Light-Reflection and Refraction
page no. 8
Q1. Define the principal focus of a concave mirror.
Ans. In a concave mirror, the principal focus is a point on the principal axis of the mirror. When the light is parallel to the principal axis, after reflection, will pass through the principal focus. The principal focus is represented by the letter F.
Q2. The radius of curvature of a spherical mirror is 20cm. What is its focal length?
Ans. Give: radius(r)=20cm.
To find: Focal length(f)=?
As we know f=r/2 (relation between f and r)
putting values f=r/2
=20/2
=10cm
therefore, the focal length (f) of the spherical mirror is 10cm.
Q3. Name a mirror that can give an erect and enlarged image of an object.
Ans. The Concave mirror is a mirror that can give an erect and enlarged image of an object.
Q4. Why do we prefer a convex mirror as a rear view mirror in vehicles?
Ans. The Convex mirror is preferred for rear view mirrors in vehicles because a convex mirror forms an erect and diminished (small size) image of the objects behind the vehicle due to which field of view is increased.
page no. 11
Q1. Find the focal length of a convex mirror whose radius of curvature is 32cm.
Ans. Give: radius(r)=32cm.
To find: Focal length(f)=?
As we know f=r/2 (relation between f and r)
putting values f=r/2
=32/2
=16cm
therefore, the focal length (f) of the spherical mirror is 16cm.
Q2. A concave mirror produces three times magnified (enlarged) real image of an object placed at 10cm in front of it. Where is the image located?
Ans. As the image is real, therefore, the height of the image is taken as negative.
Given: Magnification (m) = -3 ;
Object distance (u) = -10cm.
As we known m can be expressed as m = -v/u
putting values -3 = -(v/-10)
3 = v/-10
v = -30
therefore, the image is located at a distance of 30cm to the left side of the concave mirror.
page no. 15
Q1. A ray of light traveling in air enters obliquely into water. Does the light ray bend towards the normal or away from the normal? Why?
Ans. A ray of light will bend towards the normal because the speed of light decreases when it travels from air to water. Here the air is a rarer medium and water is a denser medium.
Q2. Light enters from air to glass having a refractive index of 1.50. What is the speed of light in the glass? The speed of light in a vacuum is 3*108 ms-1.
Ans. Given: refractive index (n) = 1.50;
speed of light in vacuum (c) = 3*108 ms-1.
To find: speed of light in the glass (v) =?
As we know n = c/v
1.50 = 3*108 ms-1/v
1.50v = 3*108 ms-1
v = 3*108 ms-1/1.50
v = 2*108 ms-1
therefore, speed of light in glass (v) = 2*108 ms-1.
Table 1.3 Absolute refractive index of some material media.
|
Material Medium |
Refractive index |
Material Medium |
Refractive index |
|
Air |
1.0003 |
Canada Basalam |
1.53 |
|
Ice |
1.31 |
Rock salt |
1.54 |
|
Water |
1.33 |
Carbon disulphide |
1.63 |
|
Alcohol |
1.36 |
Dense flint glass |
1.65 |
|
Kerosene |
1.44 |
Ruby |
1.71 |
|
Fused quartz |
1.46 |
Sapphire |
1.77 |
|
Turpentine oil |
1.47 |
Diamond |
2.42 |
|
Benzene |
1.50 |
|
|
|
Crown glass |
1.52 |
|
|
We know, n = c/v
2.42 = c/v
v = c/2.42
This means the speed of light in Diamond is 1/2.42 times the speed of light in a vacuum.
page no. 22
Q1. Define 1 dioptre of power of a lens.
Ans. Dioptre is a SI unit of power of a lens. It is denoted by the letter D.1 D or 1 dioptre is the power of a lens whose focal length is 1 meter. 1D = 1m.
The power P of a lens of focal length f is given by
P = 1/f(in m).
Q2. A convex lens forms a real and inverted image of a needle at a distance of 50cm 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.
Ans. Since it is given that size of the image is equal to the size of the object or u=-v=-50 therefore, it is clear that the needle is placed in front of the lens at a distance of 50cm.
Power of lens (P) = 1/focal length(f)
to find the Power of the lens we first need to find f
Using lens formula
1/v - 1/u = 1/f
image distance (v) = 50
object distance (u) = -50
putting values
1/50 - 1/(-50) =1/f
1/50 + 1/50 = 1/f
1+1/50 = 1/f
2/50 = 1/f
1/25 = 1/f
f = 25cm =0.25m
therefore, P = 1/0.25
P = 100/25
P = 4D
Q3. Find the power of a concave lens of focal length 2m.
Ans. Given: focal length (f) of concave lens = 2m.
To find: power (P) of concave lens =?
P = 1/f
P = 1/-2m
P = -0.5D
page no. 24 Exercises
Q1. Which one of the following materials cannot be used to make a lens?
(a) Water (b) Glass (c) Plastic (d) Clay
Ans. (c) & (d)
Q2. 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 principle focus and the center of curvature.
(b) At the center of curvature.
(c) The mirror is concave and the lens is convex.
(d) Between the pole of the mirror and its principle focus.
Ans. (d)
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 principle focus of the lens.
(b) At twice the focal length.
(c) At infinity.
(d) Between the optical center of the lens and its principle focus.
Ans. (b)
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) Beyond the center of curvature.
(d) Between the pole of the mirror and its principle focus.
Ans. (a)
Q5. 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)
Q6. 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 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.
Ans. (c)
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.
Ans. As we have to obtain an erect image of an object using concave mirror of focal length 15cm, the range of the object from the mirror must be from 0 to 15. Therefore, the object should be placed between the pole and the principal focus of the mirror. NATURE of the image: Virtual and Erect. SIZE of image will be larger than the size object.
Diagram
Name the type of mirror used in the following situations,
(a) Headlights of the car,
Ans. We use concave mirror for the head lights of the car because concave mirror will produce a strong parallel beam of light when the light source is placed at principal focus.
(b) Side/rear-view mirror of a vehicle,
Ans. We use convex mirror in vehicles for side rear-view because of its largest field of view, it forms virtual, erect and diminished image of the object.
(c) Solar furnace.
Ans. We use Concave mirror for solar furnace because the light from the sun is concentrated at the focus of the mirror which produces heat.
Support your answer with reason.
Q8. 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. Yes, it will produce a complete image of the object. This can be verified experimentally by observing the image of a distant object, When the lower half of the lens is covered with a black paper, 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 object as shown in the figure. However, the intensity or brightness of image will reduce.
Q9. An object 5cm in length is held 25cm away from a converging lens of focal length 10cm. Draw the ray diagram and find the position, size and the nature of the image formed.
Ans. Given: Object size (h1) = 5cm
Object distance (u) = -25cm
Focal length of Lens (f) = 10cm
To find: Image distance (v) & Image size (h2)
Sol: Using lens formula: 1/v - 1/u = 1/f
1/v - 1/(-25) = 1/10
1/v + 1/25 = 1/10
1/v = 1/10 - 1/25
1/v = 5-2/50
1/v = 3/50
v = 50/3
v = 16.6cm
As v is positive, the image formed will be real.
magnification (m) = v/u
= 16.6/-25
= -0.66cm
m is also equal to h2/h1
Therefore, h2/h1 = -0.66
h2/5 = -0.66
h2 = -0.66/5
h2 = -3.3cm
The negative sign shows the image is inverted.
Ans. Given : Focal length (f) = -15cm.
Image distance (v) = -10cm.
To Find : Object distance (u) = ?
Sol : Using lens formula: 1/v - 1/u = 1/f
1/-10 - 1/u = 1/-15
-1/u = 1/-15 + 1/10
-1/u = -2+3/30
-1/u = 1/30
u = -30
Q11. 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 image.
Ans. Given : Focal length (f) = 15cm.
Object distance (u) = -10cm.
To Find : Image distance (v) = ?
Sol : Using Mirror formula: 1/v + 1/u = 1/f
1/v + 1/(-10) = 1/15 1/v = 1/15 + 1/10
1/v = 2+3/30
1/v = 1/6
v = 6
here, + sign denotes the image is at the back of the mirror. It must be virtual, erect, and smaller in size than the object.
Q12. The magnification produced by a plane mirror is +1. What does this mean?
Ans. The magnification produced by a plane mirror is +1, this means h2=h1 i.e. size of image is equal to size of object + sign of m indicates that the image is erect and hence virtual.
Q13. 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.
Ans. Given : Object size (h1) = 5.0cm.
Object distance (u) = -20cm.
Radius of curvature (R) = 30cm.
To Find : Image distance (v) =?
Image size (h2) =?
Sol : 1/v+1/u = 1/f
since f = R/2
1/f = 2/R
therefore, 1/v+1/u = 2/R
b 1/v+1/(-20) = 2/30
1/v-1/20 = 2/30
1/v = 2/30+1/20
1/v =4+3/60
1/v =7/60
v =60/7
Here, positive sign of v indicates that the image is at the back of the mirror and is virtual and erect.
As m = h2/h1 = -v/u
= h2/5 = -60/7/-20
= h2/5 = 60/7 * 1/20
= h2 = 3/7 * 5/1
= h2 = 15/7
Q14. 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 focused image can be obtained? Find the size and nature of the image.
Ans.
Given : Focal length (f) = -18cm.
Object size (h1) = 7cm.
Object distance (u) = -27cm.
To Find : Image distance (v) = ?
Image size (h2) = ?
Sol : 1/v+1/u = 1/f
1/v = 1/f -1/u
1/v = 1/-18 - 1/(-27)
1/v = 1/-18 + 1/(27)
1/v = -3+2/54
1/v = -1/54
v = -54
therefore, the screen should should be held in front of the mirror at a distance of 54cm from the mirror. The image obtained on screen will be real.
As m = h2/h1 = -v/u
= h2/7 = -(-54)/-27
=h2 = 54/-27*7
=h2 = 2/-1*7
=h2 = -14cm
Here, negative sign of h2 shows that the image is inverted.
Q15. Find the focal length of a lens of power -2.0 D. What type of lens is this?
Ans.
Given : Power (p) = -2D
To Find : focal length (f) = ?
Q16. 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?
Ans.
Given : Power (p) = +1.5D
To Find : focal length (f) = ?
f = 66.6cm
There, the prescribed lens is converging.
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