Class 10 Science Chapter 10 Light-Reflection and Refraction Solutions to each chapter is provided in the list so that you can easily browse throughout different chapters NCERT Class 10 Science Chapter 10 Light-Reflection and Refraction and select need one.
Class 10 Science Chapter 10 Light-Reflection and Refraction
Also, you can read the CBSE book online in these sections Solutions by Expert Teachers as per NCERT (CBSE) Book guidelines. These solutions are part of SEBA All Subject Solutions. Here we have given Assam BoardClass 10 Science Chapter 10 Light-Reflection and Refraction Solutions for All Subjects, You can practice these here.
Light-Reflection and Refraction
Chapter – 10
GENERAL SCIENCE
Textual Questions and Answers:
Page – 168
Q.1. Define the principal focus of a concave mirror.
Ans: The principal focus of a concave mirror is a point on its principal axis to which all the parallel light rays which are close to the axis converges after reflection from the concave mirror.
Q.2. The radius of curvature of a spherical mirror is 20cm. What is its focal length?
Ans: Given that,
r = 20 cm.
f =?
We have,
f = r/2
= 20/2cm
= 10cm
Q.3. Name a mirror that can give an erect and enlarged image of an object.
Ans: Concave mirror.
Q.4. Why do we prefer a convex mirror as a rear view mirror in vehicles?
Ans: Usually, we prefer a convex mirror as a rear-view mirror in vehicles. The major reason behind this is that it provides a wider field of view. This permits the driver to view most of the traffic that is behind his vehicle.
A convex mirror always creates a virtual image. When the object is far away from the convex mirror, the image that is created is upright as well as situated at the focal point. As the object arrives near the convex mirror, the image also comes near the mirror. Further, the image develops until its height becomes equal to that of the image. Convex mirrors form images that are smaller in comparison to the object. However, the object becomes larger as it comes near the mirror.
Page – 171
Q.1. Find the focal length of a convex mirror whose radius of Curvature is 32 cm.
Ans: Given that,
r = 32 cm.
f = ?
We have,
f = r/2
= 32/2 cm
= 16 cm
Q.2. A concave mirror produces three times magnified za (enlarged) real image of an object placed at 10 cm in Inati front of it. Where is the image located?
Ans: Given that,
m = – 3
u = – 10cm
v = ?
We have,
m = – v/u
⇒ – 3 = – v/u
⇒ 3u = V
⇒ v = 3u
= 3x – 10 cm.
= – 30 cm.
Thus, the image is located at a distance of 30 cm. in front of the mirror.
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?
Ans: When a ray of light travelling in air enters obliquely into water, it bends towards the normal. This is because water is optically denser than air. On entering water, the speed of light decreases and the light bends towards normal.
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 ms⁻¹
Ans: We know that,
Refractive index of glass = Speed of light in air/Speed of light in glass
⇒ Speed of light in glass = Speed of light in air/ Refractive index of glass
= 3×10⁸/1.50 ms⁻¹
= 2×10⁸ ms⁻¹
Q.3. Find out from table 10.3 the medium having highest optical density. Also find the medium with lowest optical density.
Ans: The medium having highest refractive index has the highest optical density. Therefore diamond has the highest optical density.
The medium having lowest refractive index has the lowest optical density. Therefore air has the lowest optical density.
Q.4. You are given kerosene, turpentine and water. In which of these the light travels fastest? Use the information given in Table on page 225.
Ans: For kerosene, n = 1.44
For turpentine, n = 1.47
For water, n = 1.33
Because water has the lowest refractive index, therefore light travels fastest in this optically rarer medium than kerosene and turpentine oil.
Q.5. The refractive index of diamond is 2.42. What is the meaning of this statement?
Ans: The refractive index of diamond is 2.42. This means that the speed of light in diamond will reduce by a factor of 2.42 as compared to its speed in air.
Page – 184
Q.1. Define 1 dioptre of power of a lens.
Ans: 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.
Q.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 in size to the object? Also find the power of the lens.
Ans: Given that,
v = + 50 cm.
u = ?
We have m = v/u = – 1
⇒ v/u = – 1
⇒ u = -v
u = – 50 cm.
Again,
1/f = 1/v = 1/u
= 1/50-1/- 50
= 1/50+1/50
= 2/50
= 1/25
∴ f = 25 cm.
= 0.25m.
∴ P = 1/f
= 1/0.25m
= +4D
Q.3. Find the power of a concave lens of focal length 2m.
Ans: Given that,
f = – 2m [ For concave lens )
Power (P) = 1/f
= 1/-2m
= – 0.5 D
EXERCISES
Q.1. Which one of the following materials cannot be used to make a lens?
(a) Water.
(b) Glass.
(c) Plastic.
(d ) Clay.
Ans: (d) Clay.
Q.2. The image formed by a concave mirror is observed to the 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.
Ans: (d) Between the pole of the mirror and its principal focus.
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.
Ans: (b) At twice the focal length.
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.
Ans: (a) Both concave.
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.
Ans: (d) Either plane or convex.
Q.6. Which of the following lenses would you prefer to use To which 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 of 5 cm.
(d) A concave lens of focal length of 5 cm.
Ans: (c) A convex lens of focal length of 5 cm.
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 The smaller than the object? Draw a ray diagram to show the image formation in this case.
Ans: A concave mirror gives an erect image when the object is placed between the focus F and the pole P of the concave mirror, i.e., between 0 and 15 cm from the mirror. The image thus formed will be virtual, erect and larger than the object
Q.8. Name the type of mirror used is the following situations:
(a) Head lights of a car.
Ans: Headlights of a car- concave mirror to give parallel beam of light after reflection from concave mirror.
(b) Side/rear – view mirror of a vehicle.
Ans: A convex mirror is used as a side/rear-view mirror of a vehicle because:
- A convex mirror always forms an erect, virtual and diminished image of an object placed anywhere in front it.
- A convex mirror has a wider field of view than a plane mirror of the same size.
(c) Solar furnace Support your answer with reason.
Ans: Solar furnace- concave mirror to concentrate sunlight to produce heat in solar furnace.
Q.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.
Ans: When the lower half the convex lens is covered with a black paper, it still forms the complete image of the object as that with uncovered lens.
But the intensity of the image is reduced when one half of the convex lens of covered with a black paper.
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 nature of the image formed.
Ans: Given that,
u = – 25 cm
f = + 10 cm
v = ?
According to the lens formula,
We have,
1/v – 1/u = 1/f
⇒ 1/v – 1/-25 = 1/10
⇒ 1/v + 1/25 = 1/10
⇒ 1/v = 1/10 – 1/25
= 5-2/50
= 3/50
∴ v = + 50/3
= + 16.67cm.
Thus , the position of image is at a distance of 16.67 cm from the lens. The plus sign for image distance shows that the image is formed on the right side of lens and that the nature of image is real and inverted .
Now Magnification,
m = v/u
= 16.67/ – 25
= – 0.66
Now calculating the size of image h₂we have,
m = h2/h1
⇒ h₂ = m×h₁
= – 0.66×5
= – 3.3cm.
This the size of image is 3.3 cm. The negative sign of the size of image shows that the image is inverted.
Q.11. A concave lens of focal length 15 cm. forms an image 10 cm. from the lens. How far is the object from the lens? Draw the ray diagram.
Ans: Given that,
f = – 15 cm.
v = – 10 cm.
u = ?
Now,
1/v – 1/u = 1/f
⇒ 1/-10 – 1/u = 1/-15
⇒ 1/u = 1/15 – 1/10
= 2-3/30
= -130
∴ u = -30cm.
The position of image is at behind the mirror. The nature of image is virtual and erect.
We have,
m = -v/u = hight of image/hight of object
⇒ -v/u = h1/h
⇒ -8.57/-20 = h1/5
⇒ h1 = 8.57/20×5
= .57/4
= 2.14
∴ Height of the image = 2.14 cm.
∴ Size of image is 2.14 cm.
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 nature of the image.
Ans: Given that,
size of the object (h) = 7.0 cm.
u = 27 cm.
f = 18 cm.
Size of image (hⁱ) = ?
We have,
1/u + 1/v = 1/f
⇒ 1/v = 1/f – 1/u
= 1/18 – 1/27
= 3-2/54
= 1/54
∴ v = 54cm.
Again,
m = – v/u
⇒ = – 54/27
= – 2cm.
∴ Size of image (hⁱ) = m x h
= –2 x 7
= –14 cm.
∴ Size of image is 14 cm.
The image is read and inverted.
Q.16. Find the focal length of a lens of power,-2.0D. What type of lens is this?
Ans: Given that,
P = – 2.0 D
f = ?
We have,
P = 1/P
⇒ f = 1/P
= 1/-2D
= – ½ m
= – 1/2x100cm.
= – 50cm.
Since the power of the lens is negative therefore the lens is concave.
Q.17. A doctor has prescribed a corrective lines of power +1.5 D. Find the focal length of the lens. Is the prescribed lens diverging or converging?
Ans: Given that,
P = +1.5 D
f = ?
We have,
P = 1/f
P = 1/f
⇒ F = 1/p
= 1/+1.5D
= 1/+1.5
= 1/+1.5x100cm.
= + 66.7cm.
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