Rays of light parallel to the principal axis, after reflection from a concave mirror, meet at a point on the principal axis. This point is called the principal focus of the concave mirror.
2The radius of curvature of a spherical mirror is 20 cm. What is its focal length?
f = R/2 = 20/2 = 10 cm
3Name a mirror that can give an erect and enlarged image of an object.
Answer: Concave mirror
It gives an erect and enlarged image when the object is placed between its pole and principal focus.
4Why do we prefer a convex mirror as a rear-view mirror in vehicles?
A convex mirror always gives an erect image, though diminished. It also has a wider field of view because it is curved outwards. So the driver can see a much larger area of traffic behind.
Intext Questions (Page 145)
1Find the focal length of a convex mirror whose radius of curvature is 32 cm.
R = +32 cm (convex mirror) f = R/2 = +32/2 = +16 cm
2A concave mirror produces three times magnified (enlarged) real image of an object placed at 10 cm in front of it. Where is the image located?
A real image is inverted. So m = −3. Also u = −10 cm.
m = −v/u −3 = −v/(−10) v = −30 cm
The image is located 30 cm in front of the mirror, on the same side as the object.
Intext Questions (Page 150)
1A ray of light travelling in air enters obliquely into water. Does the light ray bend towards the normal or away from the normal? Why?
The ray bends towards the normal. Water is optically denser than air. Light slows down when it enters water. So it bends towards the normal.
2Light enters from air to glass having refractive index 1.50. What is the speed of light in the glass? The speed of light in vacuum is 3 × 108 m s-1.
n = c/v, so v = c/n v = (3 × 108)/1.50 = 2 × 108 m s-1
3Find out, from Table 9.3, the medium having highest optical density. Also find the medium with lowest optical density.
Highest optical density: Diamond (n = 2.42)
Lowest optical density: Air (n = 1.0003)
4You are given kerosene, turpentine and water. In which of these does the light travel fastest? Use the information given in Table 9.3.
Refractive indices: kerosene 1.44, turpentine 1.47, water 1.33.
Light travels fastest in the medium with the lowest refractive index. So light travels fastest in water.
5The refractive index of diamond is 2.42. What is the meaning of this statement?
It means that the speed of light in air (vacuum) is 2.42 times the speed of light in diamond. In other words, light travels 2.42 times slower in diamond than in vacuum.
Intext Questions (Page 158)
1Define 1 dioptre of power of a lens.
1 dioptre is the power of a lens whose focal length is 1 metre. 1 D = 1 m-1.
2A convex lens forms a real and inverted image of a needle at a distance of 50 cm from it. Where is the needle placed in front of the convex lens if the image is equal to the size of the object? Also, find the power of the lens.
A convex lens gives a real image of the same size when the object is at 2F. Then the image is also at 2F.
v = +50 cm = 2f, so f = +25 cm = +0.25 m u = −50 cm
The needle is placed 50 cm in front of the lens.
P = 1/f = 1/0.25 = +4 D
3Find the power of a concave lens of focal length 2 m.
f = −2 m (concave lens) P = 1/f = 1/(−2) = −0.5 D
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Exercises
1Which one of the following materials cannot be used to make a lens? (a) Water (b) Glass (c) Plastic (d) Clay
Answer: (d) Clay
Clay is not transparent.
2The image formed by a concave mirror is observed to be virtual, erect and larger than the object. Where should be the position of the object? (a) Between the principal focus and the centre of curvature (b) At the centre of curvature (c) Beyond the centre of curvature (d) Between the pole of the mirror and its principal focus.
Answer: (d) Between the pole of the mirror and its principal focus.
3Where should an object be placed in front of a convex lens to get a real image of the size of the object? (a) At the principal focus of the lens (b) At twice the focal length (c) At infinity (d) Between the optical centre of the lens and its principal focus.
Answer: (b) At twice the focal length
4A spherical mirror and a thin spherical lens have each a focal length of −15 cm. The mirror and the lens are likely to be (a) both concave. (b) both convex. (c) the mirror is concave and the lens is convex. (d) the mirror is convex, but the lens is concave.
Answer: (a) both concave.
By sign convention, a concave mirror and a concave lens have negative focal length.
5No matter how far you stand from a mirror, your image appears erect. The mirror is likely to be (a) only plane. (b) only concave. (c) only convex. (d) either plane or convex.
Answer: (d) either plane or convex.
6Which of the following lenses would you prefer to use while reading small letters found in a dictionary? (a) A convex lens of focal length 50 cm. (b) A concave lens of focal length 50 cm. (c) A convex lens of focal length 5 cm. (d) A concave lens of focal length 5 cm.
Answer: (c) A convex lens of focal length 5 cm.
A convex lens of short focal length gives a large magnified image.
7We wish to obtain an erect image of an object, using a concave mirror of focal length 15 cm. What should be the range of distance of the object from the mirror? What is the nature of the image? Is the image larger or smaller than the object? Draw a ray diagram to show the image formation in this case.
The object should be placed between the pole and the principal focus. So the distance should be more than 0 cm and less than 15 cm from the mirror.
The image is virtual and erect. It is formed behind the mirror. The image is larger than the object.
Concave mirror: object between P and F gives a virtual, erect and enlarged image behind the mirror.
8Name the type of mirror used in the following situations. (a) Headlights of a car. (b) Side/rear-view mirror of a vehicle. (c) Solar furnace. Support your answer with reason.
(a) Concave mirror. The bulb is placed at its focus. So the mirror gives a powerful parallel beam of light.
(b) Convex mirror. It always gives an erect image and has a wider field of view.
(c) Concave mirror. It collects sunlight and focuses it at one point. This produces a lot of heat.
9One-half of a convex lens is covered with a black paper. Will this lens produce a complete image of the object? Verify your answer experimentally. Explain your observations.
Yes, the lens will produce a complete image. But the image will be less bright.
Experiment: Place a lighted candle in front of a convex lens. Get its sharp image on a screen on the other side. Now cover the lower half of the lens with black paper. The image on the screen is still complete. Only its brightness decreases.
Explanation: Rays from every point of the object pass through the uncovered half of the lens. These rays still meet to form the complete image. Since fewer rays reach the screen, the image is less bright.
10An object 5 cm in length is held 25 cm away from a converging lens of focal length 10 cm. Draw the ray diagram and find the position, size and the nature of the image formed.
Given: h = +5 cm, u = −25 cm, f = +10 cm
1/v − 1/u = 1/f 1/v = 1/f + 1/u = 1/10 − 1/25 = (5 − 2)/50 = 3/50 v = 50/3 = +16.7 cm m = v/u = (50/3)/(−25) = −2/3 = −0.67 h′ = m × h = −0.67 × 5 = −3.3 cm
The image is formed 16.7 cm on the other side of the lens. It is 3.3 cm in size. It is real, inverted and diminished.
Convex lens: object beyond 2F1 gives a real, inverted and diminished image between F2 and 2F2.
11A concave lens of focal length 15 cm forms an image 10 cm from the lens. How far is the object placed from the lens? Draw the ray diagram.
Given: f = −15 cm, v = −10 cm (a concave lens forms a virtual image on the same side as the object)
The image is formed 6 cm behind the mirror. It is virtual, erect and diminished.
13The magnification produced by a plane mirror is +1. What does this mean?
m = +1 means the image is of the same size as the object. The positive sign shows that the image is virtual and erect.
14An object 5.0 cm in length is placed at a distance of 20 cm in front of a convex mirror of radius of curvature 30 cm. Find the position of the image, its nature and size.
Given: h = +5 cm, u = −20 cm, R = +30 cm, so f = +15 cm
1/v = 1/f − 1/u = 1/15 + 1/20 = (4 + 3)/60 = 7/60 v = 60/7 = +8.6 cm m = −v/u = −(60/7)/(−20) = 3/7 = +0.43 h′ = m × h = (3/7) × 5 = +2.1 cm
The image is formed 8.6 cm behind the mirror. It is virtual, erect and diminished, with a size of 2.1 cm.
15An object of size 7.0 cm is placed at 27 cm in front of a concave mirror of focal length 18 cm. At what distance from the mirror should a screen be placed, so that a sharp focussed image can be obtained? Find the size and the nature of the image.
Given: h = +7 cm, u = −27 cm, f = −18 cm
1/v = 1/f − 1/u = −1/18 + 1/27 = (−3 + 2)/54 = −1/54 v = −54 cm m = −v/u = −(−54)/(−27) = −2 h′ = m × h = −2 × 7 = −14 cm
The screen should be placed 54 cm in front of the mirror. The image is 14 cm in size. It is real, inverted and enlarged.
16Find the focal length of a lens of power −2.0 D. What type of lens is this?
f = 1/P = 1/(−2.0) = −0.5 m = −50 cm
The focal length is negative. So it is a concave lens.
17A doctor has prescribed a corrective lens of power +1.5 D. Find the focal length of the lens. Is the prescribed lens diverging or converging?
f = 1/P = 1/1.5 = 0.667 m = +66.7 cm
The focal length is positive. So it is a convex lens, which is converging.
