Page 168
Question 1. Define the principal focus of a concave mirror.
Answer 1 Principal focus or focal point: It is a point where incident rays which are parallel to principal axis get converged after reflection from the mirror.
The principal focus always lies on the principal axis, and all the rays which is parallel to principal axis will pass through principal focus after reflection.
Question 2. The radius of curvature of a spherical mirror is 20 cm. What is its focal length?
Answer 2 Radius of curvature, R = 20 cm
Radius of curvature of a spherical mirror, R = 2 × focal length (f)
R = 2f
20 = 2f
⇒ f = 20/ 2 = 10 cm
Therefore, the focal length of given spherical mirror is 10 cm.
Question 3. Name a mirror that can give an erect and enlarged image of an object.
Answer 3 A concave mirror will give an erect, enlarged and virtual image of an object. To obtain an erect, virtual and enlarged image from a concave mirror the object must be placed in between the principal focus and pole of the mirror.
Question 4. Why do we prefer a convex mirror as a rear-view mirror in vehicles?
Answer 4 Rear view mirrors should have a large field of view so that drivers can easily see most of the traffic behind them. Convex mirrors always form, diminished, erect, and virtual images of the objects.
Page 171
Question 1. Find the focal length of a convex mirror whose radius of curvature is 32 cm.
Answer 1 The relation between radius of curvature (R) and focal length (f) is:
R = 2f
⇒ 32 =2f
f= 32/2 = = 16 cm
Question 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 2 Magnification produced by a spherical mirror is given by the relation,
Magnification = height of the image / height of the object
Image distance / Object distance
m = h’/h = -v/ u
If height of object is h,
⇒height of image is -3h (image formed is real)
Object distance is u = -10 cm
Image distance v =?
Put the values in the formula
m=-3h/h = v/(-10) ⇒ v = -30 cm
The negative sign denotes that an inverted image is formed in front of the concave mirror at a distance of 30 cm from the pole.
Page 176
Question 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 1 The light ray bends towards the normal.
When a ray of light travels from an optically rarer medium to an optically denser medium, it gets bent towards the normal. Water (1.33) is optically denser than air (1.0003), a ray of light travelling from air into the water will bend towards the normal.
Question 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 2 Refractive index of a medium nm is given by,
nm = = Speed of light in vacuum / Speed of light in the medium = c / v
Speed of light in vacuum, c = 3 x 10 8 ms-1
Refractive index of glass, ng = 1.50
Speed of light in the glass,
v = c/ ng
= 3 x 108 / 1.50
= 2 x 108 ms-1
Question 3. Find out, from Table 10.3, the medium having highest optical density. Also find the medium with lowest optical density.
Table: Absolute refractive index of some material media
| Material Medium | Refractive Index | Material Medium | Refractive Index |
| Air | 1.0003 | Canada balsam | 1.53 |
| Ice | 1.31 | Rock Salt | 1.54 |
| Water | 1.33 | ||
| Alcohol | 1.36 | ||
| Kerosene | 1.44 | Carbon disulphide | 1.63 |
| Fused quartz | 1.46 | Dense flint glass | 1.65 |
| Turpentine oil | 1.47 | Ruby | 1.71 |
| Benzene | 1.50 | Sapphire | 1.77 |
| Crown gas | 1.52 | Diamond | 2.42 |
Answer 3
Optical density ∝ Refractive index
Optical density of a medium is directly related with the refractive index of that medium. A medium which has the highest refractive index will have the highest optical density and vice-versa.
It can be observed from the Table that diamond and air, respectively have the highest and lowest refractive index. Therefore, diamond has the highest optical density and air has the lowest optical density.
Question 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 4 Speed of light in a medium is given by the relation for refractive index (nm).
The relation is given as:
nm = Speed of light in vacuum / Speed of light in the medium = c/v
v = c/ nm
⇒ v ∝ 1/nm
It can be inferred from the relation that light will travel the slowest in the material which has the highest refractive index and travel the fastest in the material which has the lowest refractive index. It can be observed from the Table that the refractive indices of kerosene, turpentine, and water are 1.44, 1.47, and 1.33, respectively. Therefore, light travels the fastest in water.
Question 5. The refractive index of diamond is 2.42. What is the meaning of this statement?
Answer 5 Refractive index of a medium nm is related to the speed of light in that medium v by the relation:
nm = = Speed of light in vacuum / Speed of light in the medium =c/v
Where, c is the speed of light in vacuum/air.
The refractive index of diamond is 2.42.
This suggests that the speed of light in diamond will reduce by a factor 2.42 compared to its speed in air.
Page 184
Question 1. Define 1 dioptre of power of a lens.
Answer 1 Power of lens is defined as the reciprocal of its focal length. If P is the power of a lens of focal length F in meters, then
P = 1 /f (in meters)
The S.I. unit of power of a lens is Dioptre. It is denoted by D.
1 dioptre is defined as the power of a lens of focal length 1 metre.
Therefore, 1 D=1 m−1
Question 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 2 When an object is placed at the centre of curvature, 2F1, of a convex lens, its image is formed at the centre of curvature, 2F2, on the other side of the lens. The image formed is inverted and of the same size as the object, as shown in the given figure.
It is given that the image of the needle is formed at a distance of 50 cm from the convex lens.
Hence, 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
According to the lens formula,
1/v – 1/u = 1/f
1/f= 1/50 – 1/(-50)=1/25
f= 14cm=0.25m
power of lens
P=1/f(in meters) = 1/0.25 = +4D
Question 3. Find the power of a concave lens of focal length 2 m.
Answer 3 Focal length of concave lens, f = 2 m
Power of a lens, P = 1 /f(in meters)
=1/-2 = -0.5 D
Here, negative sign arises due to the divergent nature of concave lens. Hence, the power of the given concave lens is D.
Page 185
Question 1. Which one of the following materials cannot be used to make a lens?
(a) Water (b) Glass (c) Plastic (d) Clay
Answer 1 (d) clay
A lens is a transparent material that means light can pass through it. Since, clay is not transparent so, it cannot be used to make a lens.
Question 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 2 (d) Between the pole of the mirror and its principal focus.
Image formed by a convex mirror is always virtual, erect, and diminished in size than the object. Similarly, a plane mirror always produces a virtual and erect image of same size as that of the object.
Question 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 3 (b) At twice the focal length
When an object is placed at the centre of curvature in front of a convex lens, its image is formed at the centre of curvature on the other side of the lens. The image formed is real, inverted, and of the same size as the object.
Question 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 4 (a) both concave.
As per the sign convention, the focal length of a concave mirror and a concave lens are taken as negative. Hence, both the spherical mirror and the thin spherical lens are concave in nature.
Question 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 5 (d) either plane or convex
Image formed by a convex mirror is always virtual, erect, and diminished in size than the object. Similarly, a plane mirror always produces a virtual and erect image of same size as that of the object.
Question 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 6 (c) A convex lens of focal length 5 cm
A magnified image of an object will be obtained when it is placed between the radius of curvature and focal length of a convex lens. Magnification is also higher for convex lenses having shorter focal length. Therefore, for reading small letters, a convex lens of focal length 5 cm should be used.
Question 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 7 A concave mirror gives an erect, virtual and enlarged image if the object is placed between the pole (P) and principal focus (F). The given focal length is 15 cm. So, to obtain an erect the object must be placed anywhere in between P and F, that is, between 0 cm and 15 cm. Since all the distances are measured from pole so at pole distance is zero. The image formed will be virtual and magnified. So, range of distance of object is 0 – 15 cm.
Question 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 8
(a) Mirror used: Concave
Concave mirror produces powerful bright and parallel beam of light if the light source (bulb; in this case) is placed at their principal focus i.e., near the reflector.
(b) Mirror used: Convex
Convex mirror produces a virtual, erect, and diminished image of the objects placed in front of it. So, thereby providing a wide field of view. It enables the driver to see most of the traffic behind him/her.
(c) Mirror used: Concave
Concave mirrors are convergent mirrors. That is why they are used to construct solar furnaces. Concave mirrors converge the light incident on them at a single point known as principal focus. Hence, they can be used to produce a large amount of heat at that point.
Question 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 9 The convex lens will form complete image of an object, even if it’s one half is covered with black paper. It can be understood by the following two cases.
Case I: When the upper half of the lens is covered In this case, a ray of light coming from the object will be refracted by the lower half of the lens. These rays meet at the other side of the lens to form the image of the given object, as shown in the following figure.
Case II: When the lower half of the lens is covered In this case, a ray of light coming from the object is refracted by the upper half of the lens. These rays meet at the other side of the lens to form the image of the given object, as shown in the following figure.
Question 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 10 Given, height of object = 5 cm
Position of object, u = – 25 cm
Focal length of the lens, f = 10 cm
Hence, position of image, v =?
We know that, 1/v – 1/u = 1/f
1/v +5 = 1/10
So, 1/v = 1/10-5
So, 1/v = (5-2)/50
That is , 1/v = 3/50
So, v = 50/3 = 16.66 cm
Thus, distance of image is 16.66 cm on the opposite side of lens.
Now, magnification = v/u
m = 16.66/-25 = -0.66
Also, m= height of image/height of object
-0.66 = height of image/ 5 cm
Therefore, height of image = -3.3 cm
The negative sign of height of image shows that an inverted image is formed.
Thus, position of image = At 16.66 cm on opposite side of lens
Size of image = – 3.3 cm at the opposite side of lens.
Nature of image: Real and inverted.
Question 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 11 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= -1/10 – 1/(-15) = -1/10 + 1/15 = (-3+2)/30= -1/30
u= – 30 cm
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.
Question 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 12 Focal length of the convex mirror, f= 15 cm
Object distance, u= -10cm
According to the mirror formula,
1/v + 1/u = 1/f
1/v = 1/f -1/u= 1/15 + 1/10 = 25/ 150
v= 6 cm
The positive value of v indicates that the image is formed behind the mirror.
magnification = v/u
m=-6 /-10
m= +0.6
The positive value of magnification indicates that the image formed is virtual and erect.
Question 13. The magnification produced by a plane mirror is +1. What does this mean?
Answer 13 Magnification produced by a mirror is given by the relation
Magnification= Image height (H1) / Object height (H0)
The magnification produced by plane mirror is +1.It shows that image formed by plane mirror is of the same size as that of the object.It also shows that image formed is virtual and erect.
Question 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 14
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 = R/2 = 15 cm
According to the mirror formula
1/f= 1/v + 1/u
1/u= 1/f – 1/u =1/15 -1/-20
⇒ 1/v = 1/15 + 1/20 = 4+3/60 = 7/60
v = 8.57 cm
The positive value of v indicates that the image is formed behind the mirror.
Magnification, m = Image distance / Object distance
m = -8.57/ -20 = 0.428
Positive value of magnification shows that image formed is virtual.
Magnification, m= Height of the image/ Height of the object
m= h’ /h
h= m x h = 0.428 x 5 = 2.14 cm
The height of image is positive which shows that image formed is erect.
So, a virtual, erect and smaller than object image is formed.
Question 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 15
Object distance, u = -27 cm
Object height, h = 7 cm
Focal length f = -18 cm
Using the mirror formula
1/v + 1/u =1/f
1/v = 1/f – 1/u = 1 / (-18) – 1/ (-27)
1/v = 1/ (-18) + 1/ 27 = (-3+2)/ 54 = 1/54
v = -54 cm
The screen should be placed at a distance of 54 cm in front of the given mirror.
Magnification, m = Image distance/ Object distance = -54/-27 = -2
Negative value of magnification shows that image is real.
magnification, m = – Height of image /Height of object = h’/h
⇒h’ = m x h = (-2) x 7 = -14 cm
The height of image is negative which shows that image is inverted.
Question 16. Find the focal length of a lens of power – 2.0 D. What type of lens is this?
Answer 16 Power of lens, P = 1/ f (in meters)
P = -2D
f = -1/2 = -0.5 m
A concave lens has a negative focal length. Hence, it is a concave lens.
Question 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 17 Power of lens, P = + 1.5 D
P = 1 /F
1.5 = 1 /F
F = 1/ 1.5
F = 10/ 15
F =+0.67 m
The convex lens has a positive focal length .Hence, it is a convex lens or a converging lens.
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