Light- Reflection and Refraction
NCERT Solutions
NCERT Exercises
1. Which one of the following materials cannot be used to make a lens?
(a) Water
(b) Glass
(c) Plastic
(d) Clay
(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.
(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.
(b) 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
(b) both convex
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
(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
(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.
If we need to get an erect image with concave mirror, the object should placed at a distance less than focal length, i.e.. the object should be placed less than 15cm from mirror. The nature of the image is 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.
(a) Concave Mirror: If the light source is placed at their principal focus, concave mirrors can produce a powerful parallel beam of light.
(b) Convex Mirror: It provides a wide 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.
Yes, it will produce a complete image of the object. By using a convex lens and covering it’s half part using a black paper, place it vertically in a stand. Place a burning candle on one side. On opposite side of the lens fix a white screen. Adjust the position of candle or screen until a clear image of burning candle is formed on the screen. We observe that the image is a complete image of the object. But the intensity of light is reduced.
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.
Given
Height of the image, h_0 = 5 cm
Distance of the object from converging lens, u = -25 cm
Focal length of a converging lens, f = 10 cm
We know the lens formula
\frac{1}{v} -\frac{1}{u} = \frac{1}{f}
\frac{1}{v} = \frac{1}{f} +\frac{1}{u}
\frac{1}{v} = \frac{1}{10} -\frac{1}{25}
\frac{1}{v} = \frac{15}{250}
v = 16.66 cm
Also for converging lens
\frac{h_i}{h_0} = \frac{v}{u}
h_i =\frac{v}{u} \times h_0
h_i =\frac{50 \times 5}{3 \times(-25)} = -3.3 cm
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.
Focal length of concave lens, f = -15 cm
Image distance, v= – 10 cm
According to the lens formula,
\frac{1}{v} -\frac{1}{u} = \frac{1}{f}
\frac{1}{u} = \frac{1}{-10} -\frac{1}{-15}
v=-30cm
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.
Focal length of convex mirror (f) = +15 cm
Object distance (u) = – 10 cm
According to the mirror formula,
\frac{1}{v} -\frac{1}{u} = \frac{1}{f}
\frac{1}{v} =\frac{1}{15} – \frac{1}{-10} = 6 cm
Magnification = \frac{-v}{u} =\frac{-6}{-10} = 0.6
The image is erected and diminished.
13. The magnification produced by a plane mirror is +1. What does this mean?
The image formed is virtual, erect and of the same size as that of 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.
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,
\frac{1}{v} -\frac{1}{u} = \frac{1}{f}
\frac{1}{v} = \frac{1}{15} + \frac{1}{20} = 8.57 cm
v is positive which means image is formed behind the mirror.
Magnification = \frac{-v}{u} =\frac{-8.57}{-20} = 0.428
Magnification is positive which means image formed is virtual.
Magnification = \frac{-h^’}{h}
-h^’ = m \times h = 0.428 \times 5 = 2.14 cm
The positive value of image height indicates that the image formed is erect. So the image formed is erect, virtual, 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.
Object distance (u) = – 27 cm
Object height (h) = 7 cm
Focal length (f) = – 18 cm
From the mirror formula,
\frac{1}{v} = \frac{1}{-18} + \frac{1}{27}
v = -54 cm which means the screen should be placed at a distance of 54cm in front of the given mirror.
Magnification = \frac{-v}{u} =\frac{-54}{27} = -2
Magnification is negative which means image formed is real.
Magnification = \frac{-h^’}{h}
-h^’ = m \times h = 7 \times -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?
Power of lens P =\frac{1}{f}
f =\frac{1}{P}
f =\frac{-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?
Power of lens P =\frac{1}{f}
f =\frac{1}{P}
f =\frac{1}{1.5} = 0.66 m
A convex lens has a positive focal length. Therefore, it is a convex lens or a converging lens.
Intext Exercises
Pg. No. 168
1. Define the principal focus of a concave mirror.
The point at which light rays that are parallel to the principal axis of a concave mirror converges is known as the principal focus.
2. The radius of curvature of a spherical mirror is 20 cm. What is its focal length?
Given, radius of curvature R = 20 cm
Radius of curvature of the spherical mirror 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.
Concave Mirror.
4. Why do we prefer a convex mirror as a rear-view mirror in vehicles?
This mirror gives a wide view of state of traffic behind the driver.
Pg. No. 171
1. Find the focal length of a convex mirror whose radius of curvature is 32 cm.
Radius of curvature (R) = 32 cm
R= 2f
f = R/2 = 32/2 = 16
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?
We know that
m= \frac{h_I}{h_0} = -\frac{v}{u}
Let the height of the object h_0 = h
From the question, it is given that a concave mirror produces three times magnified (enlarged) real image of an object.
i.e. h_I = -3h
\frac{-3h}{h} = -\frac{v}{u}
,\frac{v}{u} = 3
Object 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.
Pg. No. 176
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?
The light ray bends towards the normal. When a light ray enters from a medium with low refractive index (optically rarer medium) to a medium with a high refractive index (optically denser medium) speed of the light 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.
Given
Speed of light in vacuum = 3 × 108 m/s
Refractive index of glass = 1.50
We know that
Refractive index of a medium = Speed of light in vacuum/Speed of light in the medium
Therefore, Speed of light in the glass (v) = Speed of light in vacuum/ Refractive index of glass =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.
Lowest optical density = Air
Highest optical density = Diamond
4. You are given kerosene, turpentine and water. In which of these does the light travel fastest? Use the information given in Table.
Light travel faster in water as compared to kerosene & turpentine. This is because the refractive index of water is lower than that of kerosene and turpentine.
5. The refractive index of diamond is 2.42. What is the meaning of this statement?
It means that the speed of light in diamond will reduce by a factor of 2.42 as compared to its speed in the air.
Pg. No. 184
1. Define 1 dioptre of power of a lens.
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.
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
From lens formula
\frac{1}{f}= \frac{1}{50}-\frac{1}{-50}
\frac{1}{f}= \frac{1}{25}
f=0.25m
Power of lens, P = \frac{1}{f}
P = \frac{1}{0.25}=+4D
3. Find the power of a concave lens of focal length 2 m.
Given
Focal length of concave lens f = 2 m
Power of lens P = 1/f = 1/ (-2) = -0.5D