Chapter 7 - Lenses
1. Match the columns in the following table and explain them.
| Column 1 | Column 2 | Column 3 |
|---|---|---|
Farsightedness | Nearby object can be seen clearly | Bifocal lens |
Presbyopia | Far away object can be seen clearly | Concave lens |
Nearsightedness | Problem of old age | Convex lens |
Ans:
| Column 1 | Column 2 | Column 3 |
|---|---|---|
Farsightedness | Nearby object can be seen clearly | Bifocal lens |
Presbyopia | Far away object can be seen clearly | Concave lens |
Nearsightedness | Problem of old age | Convex lens |
(i) Explanation: In farsightedness, the human eye can see faraway objects clearly but cannot see nearby objects distinctly. To correct this defect, a convex lens with the proper focal length is used.
(ii) Explanation: The ability of the muscles near the eye lens to change the focal length of the lens decreases with age. This defect is known as presbyopia. In this defect, sometimes people suffer from both nearsightedness and farsightedness. Therefore, bifocal lenses are required to correct this defect.
(iii) Explanation: In nearsightedness, the human eye can see nearby objects clearly but cannot see far away objects distinctly. To correct this defect, a concave lens with proper focal length is used.
2. Draw a figure explaining various terms related to a lens.
Ans:
O : Optical centre
C\(_{1}\), C\(_{2}\) : Centres of curvature
R\(_{1}\), R\(_{2}\) : Radii of curvature
f : Focal length
F\(_{1}\), F\(_{2}\) : Principal focii
(i) Principal axis: The imaginary line passing through both centres of curvature is called the principal axis of the lens.
(ii) Optical centre: The point inside a lens on the principal axis through which light rays pass without changing their paths is called the optical centre of a lens.
(iii) Centre of curvature: The centre of the sphere whose part forms the surface of the lens is called centre of curvature of the lens.
(iv) Radius of curvature: The radius of the sphere whose part forms the surface of the lens is called the radius of curvature of the lens.
(v) Principal focus:
a. When light rays parallel to the principal axis are incident on a convex lens, they converge to a point on the principal axis. This point is called the principal focus of the lens.
b. 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.
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.
Ans: To get a real image of the same size as the object, the object should be placed at a distance equal to twice the focal length of the lens, i.e., at a point 2F from the lens, as shown in the ray diagram.
4. Give scientific reasons:
a. Simple microscope is used for watch repairs.
Ans:
(i) When an object is placed within the focal length of the lens, one gets a magnified and erect image of the object.
(ii) Thus, the watch repairer can see the minute parts of the watch more clearly with the help of a simple microscope than with the naked eye without any strain on the eye.
(iii) A magnification of about 20 times is obtained by a simple microscope.
(iv) Hence, a simple microscope is used for watch repairs.
b. One can sense colours only in bright light.
Ans:
(i) The retina in our eyes is made of different cells, viz., rod-shaped and cone-shaped.
(ii) The rod-shaped cells respond to the intensity of the light and send the information to the brain. While cone-shaped cells send information about different colors of light to the brain.
(iii) Rod-like cells are sensitive to faint light, whereas conical cells do not respond to faint light.
(iv) Hence, one can sense colors only in bright light.
c. We can not clearly see an object kept at a distance less than 25 cm from the eye.
Ans:
(i) The muscles attached to the eye lens (ciliary muscles) help in fine adjustments of the focal length of the lens.
(ii) The capacity of these muscles to contract or relax to adjust the focal length (i.e., power of accommodation) has a limit.
(iii) The minimum distance of an object from a normal eye for which the eye lens can decrease its focal length to the least possible value is 25 cm.
(iv) Hence, we cannot clearly see an object kept at a distance less than 25 cm from the eye.
5. Explain the working of an astronomical telescope using refraction of light.
Ans: Astronomical telescopes are of two types:
(i) Refracting telescopes
(ii) Reflecting telescopes
Refracting telescopes:
(i) The telescope that works on the refraction of light is a refracting telescope.
(ii) It consists of two convex lenses called the objective (O) and eyepiece (E). The objective has a large diameter and a longer focal length. The eyepiece has a shorter focal length and a small diameter. These two lenses are mounted at the ends of a hollow metallic tube so that the distance between them can be adjusted.
(iii) Light rays from a distant object are incident on the objective and form an image AB.
(iv) This image AB acts as an object for the eyepiece; it is adjusted in such a way that the final image is deformed at infinity.
(v) To change the magnification, eyepieces of different focal lengths can be used by keeping the objective the same.
6. Distinguish between:
a. Farsightedness and Nearsightedness
Ans:
| Farsightedness | Nearsightedness |
|---|---|
(i) This problem arises due to slight flattening of the eyeball. | (i) This problem arises due to slight elongation of the eyeball. |
(ii) The curvature of the cornea and eye lens decreases so that the converging power becomes less. | (ii) The curvature of the cornea and eye lens increases so that the converging power becomes more. |
(iii) The distance between the eye lens and the retina decreases. | (iii) The distance between the eye lens and the retina increases. |
(iv) The near point of the eye shifts farther away from the eye. | (iv) The far point of the eye shifts closer to the eye. |
(v) A person suffering from this defect can see distant objects clearly but is unable to see nearby objects. | (v) A person suffering from this defect can see nearby objects clearly but is unable to see distant objects. |
(vi) It can be corrected by using spectacles having convex lenses of suitable focal length. | (vi) It can be corrected by using spectacles having concave lenses of suitable focal length. |
b. Concave lens and Convex Lens
Ans:
| Concave lens | Convex lens |
|---|---|
(i) It is thinner at the centre than at the edges. | (i) It is thicker at the centre than at the edges. |
(ii) It is called diverging lens. | (ii) It is called converging lens. |
(iii) The nature of the image formed is always diminished and erect. The image is formed between focus and optical centre (regardless of the object’s position). | (iii) The nature of the image formed depends upon the position of the object. |
(iv) It forms only virtual images. | (iv) It can form real and virtual images. |
7. What is the function of iris and the muscles connected to the lens in human eye?
Ans:
(i) Function of the iris: The iris controls the amount of light entering the eye by contracting and widening the pupil.
(ii) Function of the muscles: The muscles hold the lens such that they can provide fine adjustment of focal length of eye lens by contracting and relaxing.
8. Solve the following examples.
(i) Doctor has prescribed a lens having 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?
Given:
Power (P) = + 1.5 D
To find:
Focal length (f)
Solution:
We know that,
P = \(\large \frac {1}{f}\)
∴ f = \(\large \frac {1}{1.5}\)
∴ f = + 0.67m
The prescribed lens is convex lens as the focal length is positive. Convex lens has positive focal length and it is used for the correction of farsightedness or hypermetropia. Thus, the defect in vision is hypermetropia (farsightedness).
Ans: The prescribed lens is convex lens of focal length +0.67 m and the defect in vision is hypermetropia (farsightedness).
(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.
Given:
Focal length (f) = 10cm
Object distance (u) = – 25 cm
height of the object (\(h_{1}\)) = 5cm
To find:
Image distance (v)
Height of the image (\(h_{2}\))
Solution:
We know that,
\(\large \frac {1}{f}\) = \(\large \frac {1}{v}\) – \(\large \frac {1}{u}\)
∴ \(\large \frac {1}{10}\) = \(\large \frac {1}{v}\) – \(\large \frac {1}{-25}\)
∴ \(\large \frac {1}{10}\) = \(\large \frac {1}{v}\) + \(\large \frac {1}{25}\)
∴ \(\large \frac {1}{v}\) = \(\large \frac {1}{10}\) – \(\large \frac {1}{25}\)
∴ \(\large \frac {1}{v}\) = \(\large \frac {5\,–\,2}{50}\)
∴ \(\large \frac {1}{v}\) = \(\large \frac {-3}{50}\)
∴ v = \(\large \frac {50}{-3}\)
∴ v = 16.7 cm
As the image distance is positive, the image formed is real.
Now,
\(\large \frac {h_{2}}{h_{1}}\) = \(\large \frac {v}{u}\)
∴ \(\large \frac {h_{2}}{5}\) = \(\large \frac {16.7}{-25}\)
∴ \(h_{2}\) = \(\large \frac {-16.7\,×\,5}{-25}\)
∴ \(h_{2}\) = \(\large \frac {-16.7}{5}\)
∴ \(h_{2}\) = – 3.3 cm
The negative sign indicates that the image formed is inverted.
Ans: The real, inverted image of height 3.3 cm is formed at 16.7 cm.
(iii) Three lenses having power 2, 2.5 and 1.7 D are kept touching in a row. What is the total power of the lens combination?
Given:
P\(_{1}\) = 2 D
P\(_{2}\) = 2.5 D
P\(_{3}\) = 1.7 D
To find:
Power of combination (P)
Solution:
We know that,
P = P_{1} + P_{2} + P_{3}
∴ P = 2 + 2.5 + 1.7
∴ P = 6.2 D
Ans: The total power of the combination is 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?
Given:
Object distance (u) = – 60cm
image distance (v) = – 20 cm
To find:
Focal length (f)
Type of lens
Solution:
We know that,
\(\large \frac {1}{f}\) = \(\large \frac {1}{v}\) – \(\large \frac {1}{u}\)
∴ \(\large \frac {1}{f}\) = \(\large \frac {1}{-20}\) – \(\large \frac {1}{-60}\)
∴ \(\large \frac {1}{f}\) = \(\large \frac {3}{-60}\) – \(\large \frac {1}{-60}\)
∴ \(\large \frac {1}{f}\) = \(\large \frac {-3\,+\,1}{60}\)
∴ \(\large \frac {1}{f}\) = \(\large \frac {-2}{60}\)
∴ \(\large \frac {1}{f}\) = \(\large \frac {-1}{30}\)
∴ f = – 30 cm
Since, the focal length is negative, the lens is diverging in nature i.e., concave lens.
Ans: The focal length of the lens is -30 cm and the lens is diverging in nature i.e., concave.