NCERT Solutions Class 10th Science Chapter – 9 Light Reflection and Refraction
| Textbook | NCERT |
| Class | 10th |
| Subject | Science |
| Chapter | 9th |
| Chapter Name | Light Reflection and Refraction |
| Category | Class 10th Science |
| Medium | English |
| Source | Last Doubt |
| NCERT Solutions Class 10th Science Chapter – 9 Light Reflection and Refraction Question & Answer In This Chapter We will read about Light Reflection and Refraction, What is light reflection and refraction?, What is the reflection of the light?, What is reflection for 10th class?, What is law of reflection and refraction?, What is called refraction?, What is the Snell’s law of reflection?, What are 5 examples of reflection and refraction?, What are three laws of reflection?, What are the two types of reflection?, What are the two laws of reflection? etc. |
NCERT Solutions Class 10th Science Chapter – 9 Light Reflection and Refraction
Chapter – 9
Light Reflection and Refraction
Question & Answer
Page – 142
| 1. 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 called the principal focus of the concave mirror. |
| 2. 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 the spherical mirror = 2 × Focal length (f) R = 2f f= R/2 = 20 / 2 = 10 Therefore, the focal length of the spherical mirror is 10 cm. |
| 3. Name the mirror that can give an erect and enlarged image of an object. Answer – The mirror that can give an erect and enlarged image of an object is Concave Mirror. |
| 4. Why do we prefer a convex mirror as a rear-view mirror in vehicles? Answer – Convex mirror is preferred as a rear-view mirror in cars and vehicles as it gives a wider field of view, which helps the driver to see most of the traffic behind him. Convex mirrors always form an erect, virtual, and diminished image of the objects placed in front of it. |
Page No – 145
| 1. 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 Therefore, the focal length of the given convex mirror is 16 cm. |
| 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? Answer – Magnification produced by a spherical mirror: Magnification produced by a spherical mirror is given by the relation,m = m = h1/h0 =- u/v Let the height of the object, h0 = h Then, height of the image, h1= – 3h (image formed is real) – 3h/h = -v/u u/v = 3Object distance (u) = – 10 cm v = 3 × (- 10) = – 30 cm Therefore, the negative sign indicates that an inverted image is formed in front of the given concave mirror at a distance of 30 cm. |
Page No – 150
| 1. A ray of light travelling in air enters obliquely into water. Does the light ray bends towards the normal or away from the normal? Why? Answer – The light ray bends towards the normal. When a light ray enters from an optically rarer medium (which has low refractive index) to an optically denser medium (which has a high refractive index), its speed slows down and bends towards the normal. As water is optically denser than air, a ray of light entering from air into water will bend towards the normal. |
| 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 x 108 ms-1. Answer – Refractive index of a medium (nm) = Speed of light in vacuum/Speed of light in the medium Speed of light in vacuum (c) = 3 × 108 m/s Refractive index of glass (ng) = 1.50 Speed of light in the glass (v) = Speed of light in vacuum/ Refractive index of glass = c/ng =3 × 108/1.50 = 2x 108 ms-1. |
| 3. Find out, from Table, the medium having highest optical density. Also find the medium with lowest optical density. |
| Materialmedium | Refractive index | Material medium | Refractiveindex |
| Air | 1.0003 | Canada Balsam | 1.53 |
| Ice | 1.31 | – | – |
| Water | 1.33 | Rock salt | 1.54 |
| Alcohol | 1.36 | – | – |
| Kerosene | 1.44 | Carbon disulphide | 1.63 |
| Fusedquartz | 1.46 | Denseflint glass | 1.65 |
| Turpentine oil | 1.47 | Ruby | 1.71 |
| Benzene | 1.50 | Sapphire | 1.77 |
| Crownglass | 1.52 | Diamond | 2.42 |
| Answer – Lowest optical density = Air Highest optical density = Diamond The optical density of a medium is directly related to its refractive index. A medium with the highest refractive index will have the highest optical density and vice-versa. It can be observed from the table that air and diamond respectively have the lowest and highest refractive index. Hence, air has the lowest optical density and diamond has the highest optical density. |
| 4. You are given kerosene, turpentine and water. In which of these does the light travel fastest? Use the information given in Table. |
| Materialmedium | Refractive index | Material medium | Refractiveindex |
| Air | 1.0003 | Canada Balsam | 1.53 |
| Ice | 1.31 | – | – |
| Water | 1.33 | Rock salt | 1.54 |
| Alcohol | 1.36 | – | – |
| Kerosene | 1.44 | Carbon disulphide | 1.63 |
| Fusedquartz | 1.46 | Denseflint glass | 1.65 |
| Turpentine oil | 1.47 | Ruby | 1.71 |
| Benzene | 1.50 | Sapphire | 1.77 |
| Crownglass | 1.52 | Diamond | 2.42 |
| Answer – Light travel faster in water as compared to kerosene & turpentine as the refractive index of water is lower than that of kerosene and turpentine. The speed of light is inversely proportional to the refractive index. |
| 5. The refractive index of diamond is 2.42. What is the meaning of this statement? Answer – Diamond has a refractive index of 2.42 which means that the speed of light in diamond will reduce by a factor of 2.42 as compared to its speed in the air. In other words, the speed of light in diamond is 1/2.42 times the speed of light in vacuum. |
Page No – 158
| 1. Define 1 dioptre of power of a lens. Answer – Dioptre is the SI unit of power of lens is denoted by the letter D. 1 dioptre can be defined as the power of a lens of focal length 1 metre. |
| 2. 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 – The position of the image should be at 2F since the image is the real and same size. It is given that the image of the needle is formed at a distance of 50 cm from the convex lens. Therefore, 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 1/v – 1/u = 1/f 1/f = 1/50 – 1/-50 = 1/50 + 1/50 = 1/25 f = 25cm = 0.25m p = 1/f p = 1/0.25 = +4D |
| 3. Find the power of a concave lens of focal length 2 m. Answer – Focal length of concave lens (f) = 2 m Power of lens (P) = 1/f = 1/(-2) = -0.5D |
Exercise Question page No – 159-160
| 1. Which one of the following materials cannot be used to make a lens? (a) Water (b) Glass (c) Plastic (d) Clay Answer – (d) Clay |
| 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. Answer – (d) Between the pole of the mirror and its principal focus. |
| 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. Answer – (b) The object should be placed at twice the focal length |
| 4. A spherical mirror and a thin spherical lens have 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 are likely to be concave |
| 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 Answer – (d) either plane or convex |
| 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 Answer – (c) A convex lens of focal length 5 cm |
| 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. Answer – Range of the distance of the object = 0 to 15 cm from the pole of the mirror. Nature of the image = virtual, erect, and larger than the object. |
| 8. 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 Mirror – Because concave mirrors can produce a powerful parallel beam of light when the light source is placed at their principal focus. (b) Convex Mirror – Because of its largest field of view. (c) Concave Mirror – Because it concentrates the parallel rays of the sun at a principal focus. |
| 9. 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 – Yes, it will produce a complete image of the object, as shown in the figure. This can be verified experimentally by observing the image of a distant object like a tree on a screen when the lower half of the lens is covered with a black paper. However, the intensity or brightness of the image will reduce. |
| 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. Answer – Height of the Object, h0 = 5 cm Distance of the object from converging lens, u = -25 cm Focal length of a converging lens, f = 10 cm Using lens formula, 1/v – 1/u = 1/f 1/v = 1/f + 1/u = 1/10 – 1/25 = 15/250 v = 250/15 = 16.66 cm M = hi/h0 = v/u hi = v/u x h0 = 50 x 5/3x (-25) = 10/-3 = -3.3 Thus, the image is inverted and formed at a distance of 16.7 cm behind the lens and measures 3.3 cm. The ray diagram is shown below.  |
| 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. Answer – Focal length of concave lens (OF1), f = – 15 cm Image distance, v = – 10 cm According to the lens formula, 1/v – 1/u = 1/f 1/u = 1/v – 1/f 1/u = 1/-10 + 1/-15 = -1/10 + 1/15 u = -5/150 = -30 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.  |
| 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. Answer – Focal length of convex mirror (f) = + 15 cm Object distance (u) = – 10 cm According to the mirror formula,1/v – 1/u =1/f 1/v = 1/f – 1/u = 1/15 + 1/10 = 25/150 v = 6cmThe positive value of v indicates that the image is formed behind the mirrorm = -v/u = -6/-10 = 0.6The positive value of magnification indicates that the image formed is virtual and erect.The image is located at a distance of 6 cm from the mirror on the other side of the mirror. The positive and a value of less than 1 of magnification indicates that the image formed is virtual and erect and diminished. |
| 13. The magnification produced by a plane mirror is +1. What does this mean? Answer – The positive sign means an image formed by a plane mirror is virtual and erect. Since the magnification is 1 it means that the size of the image is equal to the size of the object. |
| 14. An object 5 cm is placed at a distance of 20 cm in front of a convex mirror of radius of curvature 30 cm. Find the position, nature and size of the image. 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,1/v – 1/u = 1/f 1/v = 1/f – 1/u = 1/15 + 1/20 = 4 + 3/60 = 7/60The positive value of v indicates that the image is formed behind the mirror. Magnification m=Image Distance Object Distance Image Distance Object Distance = -8.57/-20 = 0.428 The positive value of magnification indicates that the image formed is virtual Magnification m=Height of image Height of Object Height of image Height of Object = h/′hh’ = m x h = 0.428 x 5 = 2.14 cmThe positive value of image height indicates that the image formed is erect. Therefore, the image formed is virtual, erect, and smaller in size. |
| 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 sharply focused image can be obtained? Find the size and 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,1/v – 1/u = 1/f1/v = 1/f – 1u = -1/18 + 1/27 = -1/54v = -54cmThe screen should be placed at a distance of 54 cm in front of the given mirror.h1/h = -v/u The negative value of magnification indicates that the image formed is real. h1/7= -54/27 h1=7 x (-2) = -14 cm The negative value of image height indicates that the image formed is inverted. |
| 16. Find the focal length of a lens of power -2.0 D. What type of lens is this? Answer – Power of lens (P) = 1/f P = -2D f = -1/2 = -0.5 m A concave lens has a negative focal length. Therefore, it is a concave lens. |
| 17. 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? Answer – Power of lens (P) = 1/f P = 1.5D f = 1/1.5 = 10/15 = 0.66 m A convex lens has a positive focal length. Therefore, it is a convex lens or a converging lens. |
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