Selina Concise Class 9 Physics Solution Chapter No. 7- ‘Reflection of Light’ For ICSE Board Students.
Exercise-A Solution
1.) Solution:
Ans:
When a ray of light is incident on any surface then some part of light is reflected back in the same medium. This phenomenon is called as the reflection of light.
2.) Solution:
Ans:
The light enters from of the light the side of the polished surface and most of the incident light on plane mirror is reflected from the silvered surface.
3) Solution:
Ans:
a) Plane mirror:
The plane mirror is the mirror whose one surface is highly polished while the other surface is coated with silver. The most of the incident light from the plane mirror is reflected from the silvered surface. And plane mirror reflects all the incident light on it.
b) incident ray:
When the beam of light is striking to the surface of reflection is called the incident ray.
c) reflected ray:
When the light rays are incident on the reflecting surface then the rays which are bounced back in the same medium are called as the reflected rays.
d) angle of incidence:
It is the angle made by the incident ray with the normal at the point of incidence is called as the angle of incidence.
Angle of incidence is denoted by i.
e) angle of reflection:
It is the angle made by the reflected ray with normal at the point of incidence is called as the angle of reflection.
Angle of reflection is denoted by r.
The following diagram shows the incident ray, reflected ray, angle of incidence and angle of reflection.
4.) Solution:
Ans:
Regular reflection:
- When a beam of light is incident on the smooth polished surfaces such as the surface of plane mirror then only regular reflection occurs.
- In regular reflection, when we incident the parallel beam of light then the reflected beam also contains the parallel rays in fixed direction as shown in figure.
- To see regular reflection we have to see it through only particular direction.
Irregular reflection:
- When a beam of light is incident on the rough or normal surface such as wall of room then the reflection occurs is called as the irregular reflection.
- If we incident the parallel beam of light on the rough surface then there will be reflection at different points takes place and light rays are reflected in different directions which are not parallel to each other as shown in figure.
- As the reflected light is spreading everywhere which is not in fixed direction we can see the irregular reflection from any direction.
5.) Solution:
Ans:
- The reflection of light from the plane mirror is the regular reflection of light. As the incident parallel beam is reflected as parallel beam of light in fixed direction.
- Plane mirror has smooth reflecting surface due to which there is a regular reflection of light takes place.
- While plane of sheet paper acts as the rough surface where the incident parallel beam of light is reflected in irregular manner in different directions.
- Hence, light reflection from plane sheet of paper is the irregular reflection of light.
6.) Solution:
Ans:
When a ray of light is incident on the surface then it bounce back in the same medium which is called as the reflection of light.
There are two laws of reflection of light which are stated as follows.
1) The angle of incidence <i and the angle of reflection <r are always equal.
2) The incident ray, normal and the reflected ray at a point of incidence are lying in the same plane.
7.) Solution:
Ans:
When a ray of light is incident on the surface then it bounce back in the same medium which is called as the reflection of light.
There are two laws of reflection of light which are stated as follows.
1) The angle of incidence <i and the angle of reflection <r are always equal.
2) The incident ray, normal and the reflected ray at a point of incidence are lying in the same plane.
Following is the experimental verification of laws of reflection:
- We have fixed the white paper sheet on the drawing board and we have drawn a line PQ as shown in figure.
- We have taken a point O exactly at the middle of line PQ and a line OA is drawn in a such way that <POA must be acute angle.
- Then we have drawn a normal line ON on the line PQ at point O.
- And we have placed a plane mirror with its silvered surface on the line PQ vertically as shown in figure.
- Now, we have fixed the two points X and Y which are 5cm apart on the line OA which is on the board.
- Now we see the images of points X and Y on the other side of normal as X’ and Y’ with eye.
- Now we have fixed the two pins S and R along the line joining images X’ and Y’ as shown in figure.
- Now, by removing pins S and R we have drawn a small circles to note points and a line OB which is joining points O, S and R is drawn as shown in figure.
- From figure we can say that, OA is the incident ray and OB is the reflected ray.
- <AON is the angle of incidence and <BON is the angle of reflection.
- Now, we have recorded the angel of incidence and angle of reflection for different angle values of <POA.
- The following table shows the observations we have taken.
- From the table shown above we can say that for each angle of incidence the angle of reflection is same.
- Thus, first law of reflection is verified.
- And as the mirror is normal to the plane of paper on which we have fixed the pins which all are in the same plane. Thus, we can say that incident ray, normal and reflected ray all lie in the same plane.
Hence, second law of reflection is also verified.
8.) Solution:
Ans:
a)
When a ray of light is incident normally on the plane mirror then the angle made by incident ray with the normal is the 0°.
Hence, angle of incidence here is 0°as shown in following figure.
b)
As the incident ray is incident normally on the plane mirror then the direction of reflected ray is same as the incident ray as shown in figure.
9.) Solution:
Ans:
The following figure shows the reflection of light from the plane mirror. In which incident ray, normal, reflected ray, angle of incidence and angle of reflection are represented.
10.) Solution:
Ans:
a)
From figure, it is observed that the angle which incident ray OA makes with mirror is 30°.
That means, the angle of incidence that is the angle made by incident ray with the normal is given by,
<i= 90° – 30° = 60°
Thus, here angle of incidence is 60°.
b)
The ray OB shows the reflected ray which also makes angle 60° with the normal. Since according to first law of reflection the angle of incidence and the angle of reflection are always equal.
Hence, here <BON= r
Now, the angle between incident ray and reflected ray is given by,
<AOB= <AON + <BON
= 60° + 60°
=120°
Thus, the angle between incident ray and reflected ray is 120° here, as shown in figure.
11.) Solution:
Ans:
a) and b)
The following ray diagram shows the formation of image.
c)
As the reflected rays meets only when they are extended backwards. Hence, the image formed is virtual.
12.) Solution:
Ans:
13.) Solution:
Ans:
a)
Following are the characteristics of image formed by the plane mirror:
- The image is upright or erect.
- The image formed is virtual.
- The image is of same size as the object size.
- The image formed is laterally inverted.
b)
The position of the image formed by the plane mirror is at the same perpendicular distance behind the plane mirror as the object is placed in front of it.
14.) Solution:
Ans:
Real image:
- When there is a actual intersection of reflected rays then there will be real image is formed.
- As the image is real we can see that image on screen.
- A real image is inverted with respect to the object we have taken and placed.
For example:
The image formed in a concave mirror of distant objects is the real image.
Virtual image:
- When the reflected rays meets only when we extend the them in backward direction, then only virtual image is formed.
- As the image is virtual we can not see it on the screen.
- A virtual image is the erect with respect to the object taken and placed.
For example:
The image formed by the plane mirror is the virtual image.
15.) Solution:
Ans:
- When we see in mirror which is the plane mirror, if there is dark spot on our right side cheek then in mirror it will see on left side cheek. Also, the pocket on our shirt which is on left side we see it in mirror on the right side. This interchanging of left and right sides in the image of plane mirror is the lateral inversion.
- In simple words, the interchange of right and left sides in the image taken by plane mirror is called as the lateral inversion.
- The following figure shows the lateral inversion in case of plane mirror in detailed format.
- In figure, we see that image of letter p is seen to be q in the image in plane mirror. This explains the lateral inversion of an image in a plane mirror.
16.) Solution:
Ans:
- We know that, the image formed by the plane mirror shows lateral inversion that means there is a interchange of right and left sides in the image formed.
- Due to lateral inversion the letter are difficult to read in plane mirror. But, if write the letters laterally inverted then the image formed will be easily readable.
- Because of this reason on the front of ambulance the letters are written laterally inverted, so that the drivers in the vehicles which are moving ahead of the ambulance can read the letters fastly in the image formed in their rear view mirror and easily and they gives way to the ambulance to pass first.
17.) Solution:
Ans:
- It is difficult to read the image of a text formed by reflection due to plane mirror because the image formed in plane mirror shows lateral inversion.
- Due to lateral inversion we see the text letters inverted in the image and hence we can’t read them easily and it is more difficult to read.
Multiple choice type:
1.) According to the law of reflection
Ans: d) i = r
2.) The image formed by the plane mirror is
Ans: d) erect and of same size
3.) The image formed by the plane mirror is
Ans: c) virtual with lateral inversion
Numerical:
1.) Solution:
Ans:
From the given scenario,
The angle between incident ray and reflected ray is 90°.
But, according to laws of reflection the angle of incidence and angle of reflection must be equal.
So from figure we an write,
<i + 90° + <r = 180°
But, <i = <r
Hence, 2<i + 90° = 180°
2(<i) = 180° – 90° = 90°
Hence, <i = 90°/2 = 45°
Thus, here the angle of incidence and the angle of reflection is 90°.
2.) Solution:
Ans:
Given that, the distance between man and its image is 6m.
Hence, we can write
Distance between man and his image = distance of man from mirror + distance of image of man from mirror
But, for plane mirror the image distance and the object distance both are equal.
Hence, distance of a man from plane mirror= 6/2 = 3m
4.) Solution:
Ans:
Given that, the object is placed at 60cm in front of the plane mirror. Then the image will be formed at a distance of 60cm behind the mirror.
Thus, the distance between object and its image = 60+ 60= 120cm
Now, the mirror is moved 25 cm away from the object that means the distance of object from mirror is = 60 + 25= 85 cm
Now, in this case the distance of image formed from the plane mirror is also 85cm.
Thus, the distance between the object and the image is = 85 + 85= 170cm
Thus, the distance through which the image is shifted from its previous position= 170 – 120= 50 cm
Thus, the image is shifted by 50 cm when the mirror is moved away from the object by 25 cm.
5.) Solution:
Ans:
Given that,
The distance between man and the chart is 3m.
The distance between man and the mirror is 2m.
Hence, the distance between chart and mirror is = 3+2= 5m
As the distance of man is 2m from the mirror hence image will be formed at a distance of 2m behind the plane mirror.
Hence, according to patient the image will be formed at distance (5+2)= 7m.
Exercise-B Solution:
1.) Solution:
Ans:
When the two plane mirrors are placed making an angle a between them and if a object is placed between them then there will be many images formed. Because light reflected from one mirror incident on other and the process continues which produces many number of images.
And the number of images formed depends on the angle a between the two plane mirrors.
There will be two cases possible which are discussed as follows.
Case1)
- When the angle between the two plane mirrors is such that n= 360°/a becomes odd.
- Then, the number of images will be n, when object is placed asymmetrically between the two plane mirrors.
- And the number of images will be (n – 1), when the object is placed symmetrically between the two plane mirrors.
Case2)
- If n= 360°/a becomes even, then the number of images formed will be always (n – 1) independent of the position of the object placed between the two plane mirrors.
2.) Solution:
Ans:
- When two plane mirrors are placed making an angle a between them and a object is placed in between them.
- Then the number of images formed will be depends on the angle a between the mirrors and the number of images formed will be given by,
- n= 360°/a
- If the value of angle a is changed from 0° to 180° then the value on n here decreases.
- Hence, the number of images formed will decreases.
3.) Solution:
Ans:
- When the two plane mirrors are kept perpendicular to each other and if a object is placed between them then the number of images formed is given by,
- n= 360°/a
- Where, a is the angle between the plane mirrors which is 90° here.
- Hence, n= 360°/90= 4
- But, for the object placed in between two perpendicular mirrors the number of images formed is (n – 1) = 4 – 1= 3
- Thus, there will be three images are formed.
- The following figure shows when two plane mirrors are placed perpendicular to each other and a object P is placed between them then there will be three images P1, P2 and P3 are formed.
4.) Solution:
Ans:
- When the two plane mirrors are placed parallel with facing each other at some separation and if the point object is placed in between them then the number of images formed will be given by,
- n= 360°/a
- Where, a is the angle between the two plane mirrors.
- But, here two plane mirrors are placed parallel to each other.
- Hence, a= 0°
- Thus, number of images formed is n= 360°/0°= infinity
- Thus, the number of images of a point object placed in between the two parallel mirrors are infinite.
5.) Solution:
Ans:
Following are the uses of plane mirror:
- We all know that, we can use plane mirror mostly for looking purpose.
- In the periscope, two plane mirrors incline with an angle of 45° with the vertical wall are placed which are facing to each other.
- In barber’s shop also two plane mirrors facing with each other are placed to see hairs on the backside.
Multiple choice type:
1.) Two plane mirrors are placed making an angle 60° in between them. For an object placed in between the mirrors, the number of images formed will be
Ans: c) 5
2.) In barber’s shop two plane mirrors are placed
Ans: b) parallel to each other
Numerical:
1.) Solution:
Ans:
- Given that, the two plane mirrors are placed inclined to each other and a object is placed in between them then the number of images formed will be given by,
- n= 360°/a
- Where a is the angle between two plane mirrors.
- In first case, the angle between two plane mirrors is 90°.
- Hence, n = 360°/90°= 4
- Here, 4 is even number.
- Hence, the number of images formed will be (n – 1) = (4 – 1) = 3
- Thus, when two plane mirrors are perpendicular to each other then there will be 3 images are formed.
- In second case, the angel between the two plane mirrors is 60° then the number of images formed will be given by,
- n= 360°/60° = 6
- Here, 6 is even number.
- Hence, the number of images formed will be (n – 1)= (6 – 1) = 5
- Thus, when the two plane mirrors are placed at an angle of 60° then there will be 5 images are formed.
2.) Solution:
Ans:
Given that, two plane mirrors are placed at an angle of 50° with a object is placed in between them.
a)
When the object is placed symmetrically then the number of images formed will be given by
n= 360°/a= 360°/50° = 7.2~7
Here, 7 is the odd.
As the object is placed symmetrically then the number of images formed will be n= 7.
b)
When the object is placed asymmetrically between the two plane mirrors then the number of images formed will be,
(n -1) = (7 – 1) = 6.
Exercise-C Solution
1.) Solution:
Ans:
When the reflecting surface of mirror is in the form of sphere then it is called as the spherical mirror.
2.) Solution:
Ans:
Spherical mirrors are classified in two types on the basis of inner or outer surface is silvered as follows.
Concave mirror:
- In case of concave mirror, the outer surface of hollow sphere is silvered and hence reflection from inner surface takes place.
- The incident rays on the inner surface of concave mirror after reflection converges at a point.
- In concave mirror, the image formed may be real or virtual. When object is placed between focus and pole the image formed will be virtual and for other cases the image formed is always real.
- The image formed is diminished when object is away from the centre of curvature, the image formed will be of same size when the object is at centre of curvature and the image formed will be magnified when the object is within the centre of curvature.
Convex mirror:
- In convex mirror, the inner surface is silvered so that the reflection of light form outer bulging surface takes place.
- The light rays incident on the outer bulging surface of convex mirror get diverged after reflection.
- The image formed in convex mirror is the virtual image always for all positions of the object placed in front of the convex mirror.
- Also, the image formed is diminished always for all positions of the object placed in front of convex mirror.
3.) Solution:
Ans:
Pole:
- Pole is the geometrical centre of the spherical surface of spherical mirror.
- It is the central point of the spherical mirror and represented by P.
Principal axis:
- The line joining the pole of the mirror and its centre of curvature is called as the principal axis of the spherical mirror.
Centre of curvature:
- Centre of curvature is nothing but the centre of sphere which forms the surface of spherical mirror.
- It is represented by C.
4) Solution:
Ans:
Following figure shows that, when a beam light is incident parallel to principal axis then after reflection the ray of light passes or appears to be coming from focus in spherical mirrors.
5.) Solution:
Ans:
The following figure shows that, the convex mirror diverges the rays of light incident on it. While the concave mirror converges the rays of light incident on it as shown in figure.
6) Solution:
Ans:
- We know that, in case of concave mirror when the beam of light is incident parallel to the principal axis then after reflection these rays converges at a point on the principal axis. And this point on the principal axis of concave mirror is called as the focus of concave mirror. For concave mirror the focus is real as after reflection the rays actually meets at that point.
- It is represented by F.
- In following figure the F indicates the focus of concave mirror.
- And the distance of the focus F from the pole P of the concave mirror is called as the focal length of the concave mirror and it is represented by f.
- Here, focal length= f = PF
7) Solution:
Ans:
- We know that, when a beam of light is incident parallel to the principal axis in case of convex mirror then the rays after reflection diverges and can’t meet at a point.
- But, these reflected ray appears to be comes out from the point F which lie on the principal axis behind the convex mirror as shown in figure.
- This point is called as the focus of the convex mirror and it is represented by F.
- When these reflected rays produced backwards then only focus is obtained geometrically.
- Hence, we can say that the focus of convex mirror is virtual.
- And the distance of this focus F of convex mirror from the pole P of the convex mirror is called as the focal length of the convex mirror and it is denoted by f.
- Here, focal length f= PF
8.) Solution:
Ans:
The direction of incident ray which after reflection from spherical mirror retraces its path is directed towards the centre of curvature because here the ray incident is normal to the spherical mirror and hence angle of incidence and angle of reflection both are zero.
9) Solution:
Ans:
a)
In following figure a, the spherical mirror is the convex mirror.
And in the figure b, the spherical mirror is the concave mirror.
10) Solution:
11.) Solution:
Ans:
12) Solution:
Ans:
The following two are the convenient rays that are chosen to construct the image by a spherical mirror for given object:
1) A ray of light passing through the centre of curvature C:
When the ray of light is passed through the centre of curvature of concave mirror or when it is directed in the direction of centre of curvature of convex mirror then the ray of light after reflection is reflected back along the same path as shown in figure below.
2) When a ray of light is parallel to the principal axis:
When a ray of light is incident parallel to principal axis in concave mirror then after reflection it passes through the principal focus of the concave mirror.
And in case of convex mirror, when a ray of light is incident parallel to principal axis then it appears to be diverge from the focus of the convex mirror as shown in figure below.
13) Solution:
14) Solution:
15.) Solution:
Ans:
a)
The following figure shows the image IB formed of object OA.
b) The characteristics of image formed are erect, virtual and diminished.
16) Solution:
Ans:
The following ray diagram shows the formation of image by a concave mirror for an object placed between its pole P and focus F.
And the here the image formed is behind the mirror which is virtual, upright and magnified.
17.) Solution:
The following figure shows the formation of image by a concave mirror for an object beyond its centre of curvature C.
And hence here the image is formed between the focus F and centre of curvature C of the concave mirror and which is real, inverted and diminished.
18) Solution:
The following diagram shows the formation of image of an object kept in front of a convex mirror. Here, the image is formed between focus F and pole P of the convex mirror which is virtual, erect and diminished.
19.) Solution:
Ans:
Convex mirror always produces an erect and virtual image. And the size of image is shorter than the object size.
20.) Solution:
Ans:
a)
When the object is placed between the pole P and focus F of a concave mirror then the image will be formed magnified and erect.
b)
The image formed in part a is virtual.
21.) Solution:
a)
When the object is placed at the centre of curvature C of the concave mirror then the image formed will be of same size.
b)
And the image formed in part a is the real and inverted.
22.) Solution:
a)
The real image is the image which can be recorded on the screen or we can see it in real on screen.
b)
The concave mirror is used to obtain the real image of the object.
c)
Concave mirror does not gives real image for all positions of the object. It gives real image for all positions when the object is beyond or at focus of concave mirror.
23.) Solution:
Ans:
- When the object is moved from infinity towards the pole of the concave mirror, then the image will be also start shifting from focus of concave mirror towards th infinity.
- In case of concave mirror, when the object is at infinity the image formed will be at the focus of concave mirror.
- If we shifted the object then the image will shift and it will be formed between the focus and centre of curvature of the concave mirror.
- If we further moved the object then the image will shift at centre of curvature further moving object it shifts beyond the centre of curvature of concave mirror and finally it shifts to infinity when the object is placed exactly at the focus of concave mirror.
- But, when we place the object between focus and pole of the concave mirror the image will be formed behind the concave mirror and which is virtual.
- In this way, the position and nature of the image formed by the concave mirror changes when an object is moved from infinity towards the pole of the concave mirror.
24.) Solution:
Ans:
The position and nature of the image formed by the convex mirror when an object is moved from infinity towards the pole of the mirror will be always virtual, upright and diminished.
The image formed in convex mirror is always between focus and pole of the convex mirror and it is independent of the position of the object placed in front of the convex mirror.
25.) Solution:
Ans:
a) a real and enlarged image:
Concave mirror is used to obtain a real and enlarged image.
b) a virtual and enlarged image:
Concave mirror is used to obtain the virtual and enlarged image.
c) a virtual and diminished image:
Convex mirror is used to obtain the virtual and diminished image.
d) a real and diminished image:
Concave mirror is used to obtain real and diminished image.
26.) Solution:
The relation between the focal length and radius of curvature of spherical mirror is given by,
f= R/2
Where, f is the focal length of the spherical mirror
And R is the radius of curvature of the spherical mirror.
27.) Solution:
The mirrors formula for spherical mirror is given by,
1/u + 1/v = 1/f
Where, u is the object distance
v is the object distance
And f is the focal length of spherical mirror.
28.) Solution:
Ans:
The magnification in spherical mirror is defined as it is the ratio of length of image to the length of object.
Or
The magnification of spherical mirror is defined as the it is the ratio of distance of image v and the distance of object u.
Hence, it is given by
m = I/O = -v/u
a) for real image, both u and v are negative so the linear magnification m is also positive.
b) for virtual image, u is negative and v is positive so that the linear magnification m is also negative.
29.) Solution:
Ans:
The maximum distance from the pole where the image is obtained in convex mirror is the focal length.
And for this position of image we have to place the object at infinity.
30.) Solution:
Ans:
The maximum distance from the concave mirror where the image will be obtained is infinity.
And for this we have to place the object at the focus of the concave mirror.
31.) Solution:
Ans:
To distinguish between plane mirror, concave mirror and convex mirror we have to take it near to our face and we have to see the image of our face in it which is explained as follows.
a)
If we have taken a mirror in which we have saw our face image which is of the same size and it doesn’t changing with the moving of mirror away or near to our face. Then the mirror will be plane mirror exactly.
b)
If we have taken a mirror near and we saw our face image in it which is upright and magnified. Also if the size of image is increasing on moving the mirror away from our face then that mirror must be concave mirror.
- c)
If we have taken a mirror and kept it near to our face and we saw the image of our face in it which is upright and diminished.
Also when we move the mirror away from the our face the size of the image is decreasing then that mirror is the convex mirror exactly.
32.) Solution:
Ans:
- The concave mirror is used as a shaving mirror.
- It can be also used as reflector.
- And also used as doctors head mirror.
33.) Solution:
Ans:
- The mirror used by dentist is the concave mirror.
- And also the mirror in used in search light reflector is the concave mirror.
34.) Solution:
Ans:
a)
When a concave mirror is used as a shaving mirror then the person’s face in relation to the focus of mirror will lie between the pole and focus of the mirror.
b)
And the image obtained in part a will be erect, virtual and magnified.
35.) Solution:
Ans:
A convex mirror is used as rear view mirror in cars and vehicles because the view given by convex mirror is wide which helps the drivers to see the most of the traffic behind his vehicle.
Also, convex mirror is used as rear view mirror in cars and vehicles because the images formed will be always erect, virtual and diminished image when the object is placed in front of the convex mirror.
36.) Solution:
Ans:
- A convex mirror is used as rear view mirror in cars and vehicles because the view given by convex mirror is wide which helps the drivers to see the most of the traffic behind his vehicle.
- Also, convex mirror is used as rear view mirror in cars and vehicles because the images formed will be always erect, virtual and diminished image when the object is placed in front of the convex mirror.
- The following ray diagram illustrates the use of convex mirror as rear view mirror in cars and vehicles.
Multiple choice type:
1.) For an incident ray directed towards centre of curvature of spherical mirror, the reflected ray
Ans: a) retraces its path
2.) The image formed by the convex mirror is
Ans:
a) erect and diminished
3.) A real and enlarged image can be obtained by using a
Ans: c) concave mirror
Numerical:
1.) Solution:
Ans:
The relation between radius of curvature R and focal length f is given by,
f= R/2
Given that, R= 40cm
Hence, f= 40/2= 20cm
Thus, the focal length of given convex mirror is 20cm.
2.) Solution:
Ans:
The relation between focal length f and the radius of curvature of concave mirror is given by,
f= R/2
Given that,
f= 10cm
Hence, 10= R/2
R= 20 cm
Thus, the radius of curvature of given concave mirror is 20cm.
3.) Solution:
Ans:
Given that,
For a concave mirror,
Object height h1= 2cm
Object distance u= -20cm
Focal length f= -12cm
Let h2 be the height of the image formed.
We know that,
Mirrors formula is given by,
1/f = 1/u + 1/v
1/-12 = 1/-20 + 1/v
Thus, 1/v = -1/12 + 1/20
= (-20 + 12)/240= -8/240
= -1/30
This, 1/v = -1/30
Hence, v= -30cm
Thus, image distance will be 30cm.
Also, we know that magnification m is given by,
m= h2/h1 = -v/u
h2/2= 30/-20
Thus, h2= -3 cm
Thus, the height of image formed will be 3cm.
From above all data, we can say that the image formed will be real, inverted and magnified.
4) Solution:
Ans:
Given that,
For concave mirror,
Object distance u = -4cm
Radius of curvature R= 24cm
We know that,
The relation between focal length and the radius of curvature of concave mirror is given by,
f= R/2
f= 24/2 = 12cm
And the mirrors formula is given by,
1/f = 1/u + 1/v
1/-12 = 1/-4 + 1/v
Thus, 1/v = 1/-12 + 1/4
1/v = 2/12= 1/6
Thus, v= 6cm
This, the image distance will be 6cm and it is formed behind the mirror.
Magnification is given by,
m= -v/u = -6/-4 = 1.5
Here, the magnification produced is greater than one hence the image formed will be magnified.
5.) Solution:
Ans:
Given that,
For a concave mirror,
f= 25cm
In case of concave mirror, the image size is equal to the object size only when the object is placed at the centre of curvature of concave mirror.
Thus, we can write
Object distance= radius of curvature
Object distance= 2*focal length
Hence, object distance= 2*25= 50cm
6.) Solution:
Ans:
Given that,
For concave mirror,
Object height h1 = 5cm
Focal length f= -10cm
Object distance u= -60cm
We know that, the mirrors formula is given by,
1/f = 1/u + 1/v
1/-10= 1/-60 + 1/v
Thus, 1/v = (-6+1)/60= -5/60= -1/12
Thus, 1/v = -1/12
Hence, v= -12cm
Thus, the image distance will be 12cm.
We know that, magnification in concave mirror is given by,
m= h1/h2 = -v/u
5/h2 = 12/-60 = 1/-5
Thus, h2 = 5*(-0.2)= -1cm
Thus, we can say that the height of the image will be 1cm.
7.) Solution:
Ans:
Given that,
For convex mirror,
Object distance u= -40cm
Focal length f= 40cm
We know that, the mirrors formula is given by,
1/f = 1/u + 1/v
1/40= 1/-40 + 1/v
Thus, 1/v = 2/40= 1/20
Hence, v= 20cm
Thus, the image will be formed behind the mirror at a distance of 20cm.
8.) Solution:
Ans:
Given that,
For concave mirror,
Object height h1= 1cm
Object distance u= -4cm
Image height h2 = 1.5cm
Image distance v= 6cm
We know that, the mirrors formula is given by,
1/f = 1/u + 1/v
1/f = 1/-4 + 1/6
1/f = (2-3)/12= -1/12
Thus, f= -12cm
Hence, the focal length of the given mirror is found to be -12cm.
9.) Solution:
Ans:
Given that,
For concave mirror,
Object height h1= 4cm
Object distance u= -30cm
Focal length f= -15cm
We know that, the relation between focal length f and radius of curvature R is given by,
f= R/2
R= 15*2= 30cm
We know that, when rays of light parallel to principal axis are incident on a concave mirror then the reflected rays converges at a point in front of the mirror.
Thus, the distance between object and mirror is 30cm.
Hence, we can say that the image will be formed in front of mirror at a distance of 30cm.
Also, we have magnification formula as,
m= h1/h2= -v/u
4/h2 = -30/-30
4/h2 = 1
Thus, h2 = 4cm
Hence, length of the image formed will be 4cm.
10.) Solution:
Ans:
Given that,
For concave mirror,
Object distance u= -30cm
Magnification m= -3
We know that, the magnification is given by,
m= f/(f – u)
-3= f/(f +30)
-3f – 90= f
4f = -90
Thus, f= -90/4= -22.5cm
Thus, the focal length of the given mirror will be 22.5cm
Also we have, the mirrors formula is given by,
1/f = 1/u + 1/v
-4/90= 1/-30 + 1/v
1/v = -4/90 + 1/30
1/v = -2/45 + 1/30
1/v = (3 – 4)/90= -1/90
Hence, v = -90cm
Thus, the image will be formed at a distance of 90cm in front of mirror.
11.) Solution:
Ans:
Given that,
For concave mirror,
Object distance u = -5cm
Magnification m= 2
We know that, magnification is given by,
m= -v/u
2= -v/-5
2= v/5
Thus, v = 10cm
Hence, the image will be formed at a distance 10cm behind the mirror.
We know that, the mirror formula is given by,
1/f = 1/u + 1/v
1/f = 1/-5 + 1/10
1/f = (-2 + 1)/10
1/f = -1/10
Hence, f= -10cm
Thus, the focal length of the given mirror will be 10cm.
12.) Solution:
Ans:
Given that,
For convex mirror,
h1= h2/3
h1/h2= 1/3 = magnification m
But, we know that,
Magnification m= -v/u
1/3 = -v/u
Thus, u = -3v
This is the relation between u and v for a given mirror.
13.) Solution:
Ans:
Given that,
For concave mirror,
v/u= 2
But, we know that
m= -v/u= 2
Hence, v= -2u
This is relation between u and v for a given mirror.
14.) Solution:
Ans:
Given that,
Magnification m= -3
We know that,
Magnification m= -v/u = -3
Thus, v/u = 3
Hence, v= 3u
This is the relation between u and v for a given mirror.