Reflection of Light

incidence, normal

Reason — The angle which the incidence ray makes with the normal at the point of incidence is called angle of incidence.

Both smooth and polished surfaces

Reason — Regular reflection occurs when a beam of light falls on a smooth and polished surface, such as a plane mirror. The reflected beam is also parallel and it is in a fixed direction. It can be seen only from a particular direction.

$\text i = \text r$

Reason — According to the law of reflection the angle of incidence is equal to the angle of reflection.

real

Reason — The image which can be obtained on a screen is called a real image. It is inverted. It is formed when light rays after reflection actually intersect.

erect and of same size.

Reason — The characteristics of an image formed by a plane mirror are — upright (or erect), virtual and of the same size as the object.

virtual with lateral inversion

Reason — The characteristics of an image formed by a plane mirror are — upright (or erect), virtual and laterally inverted.

real, virtual

Reason — For a distant object, the image formed by a concave mirror is real. The image of an object formed by a plane mirror or a convex mirror is virtual.

M, T, V

Reason — Lateral inversion is the interchange of left and right sides in a plane mirror image. The letters M, T, and V do not exhibit noticeable lateral inversion in their mirror image due to their inherent symmetry about a vertical line passing through their midpoint.

d

Reason — If the object is shifted by a distance d towards a plane mirror, the image will also shift by a distance d towards the mirror.

Sun, Shooting star, Fire as they emit light by themselves.

The return of light into the same medium after striking a surface is called reflection of light.

The back silvered surface of the plane mirror reflects most of the light incident on it.

(a) When the ray of light is incident normally on a plane mirror then the angle of incidence = 0°.

(b) The direction of reflected ray is same as the incident ray i.e., it retraces the path as shown below.

The ratio of

$\bold {\frac{sin i}{sin r} = 1}$

According to the laws of reflection angle of incidence = angle of reflection, hence sin i = sin r and we get the ratio as 1.

Black polished mica sheet will produce a stronger reflected beam as highly polished and silvered surfaces reflect almost the entire light falling on them.

For a ray incident normally on a plane mirror, the angle of incidence is i = 0°, therefore the angle of reflection r = 0°. Thus, a ray of light incident normally on a mirror is reflected along the same path.

(a) Plane mirror — It is made from few mm thick glass plate. One surface of the glass plate is polished to a high degree of smoothness. This forms the front surface of the mirror and the other surface is silvered (i.e., silver mercury or some suitable material is deposited over it). The silvered surface is further coated with some opaque material so as to protect the silvering on it.

The two surfaces of the plane mirror are shown below —

(b) Incident ray — the ray of light striking a reflecting surface is called the incident ray.

(c) Reflected ray — the light ray obtained after reflection from the surface in the same medium in which the incident ray is travelling is called reflected ray.

(d) Angle of incidence — The angle which the incident ray makes with the normal at the point of incidence, is called angle of incidence.

It is denoted by the letter i.

(e) Angle of reflection — The angle which the reflected ray makes with the normal at the point of incidence, is called angle of reflection.

It is denoted by the letter r.

Below diagram illustrates the Incident ray, Reflected ray, Angle of incidence and Angle of reflection:

The two laws of reflection are —

1. The angle of incidence i is equal to the angle of reflection r (i.e., ∠i = ∠r)
   ∠AON = ∠BON
2. The incident ray, the reflected ray and the normal at the point of incidence, lie in the same plane.
   AO, ON and OB are in one plane.

Below is the labelled diagram showing the reflection of a ray of light by a plane mirror:

(a) The characteristics of an image formed by a plane mirror are —

1. Image is upright (or erect).
2. Image is virtual.
3. Image is of the same size as the object.

(b) The position of the image is situated at the same perpendicular distance behind the mirror as the object is in front of it.

The difference between a real and a virtual image are as follows —

The interchange of the left and right sides in the image of an object in a plane mirror is called lateral inversion.
Below diagram illustrates lateral inversion of an image:

The letters on the front of an ambulance are written laterally inverted like ƎƆИA⅃UꓭMA so that the driver of the vehicle moving ahead of the ambulance reads the word laterally inverted as AMBULANCE, in his rear view mirror, and gives way to the ambulance first.

Regular reflection — It occurs when a beam of light falls on a smooth and polished surface, such as a plane mirror. In the figure shown below, a parallel beam of light is incident on a plane mirror. The reflected beam is also parallel and it is in a fixed direction. It can be seen only from a particular direction.

Irregular reflection — It occurs when a beam of light falls on a rough or ordinary surface such as the walls of a room or the page of a book which appear smooth but have many small projections over it.

In the figure given below, light rays fall at different points on a rough surface and each ray gets reflected from it obeying the laws of reflection of light. Due to uneven surface at different points, light rays gets reflected in different directions and give rise to diffused or irregular reflection.

As a result, reflected light spreads over a wide area and it does not follow a particular direction. Hence, reflected light can be seen from anywhere.

The two laws of reflection are —

1. The angle of incidence i is equal to the angle of reflection r (i.e., ∠i = ∠r)
   ∠AON = ∠BON

2. The incident ray, the reflected ray and the normal at the point of incidence, lie in the same plane.
   AO, ON and OB are in one plane.

Experimental verification —

Fix a sheet of white paper on a drawing board and draw a line MM₁ as shown below. On this line, take a point O nearly at the middle of it and draw a line OA such that ∠MOA is less than 90° (say, ∠MOA = 60°). Then draw a normal ON on line MM₁ at point O, and place a small plane mirror vertically by means of a stand with it's silvered surface on the line MM₁.

Now fix two pins P and Q vertically at some distance (say 5 cm) apart on line OA, on the board. Keeping eye on the other side of the normal (but on the same side of mirror), see clearly the images P'and Q' of the pins P and Q.

Now fix a pin R such that it is in line with the images P' and Q' as observed in the mirror. Now fix one more pin S such that the pin S is also in line with the pin R as well as the images P' and Q' of pins P and Q.

Draw small circles on paper around the positions of pins as in figure. Remove the pins and draw a line OB joining the point O to the pin points S and R.

As we can observe, AO is the incident ray, OB is the reflected ray, ∠AON = i is the angle of incidence and ∠BON = r is the angle of reflection. The angles AON and BON are measured and recorded in the observation table.

The experiment is repeated for the ∠MOA equal to 50°, 40° and 30°.

Observation table —

From the above observation table, we find that in each case, angle of incidence is equal to the angle of reflection. This verifies the first law of reflection.

The experiment is being performed on a flat drawing board, with mirror normal to the plane on which white sheet of paper is being fixed. Since the lower tips of all the four pins lie on the same plane (i.e., the plane of paper), therefore, the incident ray, the reflected ray and the normal at the point of incidence, all lie in one plane. This verifies the second law of reflection.

(a) The angle of incidence = 90° - 30° = 60°

(b) Diagram showing the reflected ray is given below:

The angle between the incident ray and the reflected ray = angle of incidence + angle of reflection

As we know,

Angle of incidence = Angle of reflection = 60°

Therefore, we get,

60° + 60° = 120°

Hence, the angle between the incident ray and reflected ray = 120°

Diagram showing the image with the position of the eye marked is given below:

(c) The image formed is virtual because the reflected rays appear to meet only when they are produced backwards.

Complete diagram showing the formation of the image is given below:

According to the law of reflection,

Angle of incidence = Angle of reflection

i.e., i = r

Given,

• Reflected ray is perpendicular to incident ray

Hence,

i + r = 90°

Using above law, we get,

i + i = 90°

2i = 90°

Hence,

i = r = 45°

Therefore, angle of incidence = 45°

Given,

• Distance between the man and his image = 6 m

But for a plane mirror,

Object distance = Image distance

Therefore,

Distance between the man and mirror = Distance between mirror and image

Hence,

The distance of the man and the mirror = $\frac{6}{2}$ = 3 m

(a) For a plane mirror, the image is as far behind the mirror as the object is in front of it i.e., the perpendicular distance of image from the mirror is equal to the perpendicular distance of object from the mirror.

Hence, the image of the insect is formed 1 m behind the mirror.

(b) The distance between the insect and it's image = the perpendicular distance of image from the mirror + the perpendicular distance of object from the mirror

= 1 + 1

= 2 m

Hence, the distance between the insect and its image is 2 m.

Given,

• Initially, the distance of the object A in front of plane mirror M is AM = 60 cm.
• Therefore the image B is at a distance MB = 60 cm from the mirror M behind it.
• Distance between the object A and its image B = 60 + 60 = 120 cm

On shifting the mirror by 25 cm away from the object, the new distance of object A from the mirror M' become AM' = 60 + 25 = 85 cm .

The new image B' is now at a distance M'B' = 85 cm behind the mirror M'.

So, the current distance of the image from the object = 85 cm + 85 cm = 170 cm

Let the reference point be the position of the object,

Thus,

New distance of the image from the object - Distance of the image from the object initially = Distance between the two positions of the image

$\Longrightarrow 170 \text { cm} – 120 \text { cm} = 50 \text { cm}$

Hence, the image shifts 50 cm from its previous position.

Given,

• Distance between the man and the mirror = 2 m
• Distance between the man and the chart = 3 m

Therefore,

Distance between the chart and the mirror = Distance between the man and the mirror + Distance between the man and the chart

= 3 m + 2 m

= 5 m

As the new image is formed on the mirror which is 2 m apart from the man,

So,

Distance at which the chart can be seen by the patient = 5 m + 2 m = 7 m

Hence, the chart seen by the patient is 7 m away from him.

parallel to each other

Reason — In a barber's shop, in order to see the hairs at the back of the head two plane mirrors are placed parallel to each other at the front and at the back of the viewer.

infinity

Reason — When two mirrors are kept parallel to each other, i.e., θ = 0° then

n = $\frac{360°}{\text θ} = \frac{360°}{\text 0°}$ = infinite.

Hence, infinite number of images are formed.

3

Reason — When two mirrors are kept perpendicular to each other i.e., θ = 90°, then

n = $\frac{360°}{\text θ} = \frac{360°}{90°}$ = 4,

Since n = 4 is even so number of images will be n - 1.

Therefore,

Number of images = n - 1
= 4 - 1
= 3

Hence, for two mirrors kept perpendicular to each other, three images are formed for an object kept in between them.

second image

Reason — A thick plane mirror forms a number of images due to multiple reflections within the glass from front surface and back reflecting surface. Out of these, the second image formed due to reflection from the back reflecting surface is the brightest.

2 and 3 respectively

Reason — In a periscope two parallel plane mirrors each inclined at 45° with the vertical walls are placed facing each other. In a kaleidoscope three plane mirrors inclined with each other at 60° are used.

60°

Reason — In a kaleidoscope three plane mirrors inclined with each other at 60° are used.

45°

Reason — In a periscope two parallel plane mirrors each inclined at 45° with the vertical walls are placed facing each other.

60°

Reason — From below figure,

Reflection from mirror OM :

Angle of incidence = 30°,

So by law of reflection,

Angle of incidence = Angle of reflection = 30°

This means the reflected ray QR (say) makes an angle of 30° with the normal to OM, i.e., 60° with mirror OM.

Now ray QR moves toward mirror ON, since OM and ON are at 90° and QR makes 60° with OM then it must make 90° - 60° = 30° with mirror ON.

So, angle between incident ray and normal to ON = 60°

⟹ Angle of incidence at ON = 60° = Angle of reflection

Looking glass.

To increase the effective length of the optician's room, a plane mirror is placed on the front wall and the sign board is placed on the opposite wall, just behind the patient.

With this arrangement, the sign board is at nearly double the length of the room for the patient.

The images lie on the circumference of a circle whose centre lies at the point of intersection of the two mirrors and whose radius is equal to the distance of the object from the point of intersection

There are two cases possible when the angle θ° between the mirrors is such that —

Case 1 : If $\bold {n = \frac{360^{\circ}}{\textbf {θ°}}}$ is odd:

1. The number of images is n, when the object is placed asymmetrically between the mirrors.
2. The number of images is n - 1, when the object is placed symmetrically (i.e., on the bisector of the angle) between the mirrors.

Case 2 : If $\bold {n = \frac{360^{\circ}}{\textbf {θ°}}}$ is even :

The number of images is always n - 1 for all positions of object in between the mirrors.

When two plane mirrors are placed making an angle θ° in between them and an object placed in between the mirrors, and the angle is gradually increased from 0° to 180°, then the number of images decreases.

The uses of a plane mirror are —

1. The most common and wide use is as a looking glass.
2. In the optician's room to increase the effective length of the room.

Three images will be formed when a point object is placed between two plane mirrors that are at right angles to each other.

The ray diagram is shown below:

Infinite number of images are formed when two plane mirrors are arranged parallel and facing each other at some separation.

Below ray diagram shows the formation of images:

(a) When two mirrors are kept perpendicular to each other i.e., θ = 90°,

Then

n = $\frac{360°}{90°}$ = 4,

SInce n = 4 is even so number of images is n - 1

Therefore,

Nummber of images = n - 1
= 4 - 1
= 3

Hence, for two mirrors kept perpendicular to each other, three images are formed for an object kept in between them.

(b) When two mirrors are kept at 60° to each other,

Then

n = $\frac{360°}{60°}$ = 6

SInce n = 6 is even so number of images is n - 1,

Therefore,

Nummber of images = n - 1
= 6 - 1
= 5

Hence, for two mirrors kept at 60° to each other, five images are formed for an object kept in between them.

(i) When an object is placed asymmetrically between two plane mirrors inclined at an angle of 50° to each other then

$$\begin{aligned}\text n = \frac{360°}{50°} \\[1em] = 7.2 \\[1em] = 7\end{aligned}$$

Hence, 7 images are formed.

(ii) The number of images formed when the object is placed symmetrically is n-1,

Then,

Number of images = n - 1
= 7 - 1
= 6

Therefore, 6 images will be formed when the object is placed symmetrically, between two plane mirrors inclined at an angle of 50°.

concave, outer

Reason — A concave mirror is made by silvering the outer (or bulging) surface of a piece of a hollow sphere such that the reflection takes place from the hollow (or concave) surface.

convex, inner

Reason — A convex mirror is made by silvering the inner surface of a piece of a hollow sphere such that the reflection takes place from the outer (or bulging) surface.

(i)

Reason — For spherical mirrors :

• The radius of a sphere of which the spherical mirror is a part is called the radius of curvature.
• The geometric centre of the spherical surface of a mirror is called the pole of the mirror.
• Principal axis is the straight line joining the pole of the mirror to its centre of curvature.

principal axis, parallel

Reason — The focus of a concave mirror is a point on the principal axis through which the light rays incident parallel to the principal axis, pass after reflection from the mirror.

retraces its path

Reason — A line joining the centre of curvature to any point on the surface of mirror is normal to the mirror at that point, so a ray AD passing through the center of curvature C (or appearing to pass through through the centre of curvature C) is incident normally on the spherical mirror.
Since it's angle of incidence is zero, therefore the angle of reflection will also be zero and the ray AD gets reflected along it's own path DA as shown below.

becomes parallel to the principal axis

Reason — A ray either incident from the focus (or converging at the focus), after reflection from a spherical mirror becomes parallel to the principal axis.

real, inverted, diminished to a point

Reason — For a concave mirror, when the object is at infinity, the nature of the image formed at focus is real, inverted, diminished to a point.

at centre of curvature, real, same size as that of the object

Reason — For a concave mirror, when the object is at the centre of curvature, the image is also at the centre of curvature. It is real and same size as that of the object.

highly magnified

Reason — For a concave mirror, when the object is at focus, the image is at infinity. It is real, inverted and highly magnified.

erect and diminished

Reason — In a convex mirror, the image formed is always virtual, erect and diminished. It is always situated between it's pole and focus irrespective of the distance of object in front of the mirror.

virtual, upright and diminished

Reason — For a convex mirror, when the object is in front of the mirror, the image is between the pole and the focus. It is virtual, upright and diminished.

(i)

Reason — The rules of sign convention are :

• All distances are measured from the pole of the mirror taken as origin.
• The distances measured along the principal axis in the direction of incident light are positive while those opposite to the incident light are negative.
• The distances above the principal axis are taken positive and those below the principal axis are taken negative.

negative, positive

Reason — For a convex mirror, the value of u is always negative and the value of v is always positive.

concave mirror

Reason — The image formed by a concave mirror is real and enlarged, when the object is between centre of curvature and focus or at focus.

convex mirror

Reason — A convex polished metallic surface is used in street lamps as a reflector so as to diverge light over a larger area.

concave mirror

Reason — When a concave mirror is held near the face (such that the face is between pole and focus of the mirror), it gives an upright and magnified image so that even tiny hair can be seen.

Hence, for this concave mirror of large focal length and large aperture is used.

convex mirror

Reason — A convex mirror always forms a diminished image for all positions of the object placed in front of it.

2:30

Reason — As the mirror image of the clock shows that hour hand is pointing slightly past 9 and minute hand at 6 (i.e., 30 minutes) so the mirror image shows 9:30. Since a plane mirror always produces laterally inverted images so in this case the hour hand should actually point slightly past 2 and minute hand will suffer no inversion due to symmetry so the clock will show 2:30.

at the focus of the reflector

Reason — In a torch, we want the light rays to come out as a parallel beam as this helps the light travel a long distance in a specific direction and a concave mirror reflects light rays parallel to the principal axis if the source is placed at its focus.

So, if we want a parallel beam of light (which is the case in a torch), the bulb should be placed at the focus of the concave mirror.

A reflecting surface which is a part of a sphere is called a spherical mirror.

The centre of curvature of a mirror is the centre of the sphere of which the mirror is a part.

The radius of the sphere of which the spherical mirror is a part, is called the radius of curvature of the mirror.

The plane surface area of the mirror through which light rays enter and fall on the mirror is called its aperture.

The geometric centre of the spherical surface of mirror is called the pole of the mirror.

A convex mirror always produces an erect and virtual image. The size of the image is shorter than the size of the object.

(a) When the object is between the pole and focus of a concave mirror then the image formed is magnified and erect.

(b) The image is virtual.

(a) When the object is at the centre of curvature of a concave mirror, the image is of the same size as the object.

(b) The images thus formed is real and inverted.

(a) A real image is one that can be obtained on a screen.

(b) A concave mirror can be used to obtain a real image of an object.

(c) No, concave mirror does not form real image for all the locations of the object.

(a) Concave mirror is used to obtain a real and enlarged image.

(b) Concave mirror is used to obtain a virtual and enlarged image.

(c) Convex mirror is used to obtain a virtual and diminished image.

(d) Concave mirror is used to obtain a real and diminished image.

The focal length 'f' is related to the radius of curvature 'R' in the following way —

$\bold{f = \frac{1}{2} R}$

The formula for the spherical mirror is —

$\frac{1}{\text u} + \frac{1}{\text v} = \frac{1}{\text f}$

• u = distance of object,
• v = distance of image and
• f = focal length.

(a) Concave mirror is used by a dentist.

(b) Concave mirror is used as a search-light reflector.

A convex mirror would be preferred in comparison to a plane mirror for use as a rear view mirror in a car because it provides a much wider field view as compared to a plane mirror of the same size. The below ray diagram shows this:

$$\begin{aligned}\text{Magnification (m)} = \frac{\text{Length of image (I)}}{\text{Length of object (O)}} \\[1em] = \frac{\text{Distance of image (v)}}{\text{Distance of object (u)}} \\[1em] \text{OR} \\[1em] \text m = \frac{\text I}{\text O} = - \frac{\text v}{\text u}\end{aligned}$$

where

• 'I' is the length of the image,
• 'O' is the length of the object,
• 'v' is the distance of the image and
• 'u' is the distance of the object.

Depending on whether the inner or outer surface of the sphere is silvered, spherical mirrors are of two types —

1. Concave mirror
2. Convex mirror

Difference between the two mirrors are —

Pole — The geometric centre of the spherical surface of mirror is called the pole of the mirror.

Principal axis — It is the straight line joining the pole of the mirror to it's centre of curvature.

Centre of curvature — The centre of curvature of a mirror is the centre of sphere of which the mirror is a part.

(i) A convex mirror diverges a beam of light incident on it.

(ii) A concave mirror converges a beam of light incident on it.

Focus of a concave mirror — The focus of a concave mirror is a point on the principal axis through which the light rays incident parallel to the principal axis, pass after reflection from the mirror.

Focal length of a concave mirror — The distance of focus F from the pole P of the mirror is called the focal length of the mirror.

i.e., focal length (f) = PF

When the incident ray is directed towards the centre of curvature, after reflection from a spherical mirror, it retraces its path.

It is because the ray is normal to the spherical mirror, so ∠i (angle of incidence) = 0, therefore, ∠r (angle of reflection) = 0.

The ratio of length of the image to the length of object, is called linear magnification.

So,

$$\begin{aligned}\text{Magnification (m)} = \frac{\text{Length of image (I)}}{\text{Length of object (O)}} \\[1em] = \frac{\text{Distance of image (v)}}{\text{Distance of object (u)}} \\[1em] \text{OR} \\[1em] m = \frac{\text I}{\text O} = - \frac{\text v}{\text u}\end{aligned}$$

where

• 'I' is the length of the image,
• 'O' is the length of the object,
• 'v' is the distance of the image,
• 'u' is the distance of the object.

(a) For a real image, linear magnification m is negative.

(b) For a virtual image, linear magnification m is positive.

The maximum distance from the pole, in a convex mirror where the image can be obtained is till the focal length of the mirror. The object would then has to be at infinity.

The maximum distance from the concave mirror, where the image can be obtained is infinity. The object would then be at focus.

In order to distinguish between a plane mirror, a concave mirror and a convex mirror, the given mirror is held near the face and the image obtained is seen.

There can be the following three cases —

Case 1 — If image is upright, of same size and it does not change in size by moving the mirror towards or away from the face, then the mirror is plane.

Case 2 — If image is upright, magnified and increases in size on small movement of the mirror away from the face then the mirror is concave.

Case 3 — If image is upright, dimished and decreases in size on small movement of the mirror away from the face then the mirror is convex.

Uses of a concave mirror are as follows —

1. As a shaving mirror — When a concave mirror is held near the face (such that the face is between pole and focus of the mirror), it gives an upright and magnified image. Hence even tiny hair can be seen.
2. As a reflector — In torch, searchlight and head light of automobiles, cycles etc., a concave polished metallic surface is used to obtain a parallel beam of light.

(a) When a concave mirror is used as a shaving mirror, the person's face should be between the pole and the focus.

(b) The image formed is erect, virtual and magnified.

(i) Below ray diagram illustrates the action of concave mirror on a beam of light incident parallel to the principal axis:

(ii) Below ray diagram illustrates the action of convex mirror on a beam of light incident parallel to the principal axis:

Focus of a convex mirror — The focus of a convex mirror is a point on the principal axis from which, the light rays that are incident parallel to the principal axis, appear to come, after reflection from the mirror.

Focal length of a concave mirror — The distance of focus F from the pole P of the mirror is called the focal length of the mirror.

i.e., focal length (f) = PF.

(i) The figure (a) shows convex mirror and figure (b) shows concave mirror.

(ii) Diagrams showing reflected rays are shown below:

(a)

(b)

Below are the completed diagrams showing the reflected rays for the incident rays 1 and 2:

(a)

(b)

Below are the completed diagrams showing the reflected rays for the incident rays A and B:

(a)

(b)

The two convenient rays, chosen to construct the image by a spherical mirror are —

(i) A ray that passes through the center of curvature.

A line joining the centre of curvature to any point on the surface of mirror is normal to the mirror at that point, therefore a ray AD passing through the center of curvature C (or appearing to pass through through the centre of curvature C) is incident normally on the spherical mirror.

Since it's angle of incidence is zero, therefore the angle of reflection will also be zero and the ray AD gets reflected along it's own path DA as shown below:

(ii) A ray parallel to the principal axis.

A ray of light AD incident parallel to the principal axis, after reflection passes either through the focus F(in a concave mirror) or will appear to come from the focus F (in a convex mirror) along DB as shown below:

Ray diagram showing the formation of the image is given below:

Ray diagram showing the formation of the image is given below:

(i) Below completed ray diagram shows the position of the image of object OA:

(ii) The three characteristics of the image are virtual, upright and diminished.

Below is the ray diagram showing the formation of image by a concave mirror for an object placed between it's pole and focus:

When the object is between the focus F and the pole P, the image is formed behind the mirror. It is virtual, upright and magnified.

The image by a concave mirror for an object beyond it's center of curvature is shown below:

When object is beyond the centre of curvature C, the image is between the focus F and the centre of curvature C. It is real, inverted and diminished.

The diagram below shows the formation of image when the object is kept in front of a convex mirror.

When the object is in in front of the convex mirror, the image is between the pole P and focus F on the other side of the mirror.

The image formed is virtual, upright and diminished.

The image formed moves away from the concave mirror when an object is moved from infinity towards the pole of mirror.

The image is diminished when the object is beyond centre of curvature, but it becomes magnified as the object comes within the centre of curvature. The image is of the same size of the object when the object is at the centre of curvature.

For the object situated beyond focus, the image is always real and inverted, whereas for the object situated between the focus and pole the image is upright and virtual.

The table below shows the position, size and nature of the image formed by a concave mirror for different positions of the object.

In a convex mirror, the image formed is always virtual, upright and diminished. It is always situated between it's pole and focus irrespective of the distance of object in front of the mirror.

As the object comes closer to the mirror from infinity towards the pole, it's image shifts from focus towards the pole and increase in size.

The table below shows the position, size and nature of the image formed by a convex mirror

A convex mirror diverges the incident light beam and always forms a virtual, small and erect image behind the mirror between it's pole and focus. This fact enables the driver to use it as a rear view mirror in vehicles to see all the traffic approaching from behind.

Although a plane mirror can also be used as a rear view mirror, but a convex mirror provides a much wider field view as compared to a plane mirror of the same size.

The below diagram shows how a convex mirror provides a better field view than a plane mirror.

Focal length = $\bold{\frac{1}{2}}$ x Radius of curvature

Given,

• R = 40 cm

Substituting the values in the formula above we get,

$$\begin{aligned}\text{Focal length} = \frac{1}{2} \times 40 \\[1em] = 20 \text{ cm}\end{aligned}$$

Hence, focal length of the convex mirror is 20 cm.

Focal length = $\bold{\frac{1}{2}}$ x Radius of curvature

Given,

• f = -10 cm

Substituting the values in the formula above we get,

$$\begin{aligned}-10 = \frac{1}{2} \times \text{Radius of curvature} \\[1em] \Rightarrow \text{Radius of curvature} = -20 \text{ cm}\end{aligned}$$

Hence, radius of curvature of concave mirror is 20 cm (negative).

Given,

• Object height (O) = 2 cm
• Focal length (f) = -12 cm
• Object distance (u) = -20 cm

Using the mirror formula :

$\frac{1}{\text u} + \frac{1}{\text v} = \frac{1}{\text f}$

Substituting the values in the formula above we get,

$$\begin{aligned}\frac{1}{(-20)} + \frac{1}{\text v} = \frac{1}{(-12)} \\[1em] -\frac{1}{20} + \frac{1}{\text v} = -\frac{1}{12} \\[1em] \Rightarrow \frac{1}{\text v} = - \frac{1}{12} + \frac{1}{20} \\[1em] = \frac{ - 5 + 3 }{60} \\[1em] = - \frac{ 2}{60} \\[1em] = - \frac{1}{30} \\[1em] \Rightarrow \text v = - 30 \text{ cm}\end{aligned}$$

Hence, the image is formed at a distance of 30 cm in front of the mirror.

Now,

Magnification of a mirror is given by,

$$\begin{aligned}\text{Magnification (m)} = \frac{\text{Length of image (I)}}{\text{Length of object (O)}} \\[1em] = \frac{\text{Distance of image (v)}}{\text{Distance of object (u)}} \\[1em] \Rightarrow \frac{\text I}{2} = -\frac{(-30)}{(-20)} \\[1em] \Rightarrow \frac{\text I}{2} = -\frac{30}{20} \\[1em] \Rightarrow \text I = -\frac{30 \times 2}{20} \\[1em] = -3 \text { cm}\end{aligned}$$

Hence, the length of the image is 3 cm the image is real, inverted and magnified.

Given,

• Radius of curvature (R) = -24 cm
• Object distance (u) = - 4 cm

Focal length = $\bold {\frac{1}{2}}$ x Radius of curvature

Substituting the values in the formula above, we get,

$$\begin{aligned}\text{Focal length} = \frac{1}{2} \times (- 24) \\[1em] = -12 \text{ cm}\end{aligned}$$

Hence, focal length of the concave mirror is -12 cm.

Using the mirror formula

$\frac{1}{\text u} + \frac{1}{\text v} = \frac{1}{\text f}$

Now, substituting the values in the mirror formula , we get,

$$\begin{aligned}\frac{1}{(-4)} + \frac{1}{\text v} = \frac{1}{(-12)} \\[1em] \Rightarrow -\frac{1}{4} + \frac{1}{\text v} = - \frac{1}{12} \\[1em] \Rightarrow \frac{1}{\text v} = - \frac{1}{12} + \frac{1}{4} \\[1em] = \frac{ - 1 + 3 }{12} \\[1em] = \frac{2}{12} \\[1em] = \frac{1}{6} \\[1em] \Rightarrow \text v = 6 \text{ cm}\end{aligned}$$

Thus, the image is formed 6 cm behind the mirror.

Computing linear magnification:

$$\begin{aligned}\text{Magnification (m)} = \frac{\text{Length of image (I)}}{\text{Length of object (O)}} \\[1em] \Rightarrow \text m = \frac{\text{Distance of image (v)}}{\text{Distance of object (u)}} \\[1em] = - \frac{6}{(- 4)} \\[1em] = \frac{6}{4} \\[1em] = 1.5\end{aligned}$$

As, the length of image is 1.5 times the image of object, hence, the image is magnified.

To get an image of same size the object should be placed at the center of the curvature of a concave mirror.

Given,

• Focal length = -25 cm

As,

Centre of curvature = 2 x Focal length

Therefore,

Centre of curvature = 2 x (-25) = -50 cm

Hence, the object should be kept at 50 cm in front of the mirror so that the size of image is equal to the size of object.

Given,

• Object height (O) = 5 cm
• focal length (f) = -10 cm
• Object distance (u) = -60 cm

(i)

Using the mirror formula,

$\frac{1}{\text u} + \frac{1}{\text v} = \frac{1}{\text f}$

Substituting the values in the formula above, we get,

$$\begin{aligned}\frac{1}{(-60)} + \frac{1}{\text v} = \frac{1}{(-10)} \\[1em] \Rightarrow -\frac{1}{60} + \frac{1}{\text v} = -\frac{1}{10} \\[1em] \Rightarrow \frac{1}{\text v} = -\frac{1}{10} + \frac{1}{60} \\[1em] = \frac{- 6 + 1}{60} \\[1em] = -\frac{ 5}{60} \\[1em] = -\frac{1}{12} \\[1em] \Rightarrow \text v = -12 \text{ cm}\end{aligned}$$

Thus, the image distance (v) is 12 cm infront of the mirror.

(ii) Magnification of a mirror is given by, $$\begin{aligned}\text{Magnification (m)} = \frac{\text{Length of image (I)}}{\text{Length of object (O)}} \\[1em] = \frac{\text{Distance of image (v)}}{\text{Distance of object (u)}} \\[1em] \frac{\text I}{5} = - \frac{(-12)}{(-60)} \\[1em] \frac{\text I}{5} = - \frac{12}{60} \\[1em] \Rightarrow \text I = -\frac{12 \times 5}{60} \\[1em] = -1 \text { cm}\end{aligned}$$

Hence, the length of the image is 1 cm and negative sign shows that the image will be inverted.

Given,

• u = -40 cm
• f = 40 cm

Mirror formula:

$\frac{1}{\text u} + \frac{1}{\text v} = \frac{1}{\text f}$

Substituting the values in the formula above, we get,

$$\begin{aligned}\frac{1}{(-40)} + \frac{1}{\text v} = \frac{1}{40} \\[1em] \Rightarrow -\frac{1}{40} + \frac{1}{\text v} = \frac{1}{40} \\[1em] \Rightarrow \frac{1}{\text v} = \frac{1}{40} + \frac{1}{40} \\[1em] = \frac{1+1}{40} \\[1em] = \frac{2}{40} \\[1em] = \frac{1}{20} \\[1em] \Rightarrow \text v = 20 \text{ cm}\end{aligned}$$

Hence, the image is formed 20 cm behind the mirror.

Given,

• u = -4 cm
• v = 6 cm

Using the mirror formula,

$\frac{1}{\text u} + \frac{1}{\text v} = \frac{1}{\text f}$

Substituting the values in the formula above, we get,

$$\begin{aligned}\frac{1}{(-4)} + \frac{1}{6} = \frac{1}{\text f} \\[1em] \Rightarrow -\frac{1}{4} + \frac{1}{6} = \frac{1}{\text f} \\[1em] \Rightarrow \frac{1}{\text f} = \frac{-3+2}{12} \\[1em] = -\frac{1}{12} \\[1em] \Rightarrow \text f = -12 \text{ cm}\end{aligned}$$

Hence, the focal length of concave mirror is 12 cm.

(a) Given,

• u = -30 cm
• f = -15 cm

Using the mirror formula,

$\frac{1}{\text u} + \frac{1}{\text v} = \frac{1}{\text f}$

Substituting the values in the formula above, we get,

$$\begin{aligned}\frac{1}{(-30)} + \frac{1}{\text v} = \frac{1}{(-15)} \\[1em] \Rightarrow -\frac{1}{30} + \frac{1}{\text v} = -\frac{1}{15} \\[1em] \Rightarrow \frac{1}{\text v} = - \frac{1}{15} + \frac{1}{30} \\[1em] = \frac{ - 2 + 1 }{30} \\[1em] = -\frac{1}{30} \\[1em] \Rightarrow \text v = -30 \text{ cm}\end{aligned}$$

Thus, the image is formed at 30 cm in front of the mirror

(b) Magnification of a mirror is given by,

$$\begin{aligned}\text{Magnification (m)} = \frac{\text{Length of image (I)}}{\text{Length of object (O)}} \\[1em] = \frac{\text{Distance of image (v)}}{\text{Distance of object (u)}} \\[1em] \frac{\text I}{4} = -\frac{(-30)}{(-30)} \\[1em] \frac{\text I}{4} = -\frac{30}{30} \\[1em] \frac{\text I}{4} = -1 \\[1em] \Rightarrow \text I = -4 \text { cm}\end{aligned}$$

Hence, the length of the image is 4 cm and negative sign represents that the image is inverted.

(a) Given,

• Distance of the object (u) = -30 cm
• Image height = 3 times the height of object

So, magnification (m) = -3 (negative for the real image)

$$\begin{aligned}\text{Magnification (m)} = \frac{\text{Length of image (I)}}{\text{Length of object (O)}} \\[1em] = \frac{\text{Distance of image (v)}}{\text{Distance of object (u)}} \\[1em] -3 = - \frac{\text v}{(-30)} \\[1em] \Rightarrow \text v = 3 \times (- 30) \\[1em] = - 90 \text{ cm}\end{aligned}$$

Thus, the image is formed 90 cm in front of the mirror.

Using the mirror formula,

$\frac{1}{\text u} + \frac{1}{\text v} = \frac{1}{\text f}$

Substituting the values in the mirror formula we get,

$$\begin{aligned}\frac{1}{(-30)} + \frac{1}{(-90)} = \frac{1}{\text f} \\[1em] \Rightarrow-\frac{1}{30} - \frac{1}{90} = \frac{1}{\text f} \\[1em] \Rightarrow \frac{1}{\text f} = \frac{-3 - 1}{90} \\[1em] = -\frac{4}{90} \\[1em] \Rightarrow \text f = -22.5 \text{ cm}\end{aligned}$$

Hence, focal length of the mirror = -22.5 cm

(b) The image is formed 90 cm in front of the mirror

(a) Given,

• Distance of the object (u) = -5 cm
• Magnification (m) = 2 (positive for the virtual image)

$$\begin{aligned}\text{Magnification (m)} = \frac{\text{Length of image (I)}}{\text{Length of object (O)}} \\[1em] = \frac{\text{Distance of image (v)}}{\text{Distance of object (u)}} \\[1em] 2 = - \frac{\text v}{(-5)} \\[1em] 2 = \frac{\text v}{5} \\[1em] \Rightarrow \text v = 2 \times 5 \\[1em] = 10 \text{ cm}\end{aligned}$$

Thus, the image is formed 10 cm behind the mirror.

Using the mirror formula:

$\frac{1}{\text u} + \frac{1}{\text v} = \frac{1}{\text f}$

Substituting the values in the mirror formula, we get,

$$\begin{aligned}\frac{1}{(-5)} + \frac{1}{10} = \frac{1}{\text f} \\[1em] -\frac{1}{5} + \frac{1}{10} = \frac{1}{\text f} \\[1em] \Rightarrow \frac{1}{\text f} = \frac{ - 2 + 1 }{10} \\[1em] = -\frac{1}{10} \\[1em] \Rightarrow \text f = -10 \text{ cm}\end{aligned}$$

Hence, focal length of the mirror = -10 cm

(b) The position of the image is 10 cm behind the mirror

Given,

• Size of the image = $-\frac{1}{3} \times$ Size of the object

Magnification of a mirror is given by,

Magnification (m) = $\frac{\text{Size of image (I)}}{\text{Size of object (O)}} = \frac{\text {Image distance (v)}}{\text{Object distance (u)}} = -\frac{1}{3}$

Hence,

$$\begin{aligned}\bold {v = -\frac{1}{3}u} \\[1em] \bold {Or} \\[1em] \bold {u = -3v}\end{aligned}$$

Given,

• Size of the image = -2 x Size of the object

Magnification of a mirror is given by,

Magnification (m) = $\frac{\text{Size of image (I)}}{\text{Size of object (O)}} = \frac{\text {Image distance (v)}}{\text{Object distance (u)}} = -2$

Hence,

$$\begin{aligned}\bold {v = -2u} \\[1em] \bold {Or} \\[1em] \bold {u = -\frac{1}{2}v}\end{aligned}$$

Given,

• Magnification (m) = -3

Magnification of a mirror is given by,

$$\begin{aligned}\text m = \frac{\text {Image distance (v)}}{\text{Object distance (u)}} \\[1em] \Rightarrow -3 = -\frac{\text{v}}{\text{u}} \\[1em] \Rightarrow \text{v} = \text{3u}\end{aligned}$$

Hence, v = 3u

(a) Multiple images are formed because light undergoes repeated reflections between the two parallel plane mirrors.
Each mirror forms an image of the object as well as of the images formed by the other mirror, resulting in a large number of images (ideally infinite if mirrors are perfectly parallel).

(b) When one plane mirror is replaced by a convex mirror, predict how the number and nature of images will change.
When one plane mirror is replaced by a convex mirror, the number of images decreases greatly.
This is because a convex mirror forms only one virtual, erect, and diminished image and causes the reflected rays to diverge. Therefore, repeated reflections are greatly reduced.
Hence, the images formed become fewer, virtual, erect, and smaller in size.

(c) When two plane mirrors are placed at 90°, the number of images formed is given by

$$\begin{aligned}\text n = \frac{360°}{90°} - 1 \\[1em] = 4 - 1 \\[1em] = 3\end{aligned}$$

So, 3 images are formed.

(d) A kaleidoscope typically uses three plane mirrors inclined at 60° to each other and this arrangement produces multiple symmetrical patterns due to repeated reflections.

(a) A convex mirror should be preferred for side-view purposes because it provides a wider field of view. This allows the driver to see a larger area of the road and traffic behind the vehicle.

(b) A convex mirror forms images that are virtual, erect and diminished.

(c) No, a convex mirror cannot be used as a doctor’s head mirror because a doctor requires a bright, magnified image, which is provided by a concave mirror, not a convex mirror.

(d) Two other uses of a convex mirror are:

1. Reflector in street lamps
2. Road safety mirrors at blind turns