©2024 cbsequestions. All rights reserved.
©2024 cbsequestions. All rights reserved.
NCERT Solutions for Class 10 Science Chapter 9 Light – Reflection and Refraction : NCERT Solutions will help students to understand the concepts easily.
Q1. Define the principal focus of a concave mirror.
Sol. The principal focus of a concave mirror is a point on its principal axis to which all the light rays which are parallel and close to the axis, converge after reflection from the concave mirror.
Q2. The radius of curvature of a spherical mirror is 20cm. What is its focal length ?
Sol. Focal length (f) = radius of curvature (R) / 2
Therefore, f = 20/2 cm = 10 cm
Focal length of spherical mirror is 10cm.
Q3. Name a mirror that can give an erect and enlarged image of an object.
Sol. Concave mirror.
Q4. Why do we prefer a convex mirror as a rear view mirror in vehicles ?
Sol. We prefer a convex mirror as a rear-view mirror in vehicles because of the following reasons :
Q5. Find the focal length of a convex mirror whose radius of curvature is 32 cm.
Sol. Here, radius of curvature, R = 32 cm
As, f = R/2
f = 32/2 = 16 cm
Q6. A concave mirror produces three times magnified real image of an object placed at 10cm in front of it. Where is the image located ?
Sol. Here, linear magnification, m = -3 (negative sign for real image, which is inverted)
Object distance, u = 10 cm,
Image distance, v = ?
As m = -v/u
-3 = -v/(-10)
v = -30 cm
Q7. 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 light ray bends towards the normal because ray of light travelling from a rarer medium (air) to a denser medium (water) slows down and bends towards the normal.
Q8. 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
Sol. Refractive index of glass = Speed of light in vacuum (c) / Speed of light in glass (v)
So, 1.50 = 3 × 108 / v
v = 3 × 108 / 1.50
v = 2 × 108 m/s
Q9. Find out, from table 10.3 (NCERT), the medium having highest optical density. Also find the medium with lowest optical density.
Sol. The medium having highest optical density is diamond. The medium with lowest optical density is air.
Q10. You are given kerosene, turpentine and water. In which of these does the light travel fastest ? Use the information given in table (NCERT 10.3).
Sol. Light travels fastest in water out of kerosene, turpentine and water.
Q11. The refractive index of diamond is 2.42. What is the meaning of this statement ?
Sol. The refractive index of diamond is 2.42. It means that the ratio of the speed of light in air and the speed of light in diamond is equal to 2.42 or the speed of light in diamond is 1/2.42 times that of the speed of light in air.
Q12. Define 1 dioptre of power of a lens.
Sol. 1 dioptre is the power of a lens whose focal length is 1 meter.
Q13. A convex lens forms a real and inverted image of a needle at a distance 59cm from it. Where is the needle placed in from to the convex lens if the image is equal to the size of the object ? Also find the power of the lens.
Sol. The needle is placed at 50cm in front of the con vex lens if the image is equal to the size equal to the object. The image formed by a convex lens is of the same size as the object when the object is at 2F or center of curvature C.
C = 2F = 50cm
Focal length (f) = C/2 = 50/2 = 25 cm
Power of a lens = 1/f(m) = 100/25 = 4D
Q14. Find the power of a concave lens of focal length 2m.
Sol. Power of lens = 1/-f = 1/-2 = -0.5D
Q1. Which one of the following materials cannot be used to make a lens ?
(a) Water
(b) Glass
(c) Plastic
(d) Clay
Sol. (d) Clay
Q2. The image formed by a concave mirror is ob served 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) Between the pole of the mirror and its principal focus.
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.
Sol. (b) At twice the focal length.
Q4. A spherical mirror and a thin spherical lens have each a focal length of –15cm. 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) both concave
Q5. No matter how for you stand from a mirror, your image appears erect. The mirror is likely to be
(a) plane only
(b) concave
(c) convex only
(d) either plane or convex
Sol. (d) either plane or convex.
Q6. Which of the following lenses would you pre fer to use while reading small letters found in a dictionary ?
(a) A convex lens of focal length 50cm
(b) A concave lens of focal length 50cm.
(c) A convex lens of focal length 5cm.
(d) A concave lens of focal length 5cm.
Sol. (c) A convex lens of focal length 5cm.
Q7. We wish to obtain an erect image of an object, using a concave mirror of focal length 15cm. 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. We can obtain an erect image of the object, using a concave mirror when the object is placed be tween pole P of the mirror and focus F. Therefore, the image of distance of the object from the mirror is between 0 to 15 cm. The nature of image is virtual and erect. The image is larger than object.
Q8. Name the types 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.
Sol.
(a) Concave mirrors are used as headlights of a car because they reflect the light.
(b) Convex mirror is used in side/rear view mirror of a vehicle because it produces an erect image of the objects and gives a wide field of view of the traffic behind.
(c) Concave mirror is used in solar furnace be cause concave mirror focus sunrays on the objects to be heated.
Q9. One-half of a convex lens is covered with a black paper. Will the produce a complete image of the object ? Verify your answer experimentally. Explain your observations.
Sol. Yes, the lens will produce a complete image of the object. We observe that the rays of light object passes through half lens and produces complete image.
Q10. An object 5cm in length is held 25cm away from a converging lens focal length 10cm. Draw the ray diagram and find the position, size and the nature of the image formed.
Sol. Height of the object (h) = 5cm
Object distance (u) = – 25cm
Focal length (f) = + 10cm
since 1/v -1/u = 1/f
1/v = 1/u + 1/f
1/v = 1/(-25) + 1/10
= (-2+5)/50
= +3/50
v = +16.67cm
Magnification, m = h'/h = v/u
So, Image size (h') = v/u × h
h' = 5 × 50/3 × 1/-25
h' = -10/3 = -3.3 cm
The negative sign of h' show that the image is inverted. It is formed below the principal axis.
Thus, a real, inverted image 3.3 cm tall, is formed at a distance of 16.67cm on the right side of the lens.
Q11. A concave lens of focal length 15cm forms an image 10cm from the lens. How far is the object placed form the lens ? Draw the ray diagram.
Sol. f = – 15cm, v = – 10cm, u = ?
since 1/v -1/u = 1/f
1/u = 1/v - 1/f
1/u = 1/(-10) - 1/(-15)
= (-3+2)/30
= -1/30
u = -30
Thus, the object distance is 30cm, on the left side, from the concave lens.
Q12. An object is placed at a distance of 10cm from a convex mirror of force length 15cm. Find the position and nature of the image.
Sol. f = 15cm, u = – 10cm, v = ?
since 1/v +1/u = 1/f
1/v = 1/f - 1/u
1/v = 1/15 - 1/(-10)
1/v = (2+3)/30
1/v = 5/30
v = +6cm
The image formed is 6cm at the back of the mirror. It is virtual and erect.
Q13. The magnification produced by a plane mirror is + 1. What does this mean ?
Sol. Magnification + 1 means that the size of the image is same as (equal to ) the size of the object. Positive sign of m means that the image formed is virtual.
Q14. An object 5.0cm in length is placed at a distance of 20cm in front a convex mirror of radius of curvature 30cm. Find the position of the image, its nature and size.
Sol. R = + 30cm, u = – 20cm, h = 5.0cm v = ?, h' = ?
Focal length, f = R/2 =+30/2 cm = 15 cm
since 1/v +1/u = 1/f
1/v = 1/f - 1/u
1/v = 1/15 - 1/(-20)
1/v = (4+3)/60 = 7/60
v = 60/7 cm
The image is 60/7 cm at the back of the mirror.
Magnification, m = h'/h = v/u
So, Image size (h') = v/u × h
h' = 5 × -60/7 × 1/-20
h' = +15/7 = +2.14 cm
Q15. An object of size 7.0cm is placed at 27cm in front of a concave mirror of focal length 18cm. 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.
Sol. f = – 18cm, u = – 27cm, h = + 7.0cm v = ?, h = ?
since 1/v +1/u = 1/f
1/v = 1/f - 1/u
1/v = 1/(-18) - 1/(-27)
1/v = (-3+2)/54 = -1/54
v = -54 cm
The screen should be placed at 54cm from the mirror on the object side of the mirror. Image is real.
Also, Magnification, m = h'/h = -v/u
So, Image size (h') = -v/u × h
h' = -(-54) × 7 / (-27)
h' = -14 cm
The image is inverted, real and twice in size.
Q16. Find the focal length of a lens of power -2.0D. What type of lens is this?
Sol. Here, focal length f = ?, power, P = 2.0 D
Clearly, f = 100/P = 100/(-2.0) = -50 cm
Q17. 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. Here, P=+1.5D, f=?
From f = 100/P = 100/1.5 = 66.7 cm
The prescribed lens is converging or convex as its power is positive.