Optics: Reflection of Light

Optics is the study of the nature and behaviour of light and other electromagnetic waves. 'Light is the only thing that enable us to see.'

1.0Light

Light is a form of energy which excites our sense of sight.

Sources of light

During the day, the primary source of light is the Sun and the secondary source is the brightness of the sky. Other common sources are flames, electric bulbs, tube lights (fluorescent tubes), compact fluorescent lamps (CFLs) and light emitting diodes (LEDs). Do Light travels in a straight line in vacuum or in a homogeneous transparent medium.

Ray of light : A ray of light is the direction in which light travels. Beam of light : A bundle of light rays is called beam of light (or light beam). Convergent beam : A beam of light in which all the rays move towards a single point is called convergent beam (see figure).

Divergent beam : A beam of light in which all the rays emerge out from a single point is called divergent beam (see figure).

Parallel beam : A beam of light in which all the rays are parallel to each other is called parallel beam (see figure).

2.0Reflection of light

Reflection of light is the process in which light rays meeting the boundary between two media 'bounce back', to stay in the first medium.

• The process of sending back of light rays which fall on the surface of an object is called reflection of light.

On reflection of light from a surface, the speed, wavelength and frequency of light do not change. This is because the light stays in the same medium. But, amplitude and intensity of reflected ray are slightly less than that of incident ray as some part of energy is absorbed.

3.0Laws of reflection

First law The incident ray, the reflected ray and the normal at the point of incidence, all lie in the same plane. Second law The angle of incidence is equal to the angle of reflection. $\angle \mathbf{i}=\angle \mathbf{r}$

Some basic terms related to reflection of light Incident ray: The ray of light which falls on the mirror surface is called incident ray. Reflected ray : The ray of light which is sent back by the mirror is called reflected ray. Point of incidence : The point at which the incident ray falls on the mirror is called point of incidence. Normal : A line perpendicular to the surface of mirror passing through the point of incidence is called normal. Angle of incidence : The angle made by incident ray with the normal at the point of incidence is called angle of incidence. Angle of reflection : The angle made by reflected ray with the normal at the point of incidence is called angle of reflection.

• The plane which is discussed in first law of reflection is not the surface of mirror. It is the plane PQRS (see above figure).
• We are able to see the objects because the light gets reflected from the object and reach our eyes.

Building Concepts 1 You may have observed the image of the sun in the windows of distant buildings near the time that the sun is rising or setting. However, the image of the sun is not seen in the windows of distant building during midday. Explain it, by drawing appropriate light rays on the given diagram.

• When light incident normally over a reflecting surface angle between incident ray and reflecting surface is $90^{\circ }$ but angle of incidence and angle of reflection is equal to zero.

4.0Reflection from plane mirrors

A mirror is a highly polished surface used to reflect the light falling on it. Mirrors are usually made by depositing a thin layer of silver metal on one side of a plane glass sheet.

Some basic terms Object : Anything which gives out light rays either of its own or due to reflection is called an object. Point object : An object whose dimensions are negligibly small is called point object. Extended object : An object whose dimensions are quite large is called extended object. Point Object $\bullet$ Symbols used in Extended Object $\uparrow {\}}_{\text{Ray diagrams }}$ Image : An image of an object is formed when light rays coming from the object meet or appear to meet at a point after reflection from a mirror or refraction from a lens. Real image : A real image is formed when the light rays actually meet at a point and which can be obtained on a screen. It is always inverted. Virtual image : A virtual image is formed when the rays do not actually meet at a point but they appear to meet at a point. Such images cannot be obtained on the screen. It is always erect or upright.

Image formed by a plane mirror

The properties of image formed by a plane mirrors are (see figure $a$ and $b$ ): (1) The image is virtual and erect. (2) The distance of image from mirror is equal to distance of object from mirror. (3) The size of image is exactly equal to the size of object. (4) The image is laterally inverted.

Lateral inversion

When an asymmetric object is placed in front of a plane mirror, then the right side of the object appears to be the left side of image and the left side of the object appears to be the right side of its image. This change of sides of an object seen in the image is called left - right inversion or lateral inversion. The image is inverted side ways, thus, also called 'side ways inversion' (see figure c)

5.0Reflection from spherical mirrors

A spherical mirror, as the name suggests, has the shape of a section of a hollow sphere. A spherical mirror is a mirror whose reflecting surface is made by the part of a hollow sphere. Suppose a hollow sphere has a polished mirror surface on the inside as well outside. By removing a section of the sphere, a double-sided spherical mirror is obtained with a concave reflecting surface on one side and a convex reflecting surface on the other side (see figure).

Concave mirror

A spherical mirror in which the reflection of light takes place at bent-in surface is called 'concave mirror'. Concave mirror is also called 'converging mirror'. This is because the parallel beam of light after reflection, converge at a single point.

Convex mirror

A spherical mirror in which the reflection of light takes place at bulging-out surface is called 'convex mirror'. Convex mirror is also called 'diverging mirror'. This is because the parallel beam of light after reflection appears to diverge from a single point.

Active Physics 1

1. Take a large shining spoon and look at its inner curved surface. When your face is quite close to the spoon, you will see your erect and magnified image. Now, slowly move the spoon away from you. You will see your inverted and magnified image. As the spoon is moved further away, the inverted image gradually decreases [see figure (a)].
2. Now, look at the outer curved surface of the spoon. You will see your erect and diminished image. As the spoon is moved further away, the image remains erect and its size gradually decreases [see figure (b)]. Conclusion : The inner curved surface of the spoon acts as a concave mirror. The outer curved surface of the spoon acts as a convex mirror.
   
   (a) Looking at the inner surface of a shining spoon
   
   (b) Looking at the outer surface of a shining spoon

Some basic terms related to spherical mirrors

Centre of curvature ( $\mathbf{C}$ ) : The point in space that represents the centre of the hollow sphere from which the spherical mirror was cut is called 'centre of curvature'. The centre of the hollow sphere from which the spherical mirror is formed is called 'centre of curvature'.

Pole or vertex ( $\mathbf{P}$ ): The middle point on the surface of a spherical mirror is called 'pole'. The geometric centre of the curved mirror surface is called 'pole'.

Radius of curvature ( $\mathbf{R}$ ) : The radius of hollow sphere from which the mirror is formed is called 'radius of curvature'.

The distance between the centre of curvature and the pole of a spherical mirror is called 'radius of curvature'.

Principal axis: A line passing through the pole and the centre of curvature of the spherical mirror is called 'principal axis'. An imaginary line drawn through the pole, perpendicular to the surface of the spherical mirror at the pole is called 'principal axis'.

Principal focus (F): The point on the principal axis where all rays parallel to principal axis, either converge or appear to diverge after reflection is called 'principal focus'.

Focal length (f): The distance between the focus and pole of a spherical mirror is called 'focal length'.

Focal plane : A plane passing through focus and perpendicular to principal axis is called 'focal plane'.

Aperture : The diameter of the circular cross-section of the spherical mirror is called 'aperture' (AB). It represents the size of the mirror. More the aperture of a spherical mirror, more will be its size. Thus, it will collect more light forming brighter images after reflection.

6.0Rules to obtain images in spherical mirrors

Concave mirrors

(1) The ray parallel to the principal axis, after reflection, passes through the principal focus F of a concave mirror [see figure (a)]. (2) A ray passing through the principal focus in a concave mirror, is reflected parallel to the principal axis [see figure (b)]. (3) A ray passing through the centre of curvature in a concave mirror, is reflected back along its own path [see figure (c)].

Convex mirrors

(1) The ray parallel to the principal axis, after reflection, appears to diverge from the principal focus of a convex mirror [see figure (a)]. (2) A ray which is directed towards the principal focus in a convex mirror, is reflected parallel to the principal axis [see figure (b)]. (3) A ray directed towards the centre of curvature in a convex mirror, is reflected back along its own path [see figure (c)].

• A ray passing through the centre of curvature behave like a normal of a spherical mirror.

Building concepts 2 What happens when a ray is incident obliquely to the principal axis, towards the pole of the mirror, on a concave mirror or a convex mirror? Explanation: When a ray is incident obliquely to the principal axis, towards the pole of the mirror, on a concave mirror or a convex mirror, it is reflected obliquely such that the incident ray and the reflected ray make equal angles with the principal axis. This is because the principal axis acts as normal at the pole. The incident and reflected rays follow the laws of reflection at the point of incidence (Pole), making equal angles with the principal axis (see figure).

Active Physics 2

1. Take a concave mirror and allow the sun rays to fall on it. Take paper and move it towards the concave mirror till you obtain a bright sharp spot of light on it. The spot obtained is the image of the sun. Now, measure the distance between paper and the concave mirror. This distance is an approximate focal length of the concave mirror.
2. If this spot is kept on the paper for few minutes, the paper will start burning. This is because the light energy converts to heat energy. Important : Avoid looking at the Sun directly or its image formed by the concave mirror as the intensity of sunlight may damage the eye.
   

7.0Image formation by a concave mirror

Images formed by a concave mirror

Image formation by concave mirror

• Variation of size of image for different position of object in front of a concave mirror.
  
• If the formed image is erect, of the same size and equidistant as of the object, then the mirror is a plane mirror.
• If the image is erect, virtual but smaller in size than the object, then it is a convex mirror.
• If the formed image is erect, virtual and magnified when the mirror is close to the object, then it is a concave mirror.

Uses of concave mirrors

(1) Concave mirrors are used as shaving mirrors to see a larger image of the face.

8.0Image formation by a convex mirror

The image formed by a convex mirror is always behind the mirror that is, it is always virtual and erect. Also, the size of image is always diminished, that is, its size is always smaller than that of the object [see figure (a)].

Thok8ujk-e rays parallel to principal axis, after reflection, appears to diverge from the principal focus of the convex mirror [see figure (b)]. The image formed at the focus, behind the mirror is highly diminished. The image is virtual and erect.

Image formation by convex mirror

• Variation of size of image for different position of object in front of a convex mirror.
  

Building Concepts 3 Suppose that lower half of concave mirror's reflecting surface is covered with an opaque (non-reflecting) material. What effect will this have on the image of an object placed in front of the mirror ? Explanation: If lower half of concave mirror is obstructed, full image will be formed but with reduced brightness. This is because every part of mirror forms complete image. On obstructing the lower half, half of the light rays are obstructed, forming less bright image. In this case, the intensity (brightness) of image will be half of the initial value.

Active Physics 3

1. Take a plane mirror to observe the image of a distant tree. You may not see a full-length image of the tree. Try with plane mirrors of different sizes. You will observe that there is a certain minimum size of the plane mirror to see the full-length image of the tree. The size of the mirror depends on the distance between the mirror and the distant object. More the distance of the object from the mirror, smaller will be the size of the mirror required to see its full-length image.
2. Take a concave mirror to observe the image of a distant tree. You will not see the image of the tree in the mirror. This is because, to see an image in a concave mirror, the object should be quite close to the concave mirror (between pole and focus).
3. Now, take a convex mirror to observe the image of a distant tree. You will always see the full-length image of the distant tree. This is because, in a convex mirror, a virtual, erect and diminished image of the object is always formed for every location of the object.

Uses of convex mirrors

Rear view mirrors : Convex mirrors are commonly used as rear-view (wing) mirrors in vehicles. These mirrors are fitted on the sides of the vehicle, enabling the driver to see traffic behind him/her to facilitate safe driving.

• Convex mirrors are preferred as rear view mirrors because they always give an erect, though diminished image. Also, they have a wider field of view as they are curved outwards. Thus, convex mirrors enable the driver to view much larger area than would be possible with a plane mirror [see figure (a) and figure (b)]. Street lamps: Street lamps also use convex mirrors to diverge light over an extended area. You can see a full-length image of a tall building/tree in a small convex mirror.

9.0Sign convention for reflection by spherical mirrors

While dealing with the reflection of light by spherical mirrors, we follow a set of sign conventions called the new cartesian sign convention. In this convention, the pole ( P ) of the mirror is taken as the origin. The principal axis of the mirror is taken as the x -axis ( ${\mathrm{X}}^{\prime }\mathrm{X}$ ) of the coordinate system. The conventions are as follows [see figure (a) and (b)]: (1) The object is always placed to the left of the mirror. This implies that the light from the object falls on the mirror from the left-hand side. (2) All distances parallel to the principal axis are measured from the pole of the mirror. All the distances along XX ' axis are measured from P . (3) All the distances measured to the right of the origin (along $+x$-axis) are taken as positive while those measured to the left of the origin (along - x -axis) are taken as negative. (4) Distances measured perpendicular to and above the principal axis (along $+y$-axis) are taken as positive. Distances measured perpendicular to and below the principal axis (along - y -axis) are taken as negative.

• Distances along the direction of incident light are considered 'positive'. Distances measured opposite to the direction of incident light are considered 'negative'.
• If image is virtual and erect i.e., above principal axis, its height is taken 'positive'. If image is real and inverted i.e., below principal axis, its height is taken 'negative'.
  

10.0Formulae related to spherical mirrors

Relationship between radius of curvature and focus The focal length of a spherical mirror is equal to half of its radius of curvature. $f=\frac{R}{2}$

Mirror formula

In a spherical mirror, the distance of the object from its pole is called the object distance (u). The distance of the image from the pole of the mirror is called the image distance (v). The relationship between object distance ( $u$ ), the image distance (v) and the focal length (f) is given by mirror formula which is as given below, $\frac{1}{v}+\frac{1}{u}=\frac{1}{f}$

Magnification (m)

The ratio of height of image ( ${\mathrm{h}}_2$ ) to the height of object ( ${\mathrm{h}}_1$ ) is called 'magnification' or 'linear magnification'. $m=\frac{h_2}{h_1}$ The magnification (m) is also related to the object distance ( $u$ ) and image distance (v). It can be expressed as: $m=\frac{h_2}{h_1}=\frac{-v}{u}$ Also, magnification can be further expressed as, $\mathbf{m}=\frac{\mathbf{f}}{\mathbf{f}-\mathbf{u}}=\frac{\mathbf{f}-\mathbf{v}}{\mathbf{f}}$

• A plane mirror forms a virtual, erect and same size image as that of the object thus, the magnification of a plane mirror is +1 .

Building Concepts 4 A spherical mirror produces a magnification of +1.5 . Explain the nature and size of the image formed by it. Which type of spherical mirror is this? Explanation: Since, the sign of magnification is positive, this means the image is virtual and erect. Now, $|\mathrm{m}|=1.5$, which is greater than one, this means the image is magnified. The spherical mirror in this case is a concave mirror (a converging mirror) as it produces a virtual, erect and magnified image.

Effect on the position of the image formed when an object is moving away from a convex mirror

Effect on the position of the image formed when an object is moving towards a concave mirror

Some important points related to spherical mirrors

(a) Concave mirror (1) Object distance, $u=$ always negative. (2) Image distance, $v=$ positive, when object is placed between $P$ & $F$ (virtual and erect image). $\mathrm{v}=$ negative, all other possible cases (real and inverted image). (3) $f=$ negative, $R=$ negative (4) For concave mirror, ' $m$ ' can be positive as well as negative. Also, $|m|$ can be less than, equal to or greater than one.

(b) Convex mirror (1) Object distance, $u=$ always negative. (2) Image distance, $v=$ always positive (virtual and erect). (3) f = positive, $\mathrm{R}=$ positive. (4) Image is always diminished. (5) For convex mirror, ' $m$ ' is always positive and $|m|$ is always less than one. This is because it always forms a virtual, erect and diminished image of the object.

• We can identify the nature of image in three ways : (i) $\mathrm{v}=-\mathrm{ve}$ (Real and inverted) $\mathrm{v}=+\mathrm{ve}$ (Virtual and erect) (ii) ${\mathrm{h}}_2=+\mathrm{ve}$ (Virtual and erect) ${\mathrm{h}}_2=-\mathrm{ve}$ (Real and inverted) (iii) $\mathrm{m}=+\mathrm{ve}$ (Virtual and erect) $\mathrm{m}=-\mathrm{ve}$ (Real and inverted)
• In numerical if converging mirror is given then, mirror is concave mirror.
• In numerical if diverging mirror is given then, mirror is convex mirror.

Note: How to take LCM by division method. Example : Find LCM of 24 and 15 by the division method. Solution : Step-1 : Divide the given numbers by the least prime number. Here 2 is the least number which will divide 24 .

Step-2 : Write the quotient and the number which is not divisible by the above prime number in the second row.

In the second row, write the quotient we get after the division of 24 by 2 . Since 15 is not divisible by 2 , write 15 in the second row as it is.

Step-3 : Divide the numbers with another least prime number.

Step-4: Continue division until the remainder is a prime number or 1.

Step-5 : Multiply all the divisors and remaining prime number (if any) to obtain the LCM. LCM of 24 and $15=2\times 2\times 2\times 3\times 5=120$.

11.0Numerical Ability 1

1. A convex mirror used for rear-view on an automobile has a radius of curvature of 3.00 m . If a bus is located at 5.00 m from this mirror, find the position, nature and size of the image. Solution: Decode the problem Identify the type of mirror Concave/Convex $✓/$ Plane u will always negative for real object. For convex mirror v is always positive. For convex mirror $f$ is positive. Identify the formula $\mathrm{R}=2\mathrm{f}$ $\mathrm{f}=\mathrm{R}/2$ $\frac{1}{v}+\frac{1}{u}=\frac{1}{f}$ $\mathrm{m}=\frac{{\mathrm{h}}_2}{{\mathrm{h}}_1}=-\frac{\mathrm{v}}{\mathrm{u}}$ Given, radius of curvature, $\mathrm{R}=+3\mathrm{m}$; object distance, $\mathrm{u}=-5\mathrm{m}$; image distance, $\mathrm{v}=$ ? ; magnification, $\mathrm{m}=$ ? ; Now, focal length, $\mathrm{f}=\mathrm{R}/2=+3/2\mathrm{m}$ Mirror formula, $$\begin{gathered} \frac{1}{\mathrm{v}}+\frac{1}{\mathrm{u}}=\frac{1}{\mathrm{f}}\text{ or }\frac{1}{\mathrm{v}}+\frac{1}{(-5)}=\frac{1}{(+3/2)} \\ \text{ or }\frac{1}{\mathrm{v}}-\frac{1}{5}=+\frac{2}{3}\text{ or }\frac{1}{\mathrm{v}}=\frac{1}{5}+\frac{2}{3}=\frac{3+10}{15}=\frac{13}{15} \\ \text{ or }\mathrm{v}=\frac{15}{13}=+1.15m \end{gathered}$$ Thus, the image is 1.15 m at the back of the mirror. Now, magnification, $\mathrm{m}=\frac{{\mathrm{h}}_2}{{\mathrm{h}}_1}=-\frac{\mathrm{v}}{\mathrm{u}}=-\frac{(+15/13)}{(-5)}=+\frac{3}{13}=+0.23$ The image is virtual, erect and smaller in size by a factor of 0.23 .

2. An object, 4.0 cm in size, is placed at 25.0 cm in front of a concave mirror of focal length 15.0 cm . At what distance from the mirror should a screen be placed in order to obtain a sharp image ? Find the nature and the size of the image. Solution: Decode the problem Identify the type of mirror Concave $✓/$ Convex/Plane u will always negative for real object. For concave mirror $f$ is negative. Identify the formula $\frac{1}{v}+\frac{1}{u}=\frac{1}{f}$ $\mathrm{m}=\frac{{\mathrm{h}}_2}{{\mathrm{h}}_1}=-\frac{\mathrm{v}}{\mathrm{u}}$ Given, object size, ${\mathrm{h}}_1=+4\mathrm{cm}$; object distance, $\mathrm{u}=-25\mathrm{cm}$; focal length, $\mathrm{f}=-15\mathrm{cm}$; image distance, $\mathrm{v}=$ ? ; image size, ${\mathrm{h}}_2=$ ? Mirror formula, $\frac{1}{v}+\frac{1}{u}=\frac{1}{f}$ or $\frac{1}{v}+\frac{1}{(-25)}=\frac{1}{-15}$ or $\frac{1}{\mathrm{v}}-\frac{1}{25}=-\frac{1}{15}$ or $\frac{1}{\mathrm{v}}=\frac{1}{25}-\frac{1}{15}=\frac{3-5}{75}=\frac{-2}{75}$ or $\mathrm{v}=\frac{-75}{2}$ $\mathrm{v}=-37.5\mathrm{cm}$ Now, magnification, $\mathrm{m}=\frac{{\mathrm{h}}_2}{{\mathrm{h}}_1}=-\frac{\mathrm{v}}{\mathrm{u}}$ or $h_2=-\frac{\mathrm{v}}{\mathrm{u}}\times {\mathrm{h}}_1=-\frac{(-75/2)}{(-25)}\times (+4)=-\frac{3}{2}\times 4$ $=-6\mathrm{cm}$ The image is real, inverted and enlarged.
3. An object is placed at 10 cm in front of a converging mirror of radius of curvature 15 cm . Find the magnification of the image. Solution: Decode the problem Identify the type of mirror Concave $✓/$ Convex/Plane $u$ will always negative for real object. For concave mirror fand $R$ is negative. Identify the formula $\mathrm{R}=2\mathrm{f}$ $\mathrm{f}=\mathrm{R}/2$ $\frac{1}{v}+\frac{1}{u}=\frac{1}{f}$ $\mathrm{m}=\frac{{\mathrm{h}}_2}{{\mathrm{h}}_1}=-\frac{\mathrm{v}}{\mathrm{u}}$ Given, object distance, $\mathrm{u}=-10\mathrm{cm}$; radius of curvature, $\mathrm{R}=-15\mathrm{cm}$; image distance, $\mathrm{v}=$ ? ; magnification, $\mathrm{m}=$ ? Focal length, $f=-15/2\mathrm{cm}$ Mirror formula, $\frac{1}{v}+\frac{1}{u}=\frac{1}{f}$ or $\frac{1}{\mathrm{v}}+\frac{1}{(-10)}=\frac{1}{(-15/2)}$ or $\frac{1}{\mathrm{v}}-\frac{1}{10}=-\frac{2}{15}$ or $\frac{1}{\mathrm{v}}=\frac{1}{10}-\frac{2}{15}=\frac{3-4}{30}=-\frac{1}{30}$ or $v=-30\mathrm{cm}$ Now, magnification, $\mathrm{m}=-\frac{\mathrm{v}}{\mathrm{u}}=-\frac{(-30)}{(-10)}=-3$
4. When the distance of an object from a concave mirror is decreased from $15\mathbf{cm}$ to 9 cm, the image gets magnified 3 times than that in first case. Calculate the focal length of the mirror. Solution: Decode the problem Identify the type of mirror Concave $✓$ /Convex/Plane u will always negative for real object. For concave mirror f is negative. Identify the formula $m=\frac{f-v}{f}$ $m=\frac{f}{f-u}✓$ (since $f$ and $u$ are given) Given, in first case $\mathrm{u}=-15\mathrm{cm}$ $\therefore \mathrm{m}=\frac{\mathrm{f}}{\mathrm{f}+15}$ Given, in second case $u=-9\mathrm{cm}$ $\therefore {\mathrm{m}}^{\prime }=\frac{\mathrm{f}}{\mathrm{f}+9}$ But m' $=3\mathrm{m}$ or $\frac{f}{f+9}=\frac{3\times f}{f+15}$ or $\mathrm{f}+15=3\mathrm{f}+27$ or $f=-6\mathrm{cm}$.

12.0Some Basic Terms

1. Electromagnetic waves: A changing magnetic field will induce a changing electric field and vice-versa. These changing fields form electromagnetic waves. Electromagnetic wave do not required a medium to propagate.
2. Source of light : Any object which emits light is called source of light. To able to see anything, we need a source of light.
3. Natural source of light : A natural source of light is light that occurs without human involvement. Natural sources of light can come from objects of living species. Examples: Sun, star, fireflies etc.
4. Artificial source of light : Artificial sources of light are man made objects that emit light. Examples : Bulbs, tube lights, candles, lasers etc.
5. Medium : A medium is defined as the substance that transfers the energy or light from one substance to another substance or from one place to another. The medium act as a carrier here. The medium can transfer any form of energy, sound wave, light and heat.
6. Homogeneous medium : An optical medium which has a uniform composition throughout is called a homogeneous medium. Examples: Glass, diamond, distilled water etc.
7. Transparent: A medium that allows light to pass through it easily is called a transparent medium. Examples : Glass, air etc.
8. Waves : A wave is a disturbance that transfers energy from one place to another in a regular and organized way.
9. Wavelength : Wavelength is the distance from one crest to another, or from one trough to another of a wave (Which may be an electromagnetic wave, a sound wave or any other wave). Crest is the highest point of the wave where as the trough is the lowest. SI unit is metre (m).
10. Amplitude : It is the maximum displacement from its mean position to extreme position of a particle of the medium in which a waves propagates. SI unit is metre (m).
11. Intensity : It is the amount of light falling on a surface. It is measured in terms of lumens per square metre (lux).
12. Frequency : It is the number of complete cycles of waves passing a point in unit time. The SI unit of frequency is Hertz ( Hz ).
13. Asymmetric object : Asymmetric objects have two sides that are not mirror images of each other. If you draw a line down the middle of an asymmetric object an fold it in half, the two side will not match.
14. Symmetric object : If any object can be divided into two halves such that one half forms the mirror image of the other half.
15. Obliquely : Obliquely refers that when a light falling on the surface in slanting position neither perpendicular nor horizontal.
16. Angle : An angle is geometric shape formed by the intersection of two line segments, lines or rays.

13.0SOLVED EXAMPLES

1. A small candle, 2.5 cm in size is placed at 27 cm in front of a concave mirror of radius of curvature 36 cm . At what distance from the mirror should a screen be placed in order to obtain a sharp image? Describe the nature and size of the image. If the candle is moved closer to the mirror, how would the screen have to be moved? Solution: Given, object distance, $\mathrm{u}=-27\mathrm{cm}$; radius of curvature, $\mathrm{R}=-36\mathrm{cm}$; object size, ${\mathrm{h}}_1=$ +2.5 cm ; image distance, $\mathrm{v}=$ ? ; image size, ${\mathrm{h}}_2=$ ? Now, focal length, $\mathrm{f}=\mathrm{R}/2=-36/2\mathrm{cm}=-18\mathrm{cm}$ Mirror formula, $$\begin{gathered} \frac{1}{v}+\frac{1}{u}=\frac{1}{f} \\ \text{ or }\frac{1}{v}+\frac{1}{(-27)}=\frac{1}{(-18)} \end{gathered}$$ or $\frac{1}{v}-\frac{1}{27}=-\frac{1}{18}$ or $\frac{1}{v}=\frac{1}{27}-\frac{1}{18}=\frac{2-3}{54}=\frac{-1}{54}$ or $\mathrm{v}=-54\mathrm{cm}$ Now, magnification, or ${\mathrm{h}}_2=-\frac{\mathrm{v}}{\mathrm{u}}\times {\mathrm{h}}_1=-\frac{(-54)}{(-27)}\times (+2.5)=-5\mathbf{cm}$ The image is real, inverted and enlarged.

When the candle is moved closer to the mirror, the screen has to be moved away from the mirror. But, when candle is at a distance less than 18 cm (i.e., less than its focal length) from the mirror, image formed will be virtual which is not possible to obtain on the screen.

2. A 4.5 cm needle is placed 12 cm away from a convex mirror of focal length 15 cm . Give the location of the image and the magnification. Describe what happens as the needle is moved farther from the mirror. Solution: Given, object distance, $\mathrm{u}=-12\mathrm{cm}$; focal length, $\mathrm{f}=+15\mathrm{cm}$; object size, ${\mathrm{h}}_1=+4.5$ cm ; image distance, $\mathrm{v}=$ ? ; magnification, $\mathrm{m}=$ ?, image size, ${\mathrm{h}}_2=$ ? Mirror formula, $\frac{1}{v}+\frac{1}{u}=\frac{1}{f}$ or $\frac{1}{v}+\frac{1}{(-12)}=\frac{1}{(+15)}$ or $\frac{1}{\mathrm{v}}-\frac{1}{12}=\frac{1}{15}$ or $\frac{1}{v}=\frac{1}{15}+\frac{1}{12}=\frac{4+5}{60}=\frac{9}{60}$ or $v=+(60/9)\mathrm{cm}=+6.67\mathrm{cm}$ Now, magnification, $m=-\frac{v}{u}=-\frac{(+60/9)}{(-12)}=+\frac{5}{9}=+0.55$ Also, magnification, $m=\frac{h_2}{h_1}$ or $+\frac{5}{9}=\frac{{\mathrm{h}}_2}{(+4.5)}$ or ${\mathrm{h}}_2=+\frac{5}{9}\times (+4.5)=+2.5\mathrm{cm}$ Image is virtual, erect and diminished.

• If the needle (object) is moved farther from the mirror, its image moves away from the mirror i.e., from pole towards the focus. The image remains virtual and erect but it gradually decreases in size.
• When the object becomes infinitely far away, the image is formed at the focus and it is a point sized image. But, the image never goes beyond the focus in a convex mirror.

3. A concave mirror has a radius of curvature of 24.0 cm . An object 2.5 cm tall is placed 40.0 cm in front of the mirror. (a) At what distance from the mirror will the image be formed? (b) What is the height of the image? Solution: Given, object distance, $\mathrm{u}=-40\mathrm{cm}$; radius of curvature, $\mathrm{R}=-24\mathrm{cm}$; object height, ${\mathrm{h}}_1=2.5\mathrm{cm}$; image distance, $\mathrm{v}=$ ? ; height of image, ${\mathrm{h}}_2=$ ? (a) Focal length, $\mathrm{f}=\mathrm{R}/2=(-24)/2=-12\mathrm{cm}$ Mirror formula, $\frac{1}{v}+\frac{1}{u}=\frac{1}{f}$ or $\frac{1}{\mathrm{v}}+\frac{1}{(-40)}=\frac{1}{(-12)}$ or $\frac{1}{\mathrm{v}}=\frac{1}{40}-\frac{1}{12}=\frac{3-10}{120}=\frac{-7}{120}$ or $v=-120/7\mathrm{cm}=-17.14\mathbf{cm}$ (b) Magnification, $m=\frac{h_2}{h_1}=-\frac{v}{u}$ or $\frac{{\mathrm{h}}_2}{{\mathrm{h}}_1}=-\frac{\mathrm{v}}{\mathrm{u}}\text{ or }\frac{{\mathrm{h}}_2}{(+2.5)}=-\frac{(-120/7)}{(-40)}$ or $\frac{{\mathrm{h}}_2}{(+2.5)}=-\frac{3}{7}$ or $h_2=-\frac{3}{7}\times 2.5=-1.07\mathbf{cm}$
4. A convex supermarket surveillance mirror has a radius of curvature of 80.0 cm . A 1.7 m tall customer is standing 4.5 m in front of the mirror. (a) What is the location of the customer's image in the mirror? (b) What is the height of the customer's image? Solution: Given, object distance, $\mathrm{u}=-4.5\mathrm{m}=-450\mathrm{cm}$ ; radius of curvature, $\mathrm{R}=+80\mathrm{cm}$; object height, ${\mathrm{h}}_1=+1.7\mathrm{m}$; image distance, $\mathrm{v}=$ ? ; image height, ${\mathrm{h}}_2=$ ? (a) Focal length, $f=R/2=(+80)/2=+40\mathrm{cm}$ Mirror formula, $\frac{1}{\mathrm{v}}+\frac{1}{\mathrm{u}}=\frac{1}{\mathrm{f}}$ or $\frac{1}{\mathrm{v}}+\frac{1}{(-450)}=\frac{1}{(+40)}$ or $\frac{1}{v}=\frac{1}{450}+\frac{1}{40}=\frac{4+45}{1800}=\frac{+49}{1800}$ or $v=+1800/49\mathrm{cm}=+36.73\mathbf{cm}$ (b) Now, $\frac{{\mathrm{h}}_2}{{\mathrm{h}}_1}=-\frac{\mathrm{v}}{\mathrm{u}}$ or $\frac{{\mathrm{h}}_2}{(+1.7)}=-\frac{(+1800/49)}{(-450)}$ or $\frac{{\mathrm{h}}_2}{(+1.7)}=+\frac{4}{49}$ or ${\mathrm{h}}_2=+\frac{4}{49}\times 1.7=+0.1387\mathbf{m}$
5. What is the radius of curvature of a concave mirror that magnifies an object placed 30.0 cm from the mirror by a factor of +3.0 ? Solution: Given, object distance, $u=-30\mathrm{cm}$; magnification, $m=+3$; radius of curvature, $\mathrm{R}=$ ? Magnification, or $v=-m\times u$ or $\mathrm{v}=-(+3)\times (-30)=+90\mathrm{cm}$ Mirror formula, $\frac{1}{\mathrm{v}}+\frac{1}{\mathrm{u}}=\frac{1}{\mathrm{f}}$ or $\frac{1}{(+90)}+\frac{1}{(-30)}=\frac{1}{\mathrm{f}}$ or $\frac{1}{\mathrm{f}}=\frac{1}{90}-\frac{1}{30}=\frac{1-3}{90}=\frac{-2}{90}$ or $f=-90/2=-45\mathrm{cm}$ $R=2f=2\times (-45)=-90\mathbf{cm}$
6. A dancer is applying make-up in a concave mirror. Her face is 35 cm in front of the mirror. The image is 70 cm behind the mirror. Using the mirror equation, find the focal length of the mirror. What is the magnification of the image? Solution: Object distance, $u=-35\mathrm{cm}$; image distance, $\mathrm{v}=+70\mathrm{cm}$; Mirror formula, $\frac{1}{\mathrm{v}}+\frac{1}{\mathrm{u}}=\frac{1}{\mathrm{f}}$ or $\frac{1}{(+70)}+\frac{1}{(-35)}=\frac{1}{\mathrm{f}}$ or $\frac{1}{f}=\frac{1}{70}-\frac{1}{35}=\frac{1-2}{70}=\frac{-1}{70}$ or $f=-70\mathrm{cm}$ Magnification, $m=-\frac{v}{u}=-\frac{(+70)}{(-35)}=+2$
7. Light from a distant planet is incident on a converging mirror. The image of the planet forms on a screen 45.0 cm from the pole of the mirror. Find the focal length of the mirror. Solution: Given, object distance $=-\infty$; image distance, $\mathrm{v}=-45\mathrm{cm}$; focal length, $\mathrm{f}=$ ? Mirror formula, $\frac{1}{\mathrm{v}}+\frac{1}{\mathrm{u}}=\frac{1}{\mathrm{f}}\text{ or }$ $\frac{1}{(-45)}+\frac{1}{(-\infty )}=\frac{1}{f}$ $\mathrm{f}=-45\mathbf{cm}$
8. How far should one hold an object from a concave mirror of focal length 40 cm , so as to get an image twice the size of the object? Solution Given, focal length, $\mathrm{f}=-40\mathrm{cm}$; $|\mathrm{m}|=2$; object distance, $\mathrm{u}=$ ? Here, it is not mentioned whether the image is real or virtual. Since, in a concave mirror, we can obtain a magnified real or virtual image, thus, we have to consider both the cases. Case 1 : For real image, magnification, $\mathrm{m}=-2$ Magnification, $\mathrm{m}=\frac{-\mathrm{v}}{\mathrm{u}}$ or $\mathrm{v}=-\mathrm{m}\times \mathrm{u}$ or $\mathrm{v}=-(-2)\times \mathrm{u}$ or $\mathrm{v}=2\mathrm{u}$ Mirror formula, $\frac{1}{v}+\frac{1}{u}=\frac{1}{f}$ or [using (1)] or $\frac{1}{2\mathrm{u}}+\frac{1}{\mathrm{u}}=\frac{1}{\mathrm{f}}$ or $\frac{1+2}{2\mathrm{u}}=\frac{1}{(-40)}$ or $\frac{3}{2\mathrm{u}}=\frac{1}{(-40)}$ or $u=-\frac{40\times 3}{2}-60\mathbf{cm}$ Case 2: For virtual image, magnification, $\mathrm{m}=+2$ Magnification, $\mathrm{m}=\frac{-\mathrm{v}}{\mathrm{u}}$ or $\mathrm{v}=-\mathrm{m}\times \mathrm{u}$, or $v=-(+2)\times u$ or $\mathrm{v}=-2\mathrm{u}$ Mirror formula, $\frac{1}{\mathrm{v}}+\frac{1}{\mathrm{u}}=\frac{1}{\mathrm{f}}$ or $\frac{1}{(-2\mathrm{u})}+\frac{1}{\mathrm{u}}=\frac{1}{(-40)}[$ using (1)] or $\frac{-1+2}{2u}=\frac{1}{(-40)}$ or $\frac{1}{2u}=\frac{1}{(-40)}$ or $u=-\frac{40}{2}=-20\mathbf{cm}$
9. A concave mirror produces a magnification of $1/2$ when an object is at 60 cm from it. Where should the object be placed so that a virtual image of double the size is formed by the mirror? Solution: Given, object distance, $\mathrm{u}=-60\mathrm{cm}$; Magnification, $m=-1/2$, negative sign taken as a concave mirror forms a real image. Magnification, $m=\frac{-v}{u}$ or $v=-m\times u$ or $v=-(-1/2)\times (-60)=-30\mathrm{cm}$ Mirror formula, $$\begin{gathered} \frac{1}{\mathrm{v}}+\frac{1}{\mathrm{u}}=\frac{1}{\mathrm{f}} \\ \text{ or }\frac{1}{(-30)}+\frac{1}{(-60)}=\frac{1}{\mathrm{f}} \\ \text{ or }\frac{1}{\mathrm{f}}=\frac{-2-1}{60}=-\frac{3}{60}=-\frac{1}{20} \\ \text{ or }\mathrm{f}=-20\mathrm{cm} \end{gathered}$$ Now, we have to find the object distance ( $u$ ') when a virtual image of double size is formed i.e., the magnification, $m=+2$. Magnification, $m=\frac{-v}{u}$ or $v=-m\times u$ or $v=-(+2)\times \left(u^{\prime }\right)$ or $v=-2u^{\prime }$ Mirror formula, $\frac{1}{\mathrm{v}}+\frac{1}{\mathrm{u}}=\frac{1}{\mathrm{f}}$ or $\frac{1}{\left(-2u^{\prime }\right)}+\frac{1}{u^{\prime }}=\frac{1}{(-20)}[\mathrm{using}(1)]$ or $\frac{-1+2}{2{\mathrm{u}}^{\prime }}=\frac{1}{-20}$ or $\frac{1}{2{\mathrm{u}}^{\prime }}=\frac{1}{-20}$, or $u^{\prime }=-\frac{20}{2}=-10\mathbf{cm}$
10. The height of the real image formed by a concave mirror is four times the object height when the object is 30.0 cm in front of the mirror. What is the radius of curvature of the mirror? Solution: Given, object distance, $\mathrm{u}=-30\mathrm{cm}$ magnification $=-4$, negative sign taken as a concave mirror forms a real image; radius of curvature, $\mathrm{R}=$ ? Magnification, $\mathrm{m}=\frac{-\mathrm{v}}{\mathrm{u}}$ or $\mathrm{v}=-\mathrm{m}\times \mathrm{u}$ or $v=-(-4)\times (-30)$ or $\mathrm{v}=-120\mathrm{cm}$ Mirror formula, $$\begin{gathered} \frac{1}{\mathrm{v}}+\frac{1}{\mathrm{u}}=\frac{1}{\mathrm{f}} \\ \text{ or }\frac{1}{(-120)}+\frac{1}{(-30)}=\frac{1}{\mathrm{f}} \end{gathered}$$ or $-\frac{1}{120}-\frac{1}{30}=\frac{1}{\mathrm{f}}$ or $\frac{-1-4}{120}=\frac{1}{f}$ or $\frac{-5}{120}=\frac{1}{\mathrm{f}}$ or $f=-\frac{120}{5}=-24\mathrm{cm}$ $\mathrm{R}=2\mathrm{f}=2\times (-24)=-48\mathbf{cm}$
11. A spherical mirror is to be used to form, on a screen 5.00 m from the mirror, an image five times the size of the object. (a) Describe the type of mirror required. (b) Where should the mirror be positioned relative to the object? Solution: (a) Since, the image is obtained on a screen, it is a real image. Thus, the spherical mirror is a concave mirror (or converging mirror). (b) Image distance, $\mathrm{v}=-5\mathrm{m}$; magnification, $m=-5$, negative sign taken as a concave mirror forms a real image ; object distance, $\mathrm{u}=$ ? Magnification, or $u=-\frac{v}{m}=-\frac{(-5)}{(-5)}=-1\mathbf{m}$ Thus, the mirror should be positioned 1 m away from the object.
12. If an object of height 4 cm is placed at distance of 12 cm from a concave mirror having focal length 24 cm , find the position, nature and the height of the image. Solution: Given, object height, $h_1=4\mathrm{cm}$; object distance, $u=-12\mathrm{cm}$; focal length, $\mathrm{f}=-24\mathrm{cm}$; image distance, $\mathrm{v}=$ ?; height of image ${\mathrm{h}}_2=$ ? Mirror formula, $\frac{1}{v}+\frac{1}{u}=\frac{1}{f}$ or $\frac{1}{\mathrm{v}}+\frac{1}{(-12)}=\frac{1}{(-24)}$ or $\frac{1}{\mathrm{v}}=\frac{1}{12}-\frac{1}{24}$ or $\frac{1}{\mathrm{v}}=\frac{2-1}{24}$ or $\frac{1}{v}=\frac{1}{24}$ $\mathrm{v}=24\mathbf{cm}$ Now, magnification $\mathrm{m}=\frac{-\mathrm{v}}{\mathrm{u}}=\frac{-24}{-12}=2$ or $\frac{{\mathrm{h}}_2}{{\mathrm{h}}_1}=2$ or ${\mathrm{h}}_2=2{\mathrm{h}}_1=2\times 4=8\mathbf{cm}$ Since v is positive, the image is formed behind the mirror at the distance of 24 cm from the mirror. It is virtual and has height of 8 cm .