QUESTIONS
1. Define the principal focus of a concave mirror.
Answer:
A number of rays parallel to the principal axis are falling on a concave mirror. They are all meeting/intersecting at a point on the principal axis of 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:
Focal length = Radius of curvature ÷ 2
or, Focal length = ±20÷2
or, Focal length = ±10 cm.
3. Name a mirror that can give an erect and enlarged image of an object.
Answer:
Concave Mirror.
4. Why do we prefer a convex mirror as a rear-view mirror in vehicles?
Answer:
Convex mirrors are preferred because they always give an erect, though diminished, image. Also, they have a wider field of view as they are curved outwards. Thus, convex mirrors enable the driver to view much larger area than would be possible with a plane mirror.
1. Find the focal length of a convex mirror whose radius of curvature is 32 cm.
Answer:
Focal length = Radius of curvature = Radius of curvature÷2
or, Focal length = 32÷2
or, Focal length = 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:
Height of image÷Height of object = -(Image distance÷Object distance)
or, 3÷1 = -(Image distance÷10)
or, Image distance = -(3x10)
or, Image distance = 30 cm.
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?
Answer:
The ray of light bends towards the normal.
Water is denser with respect to air.
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 m s–1.
Answer:
Refractive index of glass with respect to air or vacuum = Speed of light in vacuum ÷ speed of light in glass.
or, 1.5 = 3×10⁸÷speed of light in glass.
or, Speed of light in glass = 3×10⁸÷1.5.
or, Speed of light in glass = 2×10⁸ m/s Ans.
3. Find out, from Table 10.3, the medium having highest optical density. Also find the medium with lowest optical density.
Answer:
Diamond having highest optical density whereas air with lowest 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 10.3.
Answer:
Water having lowest refractive index out of these three. So, the light travels fastest in the water.
5. The refractive index of diamond is 2.42. What is the meaning of this statement?
Answer:
This means that the ratio of the speed of light in air and the speed of light in diamond is equal to 2.42.
1. Define 1 dioptre of power of a lens.
Answer:
The power of a lens is the reciprocal of its focal length. The unit of power of lens is dioptre when focal length is in metre. 1 dioptre means the focal length is 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:
From question it is clear that
v = 50 cm.
Height of the image = Height of the object
So, v÷u = 1
or, u = -50 cm.
As per image formation property
Convex lens forms real and inverted image of eual size at 2f.
So, 2f = 50 cm.
or, f = 50÷2
or, f = 25 cm.
or, f = 25÷100 m.
or, Power of lens = 1÷f
or, Power of lens = 100÷25
or, Power of lens = 4 dioptre.
3. Find the power of a concave lens of focal length 2 m.
Answer:
From question it is clear that
f = -2 m.
Power of lens = 1÷f
Power of lens = 1÷(-2)
or, Power of lens = -0.5 dioptre.
EXERCISES
1. Which one of the following materials cannot be used to make a lens?
(a) Water (b) Glass (c) Plastic (d) Clay
Answer
Clay cannot be used to make a lens.
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
The correct option will be (d)
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
The correct option will be (b)
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.
Answer
The correct option will be (a)
5. No matter how far you stand from a mirror, your image appears erect. The mirror is likely to be
(a) only plane.
(b) only concave.
(c) only convex.
(d) either plane or convex.
Answer
The correct option will be (d)
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
The correct option will be (b)
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
The range of distance should be less than 15 cm.
The nature of image will be virtual.
The image is larger than object.
The ray diagram is given below
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 is used for headlights of car because it converges the ray of light in its focal length.
(b)
Convex mirror is used for side/rear-view mirror of a vehicle because it has large view range and erract and virtual image is formed by it.
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
This lens will produce a complete image when object is placed between optical centre and main focus of it.
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
From question it is clear that
u = -25 cm.
Length of object = 5 cm
f = 10 cm.
From lens formula we know that
1/v = 1/f + 1/u
1/v = 1/10 + 1/-25
or, 1/v = (5 - 2)÷50
or, 1/v = 3/50
or, v = 50/3 cm.
Length of image = Length of object×v÷u
or, Length of image = 5×50÷(25×3)
or, Length of image = 10/3 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.
Answer
From question it is clear that
f = -15 cm
v = -10 cm
u = ?
1/u = 1/v - 1/f
or, 1/u = -1/10 + 1/15
or, 1/u = (-3+ 2)/30
or, 1/u = -1/30
u = -30 cm.
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
From question it is clear that
u = -10 cm
f = 15 cm
v = ?
From mirror formula
1/v = 1/f - 1/u
or, 1/v = 1/10 + 1/15
or, 1/v = (3 + 2)/30
or, 1/v = 5/30
or, 1/v = 1/6
v = 6 cm.
Nature of image will be virtual.
13. The magnification produced by a plane mirror is +1. What does this mean?
Answer
Plane mirror makes virtual image of equal height.
14. 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
From question it is clear that
u = -20 cm
f = 30/2 = 15 cm
v = ?
From mirror formula
1/v = 1/f - 1/u
or, 1/v = 1/15 + 1/20
or, 1/v = 7/60
or, v = 60/7 cm.
Image will be virtual.
Size of image = Size of object × v/u
or, Size of image = 5×60÷20×7
or, Size of image = 15/7 cm.
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 the nature of the image.
Answer
Size of object = 7 cm
From mirror formula
1/v = 1/f - 1/u
or, 1/v = -1/18 + 1/27
or, 1/v = (-3+2)/54
or, 1/v = -1/54
or, v = -54 cm
Nature of image will be real.
Size of image = Size of object×image distance÷object distance
or, Size of image = 7×54÷27
or, Size of image = 14 cm.
16. Find the focal length of a lens of power – 2.0 D. What type of lens is this?
Answer
Focal length = 1/Power of lens
or, f = 1/-2
or, f = -0.5 m
This is type of 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
f = 1/P
or, f = 1/1.5
or, 0.66 m
The prescribed lens is converging.
