
01. Chemical Reactions
8
Lecture1.1

Lecture1.2

Lecture1.3

Lecture1.4

Lecture1.5

Lecture1.6

Lecture1.7

Lecture1.8


02. Acids, Bases and Salts
12
Lecture2.1

Lecture2.2

Lecture2.3

Lecture2.4

Lecture2.5

Lecture2.6

Lecture2.7

Lecture2.8

Lecture2.9

Lecture2.10

Lecture2.11

Lecture2.12


03. Metals and Non  metals
11
Lecture3.1

Lecture3.2

Lecture3.3

Lecture3.4

Lecture3.5

Lecture3.6

Lecture3.7

Lecture3.8

Lecture3.9

Lecture3.10

Lecture3.11


04. Periodic Classification of Elements
6
Lecture4.1

Lecture4.2

Lecture4.3

Lecture4.4

Lecture4.5

Lecture4.6


05. Life Processes  1
9
Lecture5.1

Lecture5.2

Lecture5.3

Lecture5.4

Lecture5.5

Lecture5.6

Lecture5.7

Lecture5.8

Lecture5.9


06. Life Processes  2
6
Lecture6.1

Lecture6.2

Lecture6.3

Lecture6.4

Lecture6.5

Lecture6.6


07. Control and Coordination
9
Lecture7.1

Lecture7.2

Lecture7.3

Lecture7.4

Lecture7.5

Lecture7.6

Lecture7.7

Lecture7.8

Lecture7.9


08. How do Organisms Reproduce
9
Lecture8.1

Lecture8.2

Lecture8.3

Lecture8.4

Lecture8.5

Lecture8.6

Lecture8.7

Lecture8.8

Lecture8.9


09. Heredity and Evolution
9
Lecture9.1

Lecture9.2

Lecture9.3

Lecture9.4

Lecture9.5

Lecture9.6

Lecture9.7

Lecture9.8

Lecture9.9


10. Light (Part 1) : Reflection
8
Lecture10.1

Lecture10.2

Lecture10.3

Lecture10.4

Lecture10.5

Lecture10.6

Lecture10.7

Lecture10.8


11. Light (Part 2) : Refraction
6
Lecture11.1

Lecture11.2

Lecture11.3

Lecture11.4

Lecture11.5

Lecture11.6


12. Carbon and Its Compounds
9
Lecture12.1

Lecture12.2

Lecture12.3

Lecture12.4

Lecture12.5

Lecture12.6

Lecture12.7

Lecture12.8

Lecture12.9


13. The Human Eye and The Colorful World
9
Lecture13.1

Lecture13.2

Lecture13.3

Lecture13.4

Lecture13.5

Lecture13.6

Lecture13.7

Lecture13.8

Lecture13.9


14. Electricity
8
Lecture14.1

Lecture14.2

Lecture14.3

Lecture14.4

Lecture14.5

Lecture14.6

Lecture14.7

Lecture14.8


15. Magnetic Effect of Current
10
Lecture15.1

Lecture15.2

Lecture15.3

Lecture15.4

Lecture15.5

Lecture15.6

Lecture15.7

Lecture15.8

Lecture15.9

Lecture15.10


16. Sources of Energy
5
Lecture16.1

Lecture16.2

Lecture16.3

Lecture16.4

Lecture16.5


17. Our Environment
6
Lecture17.1

Lecture17.2

Lecture17.3

Lecture17.4

Lecture17.5

Lecture17.6


18. Management of Natural Resources
4
Lecture18.1

Lecture18.2

Lecture18.3

Lecture18.4

NCERT Solutions – Light Reflection and Refraction
Intext Questions
Q.1 Define the principal focus of a concave mirror.
Sol. When rays parallel and close to the principal axis of a concave mirror meet at a point , this point is called “principal focus”.
Q.2 The radius of a curvature of a spherical mirror is 20 cm. What is its focal length?
Sol. f=R2=202=10cm.
Q.3 Name a mirror that can give an erect and enlarged imge of an object.
Sol. Concave mirror.
Q.4 Why do we prefer a convex mirror as a rearview mirror in vehicles ?
Sol. We prefer a convex mirror as rear view mirror as it gives wider view and erect images.
Page 171
Q.1 Find the focal length of a convex mirror whose radius of curvature is 32 cm.
Sol. f=R2=322=16cm.
Q.2 A 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.
Sol.
m = – 3 ; u = – 10 cm
m=−vu
−3=−v−10
Therefore, v=−30cm
Page 176
Q.1 A ray of light travelling in air enters obliquely into water. Does the light ray bend towards the normal or away from the normal ? Why ?
Sol. The ray bends towards the normal as water is optically denser than air.
Q.2 Light 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.
Sol. ref.index=speedoflightinvacuumspeedoflightinair
Therefore, Speed of light in air =3×1081.5m/s.=2×108m/s.
Q.3 Find out, from Table , the medium having highest optical density. Also find the medium with lowest optical density.
Sol. Highest optical density is of diamond , lowest optical density is of air.
Q.4 You are given kerosene, turpentine and water. In which of these does the light travel fastest ? Use the information given in Table –
Sol. Light travels faster in a medium having low optical density. Among kerosene, turpentine and water, light travels faster in water.
Q.5 The refractive index of diamond is 2.42. What is the meaning of this statement ?
Sol. It means that the speed of light in diamond is 12.42 times the speed of light in vacuum.
Page 184
Q.6 Define 1 dioptre of power of a lens.
Sol. One dioptre is the power of lens whose focal length is 1 m .
Q.7 A 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.
Sol.
v = 50 cm , m = – 1.
Therefore, u=vm=−50cm.
1f=1v−1u
=150−(−150)=125
P=10025=4D.
Exercise
Q.1 Which one of the following materials cannot be used to make a lens?
(a) Water (b) Glass (c) Plastic (d) Clay
Sol. (d) Clay.
Q.2 The 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.
Sol. (d)
Q.3 Where 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.
Sol. (b)
Q.4 A 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.
Sol. (a)
Q.5 No matter how far you stand from a mirror, your image appears erect. The mirror is likely to be
(a) plane
(b) concave
(c) convex
(d) either plane or convex
Sol. (d)
Q.6 Which 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.
Sol. (c)
Q.7 We 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.
Sol. 15 cm – range.
Virtual, Erect, Magnified
Q.8 Name the type of mirror used in the following situations.
(a) Headlights of car.
(b) Side/ rearview mirror of a vehicle
(c) Solar furnace
Support your answer with reason.
Sol. (a) Concave mirror – The bulb placed at focus throws parallel beam of light after reflection.
(b) Convex mirror – To get wider field of view and erect images.
(c) Concave mirror – To converge the rays of sun at its focus.
Q.9 Onehalf 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.
Sol. Complete image of the object will be formed. The intensity of light will be reduced.
(Only rays passing from one half of lens will converge)
Q.10 An 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.
Sol.
Image Position = 16.7 cm.
Image Size = 3.3 cm.
Real, inverted, diminished.
Q.11 A 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.
Sol.
f=−15cm,v=−10cm
1u=1v−1f
=−110−(−115)=−130
Therefore u=−30cm
Q.12 An object is placed at a distance of 10 cm from a convex mirror of focal length 15 cm. Find the position and nature of the image.
Sol.
u = – 10 cm ; f = 15 cm.
1v=1f=−1u
=115−(−110)=16
Therefore v = 6 cm.
Image is behind the mirror, virtual, erect and diminished.
Q.13 The magnification produced by a plane mirror is +1. What does this mean ?
Sol. Magnification +1 means that an erect image is formed, having same size as object.
Q.14 An 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.
Sol.
f=R2=302=15cm.
h = 5 cm ; u = – 20 cm.
1v=1f−1u
=115−(−120)=760
Therefore, v = 8.6 cm.
h′h=−vu
h′=5×60720
= 2.2 cm.
The image is at 8.6 cm behind mirror, size 2.2, erect, virtual and diminished.
Q.15 An 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.
Sol.
h = 7 cm ; u = – 27 ; f = –18 cm.
1v=1f−1u
=−118−(−127)=−154
Therefore, v = – 54 cm.
h′h=vu
Therefore h′=−v×hu
=−7×54−27
=14cm
Screen distance 54 cm, real, inverted image of size 14 cm.
Q.16 Find the focal length of a lens of power – 2.0 D. What type of lens is this ?
Sol. f=1P=−12=0.50m.
Concave lens.
Q.17 A doctor has prescribed a corrective lens of power +1.5D. Find the focal length of the lens. Is the prescribed lens diverging or converging ?
Sol. f=1p=1001.5cm=0.67m.
Lens is converging.
Exemplar
Short Answer Type Questions : –
Q.1 Identify the device used as a spherical mirror or lens in following cases, when the image formed is virtual and erect in each case.
(a) Object is placed between device and its focus, image formed is enlarged and behind it.
(b) Object is placed between the focus and device, image formed is enlarged and on the same side as that of the object.
(c) Object is placed between infinity and device, image formed is diminished and between focus and optical centre on the same side as that of the object.
(d) Object is placed between infinity and device, image formed is diminished and between pole and focus, behind it.
Sol.
(a) Concave mirror
(b) Convex lens
(c) Concave lens
(d) Convex mirror
Q.2 Why does a light ray incident on a rectangular glass slab immersed in any medium emerges parallel to itself? Explain using a diagram.
[Delhi 2013]
In a glass slab the opposite faces are parallel, so the angle of refraction by first interface, becomes
angle of incidence at the second interface i.e. ∠r1=∠r2
By the principal of reversibility, the emergent ray bends such that ∠e=∠i
Thus the incident ray and emergent ray become parallel to each other.
Q.3 A pencil when dipped in water in a glass tumbler appears to be bent at the interface of air and water. Will the pencil appears to be bent to the same extent, if instead of water we use liquids like, kerosene or turpentine. Support your answer with reason.
Sol. The pencil dipped in water appears bent due to refraction of light. The angle of refraction depends on the refractive index of the medium. Hence it will appear to be bent to different extent in different media. Greater the refractive index, more the pencil will appear to be bent.
Q.4 Refractive index of diamond with respect to glass is 1.6 and absolute refractive index of glass is 1.5 . Find out the absolute refractive index of diamond.
Sol.
ndg=1.6
ng=1.5
ndg=ndng
nd=ndg×ng = 1.5 × 1.6 = 2.3
Q.5 A convex lens of focal length 20 cm can produce a magnified virtual as well as real image. Is this a correct statement? If yes, where shall the object be placed in each case for obtaining these images?
Sol. Yes it is a correct statement.
Magnified, virtual image will be formed when object distance is less than 20 m.
Magnified, real image will be formed when object distance lies between 20 m to 40 m.
*Q.6 Sudha finds out that the sharp image of the window pane of her science laboratory is formed at a distance of 15 cm from the lens. She now tries to focus the building visible to her outside the window instead of the window pane without disturbing the lens. In which direction will she move the screen to obtain a sharp image of the building? What is the approximate focal length of this lens?
Sol. As the object distance increases, image distance decreases. Sudha should try to move the screen towards the lens to obtain sharp image of buildings.
The approximate focal length is 15 cm.
Q.7 How are power and focal length of a lens related? You are provided with two lenses of focal length 20 cm and 40 cm respectively. Which lens will you use to obtain more convergent light?
Sol. P=1f
The lens having focal length 20 cm will give more convergent light.
Q.8 Under what condition in an arrangement of two plane mirrors, incident ray and reflected ray will always be parallel to each other, whatever may be angle of incidence. Show the same with the help of diagram.
Sol. When the two plane mirror are placed perpendicular to each other.
Angle between incident ray, reflected ray = 2i
Angle between second incident ray, reflected ray = 180 – 2i
Sum of the angle =180∘
⇒ Incident ray is parallel to reflected ray.
Q.9 Draw a ray diagram showing the path of rays of light when it enters with oblique incidence.
(i) from air into water ; (ii) from water into air.
Sol. (i)
(ii)
Long Answer Type Questions : –
Q.10 Draw ray diagrams showing the image formation by a convex mirror when an object is placed
(a) at infinity
(b) at finite distance from the mirror
Sol. (a)
(b)
Q.11 The image of a candle flame formed by a lens is obtained on a screen placed on the other side of the lens. If the image is three times the size of the flame and the distance between lens and image is 80 cm, at what distance should the candle be placed from the lens? What is the nature of the image at a distance of 80 cm and the lens?
Sol.
m = – 3
v = 80 cm.
m=vu
u=−803cm
=−26.67cm.
The image is real and lens is convex lens.
Q.12 Size of image of an object by a mirror having a focal length of 20 cm is observed to be reduced to 1/3rd of its size. At what distance the object has been placed from the mirror? What is the nature of the image and the mirror?
Sol. As image could be real or virtual i.e inverted or erect.
Case I
m=−13 (In case of concave mirror)
f = – 20 cm.
m=−13=−vu
v=u3
1f=1v+1u
−120=3u+1u=4u
u=−80cm.
Case II
m=+13 (In case of convex mirror)
f = + 20 cm.
m=+13=−vu
v=−u3
1f=1v+1u
120=−34+14=−2u
Therefore, u = – 40 cm.
Q.13 Define power of a lens. What is its unit? One student uses a lens of focal length 50 cm and another of 50 cm. What is the nature of the lens and its power used by each of them?
Sol.
Power of a lens is the degree of its convergence or divergence. It is the reciprocal of it’s focal length in meters. It’s unit is dioptre.
P1=1f=10050=+2D−Convexlens.
P2=1f=−10050=−2D−Concavelens.