NCERT solution of Class 10 Science Chapter 10 Light Reflection and Refraction
Light Reflection and Refraction
NCERT solutions of chapter 10 ‘Light reflection and refraction’ are created by an expert of Future Study Point, these NCERT solutions of chapter 10 ‘Light reflection and refraction are prepared for helping the class 10 students to boost their preparation of exams and complete their school assignments. All NCERT solutions of chapter 10 are the solutions of the questions mentioned within the pages and of back exercises of chapter 10 of class 10 science NCERT textbook prescribed by CBSE the most reputed board of India. Here each question is explained by step by step way so all students can understand these NCERT solutions easily, however, if you face any type of problem then you can write it in the comment box, you will get its reply without any delay.
Chapter 1-Light reflection and refraction is the part of the physics portion of the NCERT science textbook of 10 class. The NCERT solutions of chapter 10, light reflection and refraction provide you complete knowledge of the topics. All NCERT solutions of chapter 10 of science NCERT textbook are explained here readily and scientifically. The NCERT solutions of chapter-10 of class 10 NCERT science textbook are based on characteristics of light i.e laws of reflection and refraction, the image formation by the spherical mirrors and lenses, refractive index, power of lense and magnification of lenses and mirrors.
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NCERT Solutions of class 10 science chapter 10 -Light Reflection and Refraction
Q1. Define the principal focus of a concave mirror.
Answer. Light rays that are parallel to the principal axis of a concave mirror converge at a specific point on its principal axis after reflecting from the mirror. This point is known as the principal focus of the concave mirror.
Q2.The radius of curvature of a spherical mirror is 20 cm. What is its focal length?
Answer. Radius of curvature, R = 20 cm
Radius of curvature of a spherical mirror = 2 × Focal length (f)
⇒f = R/2 = 20/2 = 10
Hence, the focal length of the given spherical mirror is 10 cm.
Q3.Name the mirror that can give an erect and enlarged image of an object.
Answer. When an object is placed between the pole and the principal focus of a concave mirror, the image formed is virtual, erect, and enlarged.
Q4. Why do we prefer a convex mirror as a rear-view mirror in vehicles?
Answer. Convex mirrors give a virtual, erect, and diminished image of the objects placed in front of them. They are preferred as a rear-view mirror in vehicles because they give a wider field of view, which allows the driver to see most of the traffic behind him.
Page no 171
Q1.Find the focal length of a convex mirror whose radius of curvature is 32 cm.
Answer.Radius of curvature, R = 32 cm
Radius of curvature = 2 × Focal length (f)
R = 2f
⇒f = R/2 = 32/2 = 16
Hence, the focal length of the given convex mirror is 16 cm.
Q2. A concave mirror produces three times a magnified (enlarged) real image of the object placed at 10 cm in front of it. Where is the image located?
Answer. Magnification produced by a spherical mirror is given by the relation,
Let the height of the object, ho = h
Then, the height of the image, hI = −3h (Image formed is real)
hI/ ho = -v/u
-3h/h = -v/u
v/u = 3
Object distance, u = −10 cm
v = 3 × (−10) = −30 cm
Here, the negative sign indicates that an inverted image is formed at a distance of 30 cm in front of the given concave mirror.
Page no 176
Q1. A ray of light traveling in air enters obliquely into water. Does the light ray bend towards the normal or away from the normal? Why?
Answer. The light ray bends towards the normal. When a ray of light travels from an optically rarer medium to an optically denser medium, it gets bent towards the normal. Since water is optically denser than air, a ray of light traveling from air into the water will bend towards the normal.
Q2.Light enters from air to glass having a refractive index 1.50. What is the speed of light in the glass? The speed of light in a vacuum is 3 × 108 m s−1.
Answer.Refractive index of a medium nm is given by,
⇒nm = C/v
⇒1.50 = ( 3 x 108)/v
⇒v = (3 x 108)/1.50
⇒v =2 x 108
S0, the speed of light in the glass is 2 x 108 m/s.
Q3.Find out, from Table, the medium having the highest optical density. Also, find the medium with the lowest optical density
Answer.Highest optical density = Diamond
Lowest optical density = Air
The optical density of a medium is directly related to the refractive index of that medium. A medium which has the highest refractive index will have the highest optical density and vice-versa.
It can be observed from table 10.3 that diamond and air respectively have the highest and lowest refractive index. Therefore, diamond has the highest optical density and air has the lowest optical density.
Q4.You are given kerosene, turpentine and water. In which of these does the light travel fastest? Use the information given in Table.(Refer to question no-3).
Answer.The speed of light in a medium is given by the relation for refractive index. The relation is given as
Let refractive index of the medium is , velocity in air is ‘v’ and velocity in the medium is
is inversely proportional to
It can be inferred from the relation that light travels the slowest in the material which has the highest refractive index and travels the fastest in the material which has the lowest refractive index.
It can be observed from table 10.3 that the refractive indices of kerosene, turpentine, and water are 1.44, 1.47, and 1.33 respectively. Therefore, light travels the fastest in water.
Q5. The refractive index of diamond is 2.42. What is the meaning of this statement?
Answer. Refractive index of a medium nm is related to the speed of light in that medium v by the relation:
⇒nm = C/v
Where, c is the speed of light in vacuum and v in the air
The refractive index of diamond is 2.42. This suggests that the speed of light in the diamond will reduce by a factor 2.42 compared to its speed in air.
Q1.Define 1 dioptre of power of a lens.
Answer.The power of lens is defined as the reciprocal of its focal length. If P is the power of a lens of focal length F in meters, then
F = 1/P
P = 1/F, if F = 1m, then P = 1D
The S.I. unit of power of a lens is Dioptre. It is denoted by D.
1 dioptre is defined as the power of a lens of focal length 1 meter.
Q2.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.
Answer. When an object is placed at the center of curvature, 2F1, of a convex lens, its image is formed at the center of curvature, 2F2, on the other side of the lens. The image formed is inverted and of the same size as the object, as shown in the given figure.
It is given that the image of the needle is formed at a distance of 50 cm from the convex lens. Hence, the needle is placed in front of the lens at a distance of 50 cm.
Object distance, u = −50 cm
Image distance, v = 50 cm
Focal length = f
Applying the lens formula
f = 25 cm = 0.25 m
The relationship between power and focal length of the lens is
Hence, the power of the given lens is +4 D.
Q3.Find the power of a concave lens of focal length 2 m.
Answer. Focal length of a concave lens, f = -2 m
The relationship between power of the lens and its focal length is following
Here, a negative sign arises due to the divergent nature of concave lens.
Hence, the power of the given concave lens is −0.5 D.
Q1. Which one of the following materials cannot be used to make a lens?
Answer.(d) A lens allows light to pass through it. Since clay does not show such property, it cannot be used to make a lens.
Q2. 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 center of curvature
(b) At the center of curvature
(c) Beyond the center of curvature
(d) Between the pole of the mirror and its principal focus.
Answer.(d) When an object is placed between the pole and principal focus of a concave mirror, the image formed is virtual, erect, and larger than the object.
Q3.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.
Answer.(b) When an object is placed at the center of curvature in front of a convex lens, its image is formed at the center of curvature on the other side of the lens. The image formed is real, inverted, and of the same size as the object.
Q4.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
Answer. (a) By convention, the focal length of a concave mirror and a concave lens are taken as negative. Hence, both the spherical mirror and the thin spherical lens are concave in nature.
Q5.No matter how far you stand from a mirror, your image appears erect. The mirror is likely to be
(d) either plane or convex
Answer.(d) A convex mirror always gives a virtual and erect image of smaller size of the object placed in front of it. Similarly, a plane mirror will always give a virtual and erect image of same size as that of the object placed in front of it. Therefore, the given mirror could be either plane or convex.
Q6. 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
Answer.(c) A convex lens gives a magnified image of an object when it is placed between the radius of curvature and focal length. Also, magnification is more for convex lenses having a shorter focal length. Therefore, for reading small letters, a convex lens of focal length 5 cm should be used.
Q7.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.
Answer.Range of object distance = 0 cm to15 cm A concave mirror gives an erect image when an object is placed between its pole (P) and the principal focus (F).
Hence, to obtain an erect image of an object from a concave mirror of focal length 15 cm, the object must be placed anywhere between the pole and the focus. The image formed will be virtual, erect, and magnified in nature, as shown in the given figure.
Q8.Name 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.
Answer.(a) Concave (b) Convex (c) Concave
(a) Concave mirror is used in the headlights of a car. This is because concave mirrors can produce a powerful parallel beam of light when the light source is placed at their principal focus.
(b) Convex mirror is used inside/rearview mirror of a vehicle. Convex mirrors give a virtual, erect, and diminished image of the objects placed in front of it. Because of this, they have a wide field of view. It enables the driver to see most of the traffic behind him/her.
(c) Concave mirrors are convergent mirrors. That is why they are used to construct solar furnaces. Concave mirrors converge the light incident on them at a single point known as the principal focus. Hence, they can be used to produce a large amount of heat at that point.
Q9.One-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.
Answer. The convex lens will form a complete image of an object, even if its one half is covered with black paper. It can be understood as the following fig.
When the lower half of the lens has covered this case, two rays of light coming from the object is refracted by the upper half of the lens. These rays meet at the other side of the lens to form the image of the given object, as shown in the figure.
Q10. 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 nature of the image formed.
Answer.Object distance, u = −25 cm
Object height, h = 5 cm
Focal length, f = +10 cm
According to the lens formula,
Position of the image is at 16.67 cm in front of the lense
Magnification’ of the lens is given
The negative value of image height indicates that the image formed is inverted. Therefore image hight is 3.3 cm,position of image is at 16.67 cm in front of the lens and the image form is inverted
The position, size, and nature of the image are shown in the following ray diagram.
Q11. 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.
Focal length of concave lens (OF1), f = −15 cm
Image distance, v = −10 cm
According to the lens formula,
The negative value of u indicates that the object is placed 30 cm in front of the lens. This is shown in the following ray diagram.
Q12.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.
Answer.Focal length of convex mirror, f = +15 cm
Object distance, u = −10 cm
According to the mirror formula,
Q13.The magnification produced by a plane mirror is +1. What does this mean?
Answer.Magnification produced by a mirror is given by the relation
The magnification produced by a plane mirror is +1. It shows that the image formed by the plane mirror is of the same size as that of the object. The positive sign shows that the image formed is virtual and erect.
Q14.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.
Answer.Object distance, u = −20 cm
Object height, h = 5 cm
Radius of curvature, R = 30 cm
Radius of curvature = 2 × Focal length
R = 2f
f = 15 cm
According to the mirror formula,
The positive value of image height indicates that the image formed is erect. Therefore, the image formed is virtual, erect, and smaller in size.
Q15.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 focused image can be obtained? Find the size and the nature of the image.
Answer.Object distance, u = −27 cm
Object height, h = 7 cm
Focal length, f = −18 cm
According to the mirror formula,
The negative value of magnification indicates that the image formed is real.
The negative value of image height indicates that the image formed is inverted.
Q16.Find the focal length of a lens of power −2.0 D. What type of lens is this?
The negative sign of its focal length shows that it is a concave lense.
Q17.A 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?
Therefore the focal length of the lens is 0.67 m
The positive sign of its focal length shows that the given lens is converging(convex).
⇒In the chapter Light reflection and refraction chapter, 10 of class 10 is very important for class 10 students. The whole of the chapter is oriented on the properties of light ray (i) Reflection of light (ii) Refraction of light.
Light: Light is energy which when falls on our eyes gives us a visual effect of anything we see.Light energy is used by the plants for the process of photosynthesis,so we can say that the biggest source of light on the earth is the Sun.
Reflection of light: When a light ray falls on an opaque object, it is bounced back by the object to another direction, known as the reflection of light.
Laws of reflection of the light: (i) Incident angle is equal to the reflected angle.
(ii) The incident ray, normal and the reflected rays, all lie on the same plane.
Refraction of light ray: When light rays passes from one medium to another medium it deviates from its path,this phenomena of light is known as refraction of light.
The Laws of refraction: There are two laws of refraction of light.
(i) The incident ray, incident point and reflected ray all lie on the same plane.
(ii) The ratio between sine of incident angle and reflected angle is constant.
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