Answer:
At 2F₁.

7 Lenses Exercise

Question 4.

Give scientific reasons:
a. A simple microscope is used for watch repairs.

Answer:
(1) 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.

(2) 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, watch repairers use a magnifying glass (a simple microscope) while repairing the watches.

b. One can sense colours only in bright light.

Answer:
(1) The retina in the eye is made of many light sensitive cells. The rod-shaped cells respond to the intensity of light while the cone-shaped cells j respond to various colours.
(2) The cone-shaped cells do not respond to faint light. They function only in bright light. Hence, one can sense colours only in bright, light.

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

Answer:
(1) When we try to see a nearby object, the eye lens becomes more rounded and its focal length decreases. Then a clear image of the object is formed on the retina of the eye.
(2) The focal length of the eye lens cannot be decreased beyond some limit. Therefore we cannot clearly see an object kept at a distance less than 25 cm from the eye.

7 Lenses Exercise

Question No.5]

Explain the working of an astronomical telescope using refraction of light.

Answer:
Construction of a refracting telescope: It consists of two convex lenses called the objective lens (directed towards the object) and the eyepiece (directed towards the eye). The focal length and diameter of the objective lens are respectively greater than the focal length and diameter of the eyepiece. The objective lens is fitted at one end of a long metal tube.

A metal tube of smaller diameter is fitted in this metal tube and the eyepiece is fitted at the outer end of the smaller tube. With the help of a screw it is possible to change the distance between the eyepiece and the objective lens by sliding the tube fitted with the eyepiece. The principal axes of the objective lens and the eyepiece are along the same line. A telescope is usually mounted on a stand.

Working: When the objective lens is pointed towards the distant object to be observed, the rays of light from the distant object, which are almost parallel to each other, pass through the objective lens. The objective lens collects maximum amount of light as it is large in size. It forms a real, inverted and diminished image in the focal plane of the objective lens. Now, the position of the eyepiece is adjusted such that this image falls just within the focal length of the eyepiece and serves as the object for the eyepiece which works as a simple microscope.

The final image is highly magnified, virtual, on the same side as that of the object and inverted with respect to the original object. The final image can be observed by keeping the eye close to the eyepiece. If the image formed by the objective lens lies in the focal plane of the eyepiece, the final image is formed at infinity.

7 Lenses Exercise

Question 6.

Distinguish between the following:
a. Farsightedness (Hypermetropia) and Nearsightedness (Myopia).

Answer:
Farsightedness:

1. In hypermetropia, a human eye can see distant distinctly but is unable to see nearby objects clearly.
2. In this case, the image of a nearby object would be formed behind the retina.
3. This defect can be corrected using a convex lens of appropriate power.

Nearsightedness:

1. In myopia, a human eye can see near objects distinctly, but is unable to see distant objects clearly.
2. In this case, the image of a distant object is formed in front of the retina.
3. This defect can be corrected using a concave lens of appropriate power.

b. Concave lens and Convex lens.

Answer:
Concave lens:

1. A concave lens has its surfaces curved inwards.
2. It is thicker at the edges than in the middle.
3. It can form only a virtual image.
4. It can form only a diminished image.

Convex lens:

1. A convex lens has its surfaces puffed up outwards.
2. It is thicker in the middle than at the edges.
3. It can form a real image as well as a % virtual image.
4. It can form a magnified, diminished or the same sized image (relative to the object) depending on the position of the object.

7 Lenses Exercise

Question No.7]

What is the function of the iris and the muscles connected to the lens in the human eye?

Answer:

When the incident light is very bright, the muscles of the iris stretch to reduce the size of the pupil. When the incident light is dim, the muscles of the iris relax to increase the size of the pupil. Thus, the iris controls the size of the pupil and thereby regulates the amount of light entering the eye.

When a distant object is to be observed, the ciliary muscles relax so that the eye lens becomes flat. This increases the focal length of the lens. Therefore, a sharp image of the distant object is formed on the retina.
Thus, we can see a distant object clearly.

7 Lenses Exercise

Question No.8]

Solve the following examples:

i. Doctor has prescribed a lens having I power + 1.5 D. What will be the focal length of the lens? What is the type of the lens and what must be the defect of vision?

Solution:
Data: P = + 1.5 D, f = ?
Focal length of the lens, f = 1/P

=1/1.5D
= 10/15 m

= 0.6667 m

= 0.67 m
P is positive. This shows that the lens is convex. The defect of vision is farsightedness (hypermetropia).

ii. 5 cm high object is placed at a distance of 25 cm from a converging lens of focal length of 10 cm. Determine the position, size and type of the image.

Solution:
Data: Converging lens, f = 10 cm
u = – 25 cm, h₁ = 5 cm, v = ?, h₂ = ?

The height of the image = -3.3 cm (inverted image ∴ minus sign).
(iii) The image is real, inverted and smaller than the object.

iii. Three lenses having powers 2, 2.5 and 1.7 D are kept touching in a row. What is the total power of the lens combination?

Solution:
Data: P₁ = 2 D, P₂ = 2.5 D, P₃ = 1.7 D, P = ?
Total power of the lens combination,
P = P₁ +P₂ + P₃
= 2 D + 2.5 D + 1.7 D
= 6.2 D.

iv. An object kept 60 cm from a lens gives a virtual image 20 cm in front of the lens. What is the focal length of the lens? Is it a converging lens or diverging lens?

Solution:
Data: u = -60 cm, v = -20 cm, f = ?

∴ The focal length of the lens, f = – 30 cm. As f is negative, it is a diverging lens.

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