Class 10 Science Light Reflection Exam Notes

Class 10 Science Light Reflection Exam Notes. Please refer to the examination notes which you can use for preparing and revising for exams. These notes will help you to revise the concepts quickly and get good marks.

LIGHT: REFLECTION

THE NATURE OF LIGHT

Light is a form of energy (optical energy) which helps us in seeing objects by its presence.

(A) THEORIES ABOUT THE NATURE OF LIGHT

(i) Particle nature of light (Newton’s corpuscular theory). According to Newton, light travels in space with a great speed as a stream of very small particles called corpuscles. This theory failed to explain interference and diffraction of light, so the wave theory of light was put forward.

(ii) Wave nature of light. Light waves are electromagnetic waves, so they do not need a medium for propagation and can travel in vacuum also. The speed of these waves in air or in vacuum is the maximum possible, 3 × 10⁸ m/s. The photoelectric effect could not be explained with the help of the wave theory, so Planck gave a new theory known as the quantum theory of light.

(iii) Quantum theory of light. When light falls on the surface of metals like caesium and potassium, electrons are given out. These electrons are called photoelectrons and the phenomenon is called the photoelectric effect. It was explained by Einstein. According to Planck, light consists of packets or quanta of energy called photons. The rest mass of a photon is zero. Each quantum carries energy

\[E=h\nu\]

where h is Planck’s constant (6.6 × 10⁻³⁴ J s) and \(\nu\) is the frequency of the light.

Some phenomena, such as interference and diffraction of light, are explained with the help of the wave theory, but the wave theory fails to explain the photoelectric effect, which is explained by the quantum theory. So light has a dual nature: (i) wave nature and (ii) particle nature.

(B) SOURCES OF LIGHT

• Objects which emit (give out) light are called luminous objects. They may be natural or man-made. The Sun is a natural source of light; an electric lamp and an oil lamp are man-made sources of light.
• Non-luminous objects do not emit light. However, such objects become visible due to the reflection of the light falling on them. The Moon does not emit light; it becomes visible due to the reflection of the sunlight falling on it.

(C) PROPAGATION OF LIGHT

Light travels along straight lines in a medium or in vacuum. The path of light changes only when there is an object in its path or where the medium changes. Apart from vacuum and gases, light can travel through some liquids and solids as well.

• Transparent medium: a medium in which light can travel freely over large distances. Examples: water, glycerin, glass and clear plastics.
• Opaque medium: a medium in which light cannot travel. Examples: wood, metals, bricks.
• Translucent medium: a medium in which light can travel some distance, but its intensity reduces rapidly. Examples: oil (also oil paper, frosted glass).

(D) THE CHARACTERISTICS OF LIGHT

• Light is an electromagnetic wave, and it is a transverse wave. It does not need any medium to travel and can travel through vacuum.
• Light travels in a straight line.
• The speed of light in vacuum is 3 × 10⁸ m/s. (Scientists have assigned the value 299 792 458 m/s to the speed of light in vacuum.)
• According to current scientific theories, no material particle can travel at a speed greater than that of light in vacuum.
• The velocity of light changes when it travels from one medium to another.
• The wavelength (λ) of light changes when it goes from one medium to another.
• The frequency (f) of a light wave remains the same in all media.
• Light gets reflected back from polished surfaces such as mirrors and polished metal surfaces.
• Light undergoes refraction (bending) when it travels from one transparent medium to another.

REFLECTION OF LIGHT

When light rays are incident on an opaque, polished surface (medium), they are returned back into the same medium. This phenomenon of the returning of a ray of light into the same medium is called reflection of light.

Terms associated with reflection. Consider a ray PO incident at the point O on a reflecting surface XY, with the normal ON and the reflected ray OQ.

• Reflecting surface: the surface from which the light is reflected (XY).
• Point of incidence: the point on the reflecting surface at which a ray of light strikes (O).
• Normal: the perpendicular drawn on the reflecting surface at the point of incidence (ON).
• Incident ray: the ray of light which strikes the reflecting surface at the point of incidence (PO).
• Reflected ray: the ray of light reflected from the reflecting surface at the point of incidence (OQ).
• Angle of incidence (i): the angle that the incident ray makes with the normal (angle PON).
• Angle of reflection (r): the angle that the reflected ray makes with the normal (angle QON).
• Plane of incidence: the plane in which the normal and the incident ray lie (the plane of the book page in the usual diagram).
• Plane of reflection: the plane in which the normal and the reflected ray lie.

LAWS OF REFLECTION OF LIGHT

• 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 reflection (r) is always equal to the angle of incidence (i), i.e. r = i.

(For normal incidence, i = 0 and r = 0: the ray is reflected back along the normal.)

1. A ray of light striking the surface normally retraces its path. When a ray strikes a surface normally, the angle of incidence is zero. By the law of reflection r = i = 0, so the reflected ray is also perpendicular to the surface; an incident ray normal to the surface retraces its path.
2. The laws of reflection are also obeyed when light is reflected from spherical or curved surfaces.
3. Regular and irregular reflection. Regular reflection: a parallel beam of light travelling through a medium, on striking a smooth, polished surface, bounces off as a parallel beam in some other fixed direction. It takes place from a looking glass, still water, oil and highly polished metals. Regular reflection is useful in the formation of images, e.g. we can see our face in a mirror only because of regular reflection; however, it causes a very strong glare in our eyes. Irregular (diffused) reflection: a parallel beam of light, on striking a rough surface, gets reflected in various directions. It takes place from the ground, walls, trees, suspended particles in air and a variety of other objects that are not very smooth. Irregular reflection spreads light energy over a vast region and decreases its intensity; it helps in the general illumination of places and helps us to see things around us.

Note: the laws of reflection are valid whether the reflection is regular or irregular.

NATURE OF IMAGE

Incident rays starting from a point object and reflected from a mirror either actually meet at a point or appear to come from a point; that point is called the image of the point object.

REFLECTION FROM A PLANE MIRROR

The distance of the object from a plane mirror and the distance of its image from the mirror are equal. To verify this, consider rays OP and OD starting from the object O and falling on the mirror. The ray OP is perpendicular to the mirror and is reflected back along PO. The incident ray OD and the reflected ray DE make equal angles with the normal DG. The two reflected rays, when produced backwards, meet at I, producing a virtual image there.

Now ∠EDG = ∠DIO (DG ∥ IO), ∠EDG = ∠GDO (law of reflection), and ∠GDO = ∠DOI (DG ∥ IO). Hence ∠DIO = ∠DOI, so OD = DI. Also \(OP^2=OD^2-DP^2\) and \(PI^2=DI^2-DP^2\). Since OD = DI, \(OP^2=PI^2\), i.e. OP = PI. So in a plane mirror the image is formed as far behind the mirror as the object is in front of it.

SOME IMPORTANT RESULTS ABOUT REFLECTION FROM PLANE SURFACES

Lateral inversion. When you see your image in a vertical plane mirror, the head in the image is up and the feet are down, the same way as you actually stand; such an image is called an erect image. However, if you move your right hand, it appears as if the left hand of your image is moving. If you keep a printed page in front of a plane mirror, the letters of the image appear erect but inverted sideways. This inversion is called lateral inversion.

Relative motion of object and image. Case I: if an object moves towards (or away from) a plane mirror at speed v, the image also approaches (or recedes) at speed v, so the speed of the image relative to the object is v − (−v) = 2v. Case II: if the mirror is moved towards (or away from) the object with speed v, the image moves towards (or away from) the object with speed 2v.

Multiple reflection. The number of images formed by a combination of plane mirrors depends on the angle between the mirrors. If two plane mirrors are inclined at 90° to each other, the number of images of a point object is 3. (In general, the number of images is \(\dfrac{360^\circ}{\theta}-1\).)

Deviation. The deviation δ is the angle between the directions of the incident ray and the emergent ray. If light is incident at an angle of incidence i:

\[\delta=180^\circ-(i+r)=180^\circ-2i\quad(\text{as } i=r)\]

So if light is incident at 30°, \(\delta=180^\circ-2\times30^\circ=120^\circ\); for normal incidence i = 0° and δ = 180°.

Characteristics of the image formed by a plane mirror

1. The image is virtual.
2. The image is erect.
3. The size of the image is the same as that of the object. If the object is 10 cm high, the image is also 10 cm high.
4. The image is formed as far behind the mirror as the object is in front of it. If an object is 5 cm in front of a plane mirror, its image is 5 cm behind the mirror.
5. The image is laterally inverted: the right side of the object appears as the left side of the image and vice versa.

REFLECTION FROM A SPHERICAL MIRROR

There are two types of spherical mirrors: (i) concave mirror, whose inner hollow surface is the reflecting surface, and (ii) convex mirror, whose outer bulging surface is the reflecting surface.

Some terms associated with spherical mirrors

• Aperture: the diameter of the circular rim of the mirror (AB in the usual diagram).
• Pole (P): the centre of the spherical surface of the mirror; it lies on the surface of the mirror.
• Centre of curvature (C): the centre of the spherical shell of which the mirror is a section. It lies outside the mirror surface, and every point on the mirror surface is at the same distance from it.
• Principal axis: the straight line passing through the pole and the centre of curvature of the mirror.
• Principal focus (F): a point on the principal axis such that rays incident on the mirror parallel to the principal axis, after reflection, actually meet at this point (concave mirror) or appear to come from it (convex mirror).
• Radius of curvature (R): the distance between the pole and the centre of curvature; it equals the radius of the spherical shell of which the mirror is a section.
• Focal length (f): the distance between the pole and the principal focus. Magnitude \(f=\dfrac{R}{2}\); with the sign convention, \(f=+\dfrac{R}{2}\) for a convex mirror and \(f=-\dfrac{R}{2}\) for a concave mirror.
• Principal section: a section of the spherical mirror cut by a plane passing through its centre of curvature and its pole. It contains the principal axis.

RULES FOR IMAGE FORMATION BY THE RAY DIAGRAM METHOD

Concave mirror

1. A ray of light incident parallel to the principal axis is reflected through the principal focus F.
2. A ray of light incident towards (or passing through) the principal focus is reflected parallel to the principal axis.
3. A ray of light incident towards (or passing through) the centre of curvature is reflected back along the same path.
4. A ray of light incident on the pole of the mirror is reflected making the same angle with the principal axis on the other side (i = r).

Convex mirror

1. A ray of light incident parallel to the principal axis is reflected so that it appears to come from the principal focus F behind the mirror.
2. A ray of light incident on the mirror directed towards the centre of curvature C (behind the mirror) is reflected back along the same path.

(The ray diagrams are not reproduced here.)

SIGN CONVENTION

The New Cartesian sign convention fixes the signs of the different distances:

1. All distances are measured from the pole of the mirror.
2. Distances measured in the same direction as the incident light from the pole are taken as positive.
3. Distances measured in the direction opposite to the incident light from the pole are taken as negative.
4. Distances measured upward and perpendicular to the principal axis are taken as positive.
5. Distances measured downward and perpendicular to the principal axis are taken as negative.

IMAGE FORMATION BY A SPHERICAL MIRROR IN DIFFERENT CASES

From the mirror formula, for a mirror of fixed focal length f, as the object distance u changes, the image distance v also changes.

(A) Concave mirror

1. Object at infinity: rays come parallel to the principal axis and, after reflection, actually meet at the focus F. The image is formed at F; it is real, inverted and point-sized (highly diminished).
2. Object beyond the centre of curvature: the image is formed between the focus and the centre of curvature; it is real, inverted and diminished.
3. Object at the centre of curvature: the image is formed at the centre of curvature; it is real, inverted and of the same size as the object.
4. Object between the centre of curvature and the focus: the image is formed beyond the centre of curvature; it is real, inverted and enlarged.
5. Object at the focus: the image is formed at infinity (the reflected rays are parallel); it is real, inverted and highly enlarged.
6. Object between the focus and the pole: the image is formed behind the mirror; it is virtual, erect and enlarged.

(B) Convex mirror

1. Object at infinity: rays come parallel to the principal axis and, after reflection, appear to diverge from the focus F behind the mirror. The image is formed at F; it is virtual and point-sized (highly diminished).
2. Object anywhere between infinity and the pole: the image is formed behind the mirror between the pole and the focus; it is virtual, erect and diminished.

(The corresponding ray diagrams are not reproduced here. The source described the image for an object at the focus of a concave mirror as “imaginary”; the image is real, inverted and formed at infinity, as stated above.)

NUMERICAL METHOD IN SPHERICAL MIRRORS

(A) Mirror formula. The equation relating the object distance (u), the image distance (v) and the focal length (f) of a mirror is called the mirror formula:

\[\dfrac1v+\dfrac1u=\dfrac1f\]

Assumptions made: (i) the mirror has a small aperture; (ii) the object lies close to the principal axis of the mirror; (iii) the incident rays make small angles with the mirror surface or the principal axis.

(B) Linear magnification. The ratio of the size (height) of the image formed by the mirror to the size (height) of the object is called the linear magnification, m:

\[m=\dfrac{h'}{h}=-\dfrac{v}{u}\]

(C) Power of a mirror (in dioptres) \(=\dfrac{1}{f\ (\text{in metre})}\).

SUMMARY OF IMAGES BY SPHERICAL MIRRORS

SOLVED EXAMPLES

Example 1

An object is placed in front of a plane mirror. If the mirror is moved away from the object through a distance x, by how much distance will the image move?

Answer: The image moves through a distance 2x.

Let the object O be at a distance d from the mirror M. The image O′ is at a distance d behind the mirror, so OO′ = 2d. The mirror is shifted by a distance x to M′, so the object is now at a distance d + x from the mirror, and the image O″ is at a distance d + x behind M′, so OO″ = 2(d + x). The shift of the image is O′O″ = 2(d + x) − 2d = 2x.

Example 2

An insect is at a distance of 1.5 m from a plane mirror. Calculate (i) the distance at which the image of the insect is formed, (ii) the distance between the insect and its image.

Answer: (i) 1.5 m behind the mirror; (ii) 3 m.

(i) The insect is 1.5 m from the mirror, so its image is formed 1.5 m behind the mirror. (ii) The distance between the insect and its image = 1.5 + 1.5 = 3 m.

Example 3

A concave mirror is made by cutting a portion of a hollow glass sphere of radius 30 cm. Calculate the focal length of the mirror.

Answer: 15 cm.

Radius of curvature R = 30 cm, so \(f=\dfrac R2=\dfrac{30}{2}=15\) cm (the focal length of a concave mirror is −15 cm by the sign convention).

Example 4

An object is placed at a distance of 15 cm from a concave mirror of focal length 10 cm. Find the position of the image.

Answer: The image is formed 30 cm in front of the mirror.

u = −15 cm, f = −10 cm. \(\dfrac1v+\dfrac1u=\dfrac1f\) gives \(\dfrac1v-\dfrac1{15}=-\dfrac1{10}\), so \(\dfrac1v=\dfrac1{15}-\dfrac1{10}=-\dfrac1{30}\) and v = −30 cm. The negative sign shows that the image is formed in front of the mirror, 30 cm from it.

Example 5

A 3 cm long object is placed perpendicular to the principal axis of a concave mirror. The distance of the object from the mirror is 15 cm, and its image is formed 30 cm from the mirror on the same side of the mirror as the object. Calculate the height of the image formed.

Answer: 6 cm, inverted.

u = −15 cm, v = −30 cm, h = 3 cm. \(m=\dfrac{h'}{h}=-\dfrac vu=-\dfrac{-30}{-15}=-2\), so h′ = −2 × 3 = −6 cm. The height of the image is 6 cm; the minus sign shows that it is below the principal axis, i.e. inverted.

(The source wrote the size of the object as 2 cm in one line; the question gives 3 cm.)

Example 6

A 1.4 cm long object is placed perpendicular to the principal axis of a convex mirror of focal length 15 cm at a distance of 10 cm from it. Calculate (i) the location of the image, (ii) the height of the image, (iii) the nature of the image.

Answer: (i) 6 cm behind the mirror; (ii) 0.84 cm; (iii) virtual, erect and diminished.

(i) For a convex mirror f = +15 cm; u = −10 cm. \(\dfrac1v=\dfrac1f-\dfrac1u=\dfrac1{15}+\dfrac1{10}=\dfrac16\), so v = +6 cm. The image is formed behind the mirror, 6 cm from it.

(ii) \(m=-\dfrac vu=-\dfrac{6}{-10}=+0.6\), so \(h'=0.6\times1.4=0.84\) cm.

(iii) Since h′ is positive, the image is on the same side of the principal axis as the object, i.e. erect. It is virtual, erect and diminished.

Example 7

An object is placed at a distance of 40 cm from a convex mirror of focal length 30 cm. Find the position of the image and its nature.

Answer: 17.1 cm behind the mirror; virtual, erect and diminished.

u = −40 cm, f = +30 cm. \(\dfrac1v=\dfrac1{30}+\dfrac1{40}=\dfrac{7}{120}\), so \(v=\dfrac{120}{7}=+17.1\) cm. The positive sign shows that the image is formed behind the mirror. Magnification \(m=-\dfrac vu=\dfrac{120/7}{40}=+\dfrac37\); the positive value shows that the image is erect, and since m < 1 it is diminished.

Example 8

A 3 cm high object is placed at a distance of 30 cm from a concave mirror. A real image is formed 60 cm from the mirror. Calculate the focal length of the mirror and the size of the image.

Answer: f = −20 cm (focal length 20 cm); the image is 6 cm high and inverted.

u = −30 cm, v = −60 cm (a real image is formed on the same side). \(\dfrac1f=\dfrac1v+\dfrac1u=-\dfrac1{60}-\dfrac1{30}=-\dfrac3{60}=-\dfrac1{20}\), so f = −20 cm. Magnification \(m=-\dfrac vu=-\dfrac{-60}{-30}=-2\), so h′ = −2 × 3 = −6 cm. The image is 6 cm high; the negative sign shows it is inverted.

Example 9

A 1 cm high object is placed 20 cm in front of a concave mirror of focal length 15 cm. Find the position and nature of the image.

Answer: The image is formed 60 cm in front of the mirror; it is real, inverted and 3 cm high.

u = −20 cm, f = −15 cm, h = 1 cm. \(\dfrac1v=\dfrac1f-\dfrac1u=-\dfrac1{15}+\dfrac1{20}=-\dfrac1{60}\), so v = −60 cm. The signs of u and v are the same, so the object and the image are on the same side of the mirror and the image is real. \(m=-\dfrac vu=-\dfrac{-60}{-20}=-3\), so h′ = −3 × 1 = −3 cm. The image is real, inverted, 3 cm in size and formed 60 cm in front of the mirror.

Example 10

An object 4 cm high is placed 25 cm in front of a concave mirror of focal length 15 cm. At what distance from the mirror should a screen be placed in order to obtain a sharp image? Find the nature and size of the image.

Answer: Screen at 37.5 cm in front of the mirror; the image is real, inverted and 6 cm high.

u = −25 cm, f = −15 cm, h = +4 cm. \(\dfrac1v=-\dfrac1{15}+\dfrac1{25}=-\dfrac{2}{75}\), so v = −37.5 cm. The screen must be placed 37.5 cm from the mirror, on the same side as the object. \(m=-\dfrac vu=-\dfrac{-37.5}{-25}=-1.5\), so h′ = −1.5 × 4 = −6 cm. The image is real, inverted and 6 cm high.

Example 11

An object 5 cm high is placed at a distance of 20 cm from a convex mirror of radius of curvature 30 cm. Find the position, nature and size of the image.

Answer: The image is formed 8.57 cm behind the mirror; it is virtual, erect and 2.1 cm high.

u = −20 cm, R = +30 cm, so f = R/2 = +15 cm. \(\dfrac1v=\dfrac1{15}+\dfrac1{20}=\dfrac7{60}\), so \(v=\dfrac{60}{7}=+8.57\) cm. The positive sign shows that the image is formed behind the mirror, so it is virtual. \(m=-\dfrac vu=\dfrac{60/7}{20}=+\dfrac37\), so \(h'=5\times\dfrac37=2.14\approx2.1\) cm. The positive sign shows that the image is erect.

Example 12

A convex mirror used on an automobile has a 3 m radius of curvature. If a bus is located 5 m from this mirror, find the position, nature and size of the image.

Answer: The image is 1.15 m behind the mirror; it is virtual, erect and smaller in size (m = +0.23).

u = −5 m, R = +3 m, so f = +1.5 m. \(\dfrac1v=\dfrac1f-\dfrac1u=\dfrac1{1.5}+\dfrac15=0.867\) m⁻¹, so v = +1.15 m. The image is 1.15 m behind the mirror. \(m=-\dfrac vu=\dfrac{1.15}{5}=+0.23\). The image is virtual, erect and smaller than the object (about 0.23 times its size).

IMPORTANT POINTS TO REMEMBER

• Laws of reflection: (a) the angle of incidence is equal to the angle of reflection; (b) the incident ray, the reflected ray and the normal all lie in the same plane.
• Mirror: a smooth, highly polished reflecting surface. The two types are plane mirrors and curved mirrors. Curved mirrors are spherical or parabolic, depending on their curvature.
• Concave mirror: a spherical mirror whose inner hollow surface is the reflecting surface. It forms a real, inverted image of an object placed anywhere beyond F. When the object lies between F and P, the image is erect and virtual.
• Convex mirror: a spherical mirror whose outer bulging surface is the reflecting surface. The image formed is always erect, virtual and smaller than the object, wherever the object is placed. A convex mirror is used as the side mirror (rear-view mirror) on automobiles.
• Aperture: the effective width of a spherical mirror from which reflection can take place.
• Pole: the centre of a curved mirror.
• Centre of curvature: the centre of the hollow sphere of which the mirror is a part. For a concave mirror it is in front of the mirror; for a convex mirror it is behind it.
• Radius of curvature: the radius of the hollow sphere of which the mirror is a part.
• Principal axis: the straight line passing through the centre of curvature and the pole.
• Focus: the point on the principal axis at which rays coming from infinity meet or appear to meet.
• Focal length: the distance between the pole and the focus.
• Real image: an image that can be obtained on a screen; formed only when the reflected or refracted rays actually meet at a point.
• Virtual image: an image that can be seen in a mirror but cannot be obtained on a screen; formed when the rays appear to come from a point.
• Sign convention: the focal length of a concave mirror is negative and that of a convex mirror is positive.

Mirror formula:

\[\dfrac1f=\dfrac1v+\dfrac1u\]

Magnification: the ratio of the size of the image to that of the object:

\[M=-\dfrac vu=\dfrac{h_i}{h_o}\]

Power (in dioptres) \(=\dfrac1{f\ (\text{in metre})}\).

CONCLUSIONS FROM THE SIGN CONVENTION (SPHERICAL MIRRORS)

EXERCISE 1

The source gave only part of the answer key for this exercise; the missing answers were written for this document.

A. Very Short Answer Type Questions

A ray of light is incident on a plane mirror, i being the angle of incidence. What is the deviation suffered by the ray of light?

Answer: δ = 180° − 2i.

The incident and the reflected rays each make angle i with the normal, so the ray is turned through 180° − (i + r) = 180° − 2i.

(The source key printed “– 2i”; the full deviation is 180° − 2i.)

A plane mirror reflects a pencil of light to form a real image. What is the nature of the pencil of light incident on the mirror?

Answer: The incident pencil of light is converging (it is directed towards a point behind the mirror, i.e. the object is a virtual object).

Define the principal axis of a spherical mirror.

Answer: The principal axis is the straight line passing through the pole and the centre of curvature of the mirror.

What is the focal length of a plane mirror?

Answer: Infinity. A plane mirror is a spherical mirror of infinite radius of curvature, so f = R/2 is infinite.

Two perpendicular plane mirrors form ______ number of images of a point source of light.

Answer: 3. The number of images is 360°/90° − 1 = 3.

What is the magnification produced by a plane mirror?

Answer: +1. The image is the same size as the object and erect.

Which mirror would you use for shaving?

Answer: A concave mirror. When the face is held close to it (between the pole and the focus), it forms an erect, enlarged, virtual image.

(The source key said “plane mirror”; a concave mirror is the one used for shaving because it gives an enlarged image.)

Suppose x and y are the distances of the object and the image respectively from a mirror. What shall be the shape of the graph between 1/x and 1/y for a concave mirror?

Answer: A straight line.

The mirror formula \(\dfrac1x+\dfrac1y=\dfrac1f\) (distances as magnitudes) is the equation of a straight line in 1/x and 1/y, with slope −1 and equal intercepts 1/f on both axes.

B. Short Answer Type Questions

An object is placed between two plane parallel mirrors. Why do the distant images get fainter and fainter?

Answer: Each reflection at a mirror absorbs a little of the light, and the light spreads out. The more distant images are formed after a larger number of reflections, so less light reaches the eye from them, and they get fainter and fainter.

Why are mirrors used in a search light parabolic and not concave spherical?

Answer: A source of light placed at the focus of a parabolic mirror gives a truly parallel beam of light. A concave spherical mirror suffers from spherical aberration: rays falling far from the axis do not become parallel after reflection, so the beam spreads out. A parabolic mirror therefore throws a strong, parallel beam to a long distance.

You read a newspaper because of the light that it reflects. Then why do you not see even a faint image of yourself in the newspaper?

Answer: The surface of a newspaper is rough, so light is reflected from it irregularly (diffused reflection) in all directions. For an image to form, regular reflection from a smooth, polished surface is needed. So no image of yourself is formed, although you can see the printed page.

If you were driving a car, what type of mirror would you prefer to use for observing traffic at your back and why?

Answer: A convex mirror. It always forms an erect, virtual, diminished image and has a wide field of view, so a large area of traffic behind can be seen at a time.

We know that plane and convex mirrors produce virtual images of objects. Can they produce real images under some circumstances? Explain.

Answer: Yes. If a converging beam of light falls on a plane or convex mirror (a beam directed towards a point behind the mirror, i.e. a virtual object), the mirror reflects it so that the rays actually meet in front of the mirror. A real image is then formed.

The wall of a room is covered with a perfect plane mirror. Two movie films are made, one recording the movement of a man and the other of his mirror image. From viewing the films later, can an outsider tell which is which?

Answer: Yes. The image is laterally inverted: if the man uses his right hand, the image appears to use its left hand, and any writing or the parting of the hair appears reversed. By noticing this left–right reversal, an outsider can tell which film shows the image.

A concave mirror is held in water. What would be the change in the focal length of the mirror?

Answer: No change. The focal length of a mirror depends only on its radius of curvature (f = R/2) and not on the medium, because the light is reflected and does not enter the mirror.

What is the difference between the virtual images produced by (i) a plane mirror, (ii) a concave mirror, (iii) a convex mirror?

Answer: The differences are:

Show that if a ray of light is reflected successively from two mirrors inclined at an angle θ, the deviation of the ray does not depend upon the angle of incidence.

Answer: The total deviation is 360° − 2θ, which does not contain the angle of incidence.

Let the ray strike the first mirror at angle of incidence i₁ and the second at angle i₂. The deviation at the first mirror is 180° − 2i₁ and at the second is 180° − 2i₂. The ray between the two mirrors, together with the two mirrors, forms a triangle in which the angles are θ, (90° − i₁) and (90° − i₂) (the angles the ray makes with the mirrors). Their sum is 180°, so θ + 180° − i₁ − i₂ = 180°, giving i₁ + i₂ = θ.

Total deviation = (180° − 2i₁) + (180° − 2i₂) = 360° − 2(i₁ + i₂) = 360° − 2θ. It depends only on the angle θ between the mirrors and not on the angle of incidence.

Use the mirror equation to deduce that an object placed between f and 2f of a concave mirror produces a real image beyond 2f.

Answer: Let the focal length be f (a magnitude) and the object distance x, with f < x < 2f. By the sign convention u = −x and the focal length is −f.

\(\dfrac1v=\dfrac1{f_{\text{mirror}}}-\dfrac1u=-\dfrac1f+\dfrac1x=\dfrac{f-x}{fx}\). Since x > f, this is negative, so v is negative: the image is real (in front of the mirror). Its magnitude is \(|v|=\dfrac{fx}{x-f}\). The condition |v| > 2f means \(\dfrac{x}{x-f}>2\), i.e. x > 2x − 2f, i.e. x < 2f, which is true. So the image is real and lies beyond 2f.

Show that a convex mirror always produces a virtual image independent of the location of the object.

Answer: For a convex mirror f is positive and the object distance u is always negative (a real object).

\(\dfrac1v=\dfrac1f-\dfrac1u=\dfrac1f+\dfrac1{|u|}\). Both terms are positive, so 1/v is positive and v is positive for every position of the object. A positive v means the image is behind the mirror, so it is always virtual.

Prove that the virtual image produced by a convex mirror is always diminished in size and is located between the focus and the pole.

Answer: From the previous result, \(\dfrac1v=\dfrac1f+\dfrac1{|u|}\), so \(v=\dfrac{f|u|}{f+|u|}\).

Since \(f+|u|>|u|\), we get v < f; and v is positive, so 0 < v < f: the image lies between the pole and the focus. The magnification is \(m=-\dfrac vu=\dfrac{v}{|u|}=\dfrac{f}{f+|u|}\), which is less than 1 because |u| is positive. So the image is diminished (and erect).

Show analytically that an object placed between the pole and the focus of a concave mirror produces a virtual and enlarged image.

Answer: Let the focal length be f (magnitude) and the object distance x with x < f. Then u = −x and the focal length is −f.

\(\dfrac1v=-\dfrac1f+\dfrac1x=\dfrac{f-x}{fx}\). Since x < f, this is positive, so v is positive (behind the mirror): the image is virtual. The magnification \(m=-\dfrac vu=\dfrac vx=\dfrac{f}{f-x}\) is positive (erect) and greater than 1 because f > f − x. So the image is virtual, erect and enlarged.

We know that a virtual image cannot be obtained on a screen. But when we see a virtual image, we are obviously bringing it on the retina (which may be regarded as a screen) of the eye. Point out the contradiction, if any.

Answer: There is no contradiction.

A virtual image is formed where the reflected rays only appear to meet; the rays do not actually pass through it, so it cannot be caught on a screen placed at the image position. The eye lens converges the diverging rays that enter the eye and forms a real image on the retina. The retina therefore shows a real image of the virtual image; the virtual image itself is not on the retina.

Why does a concave mirror of small aperture form a sharper image?

Answer: With a small aperture, the rays fall on the mirror close to the principal axis and make small angles with it. Such rays, after reflection, meet very nearly at a single point, so the image is sharp. With a large aperture, rays far from the axis meet at points different from those of the rays near the axis (spherical aberration), and the image becomes blurred.

What do you understand by the term “parallax”?

Answer: Parallax is the apparent shift in the position of an object with respect to another object (or the background) when the observer’s position is changed. In experiments with mirrors, the image position is found by moving an object pin until there is no parallax between the pin and the image.

How can you distinguish between three different mirrors just by looking at them?

Answer: Hold each mirror close to your face and look at the image. A plane mirror shows an erect image of the same size. A concave mirror shows an erect, enlarged image. A convex mirror shows an erect, smaller (diminished) image, with a wider field of view. (If a concave mirror is moved far away, its image becomes inverted.)

What is the effect of the size of the mirror on the nature of the image?

Answer: The nature (real or virtual) and the position of the image do not change with the size of the mirror. A larger mirror (larger aperture) collects more light, so the image becomes brighter; a smaller mirror gives a fainter image.

Does irregular reflection follow the laws of reflection or not?

Answer: Yes. The laws of reflection are valid for every reflection. In irregular reflection the surface is rough, so the normal is in a different direction at each point, and the reflected rays go in different directions. At every point, however, the angle of incidence equals the angle of reflection.

C. Long Answer Type Questions

Prove that the radius of curvature of a spherical mirror is equal to twice the focal length of the mirror.

Answer: R = 2f.

Take a concave mirror with pole P, centre of curvature C and principal focus F. A ray MA parallel to the principal axis falls on the mirror at A (close to P) and is reflected through F. CA is the normal at A. Let ∠ACP = θ. By the law of reflection, the angle of incidence ∠MAC = ∠CAF = θ (the angle MAC equals ∠ACP, as they are alternate angles). So in triangle CAF, ∠AFP (exterior angle) = ∠FAC + ∠ACF = 2θ.

For a small aperture, tan θ ≈ θ = AP/CP and tan 2θ ≈ 2θ = AP/FP. Therefore 2 × AP/CP = AP/FP, i.e. FP = CP/2. That is, f = R/2, or R = 2f.

Derive the mirror formula for a concave mirror when the image formed is (i) real, (ii) virtual. Also give the sign convention used.

Answer: The mirror formula is \(\dfrac1v+\dfrac1u=\dfrac1f\), derived below. (The New Cartesian sign convention is given in the section on sign convention above.)

(i) Real image. Let AB be an object in front of a concave mirror beyond F, and A′B′ its real image. Take magnitudes u = PB, v = PB′, f = PF. Draw BE from B parallel to the principal axis (EP = AB). Triangles ABP and A′B′P are similar, so \(\dfrac{A'B'}{AB}=\dfrac vu\). Triangles A′B′F and EPF are similar, so \(\dfrac{A'B'}{EP}=\dfrac{B'F}{PF}=\dfrac{v-f}{f}\). Since EP = AB, \(\dfrac vu=\dfrac{v-f}f\), so vf = uv − uf. Dividing by uvf: \(\dfrac1u=\dfrac1f-\dfrac1v\), i.e. \(\dfrac1u+\dfrac1v=\dfrac1f\) (for magnitudes). Applying the sign convention (u, v and f of a concave mirror are all negative) gives \(\dfrac1v+\dfrac1u=\dfrac1f\).

(ii) Virtual image. Let the object AB lie between P and F, with the virtual image A′B′ behind the mirror. Similar triangles A′B′P and ABP give \(\dfrac{A'B'}{AB}=\dfrac vu\), and similar triangles A′B′F and EPF give \(\dfrac{A'B'}{EP}=\dfrac{f+v}{f}\). So \(\dfrac vu=\dfrac{f+v}f\), i.e. vf = uf + uv. Dividing by uvf: \(\dfrac1u=\dfrac1v+\dfrac1f\) in magnitudes. With the sign convention (u and f negative, v positive) this becomes the same formula \(\dfrac1v+\dfrac1u=\dfrac1f\).

Find formulae for magnification produced in the following cases: (i) concave mirror, when the image formed is real, (ii) concave mirror, when the image formed is virtual, (iii) convex mirror.

Answer: In all three cases the magnification is \(m=\dfrac{h'}{h}=-\dfrac vu\) (with the sign convention).

(i) Concave mirror, real image: from the similar triangles ABP and A′B′P, \(\dfrac{A'B'}{AB}=\dfrac{v}{u}\) (magnitudes). With signs, u and v are both negative and the image is inverted (h′ negative), so \(m=-\dfrac vu\), which is negative.

(ii) Concave mirror, virtual image: u is negative and v is positive. The same similar triangles give \(m=-\dfrac vu\), which is positive (erect image) and greater than 1 (enlarged).

(iii) Convex mirror: u is negative and v is positive, so \(m=-\dfrac vu\) is positive (erect image) and less than 1 (diminished).

Draw a ray diagram to show the formation of the image of an object placed between the pole and the centre of curvature of a concave mirror. Derive the formula connecting the object distance (u), the image distance (v) and the focal length (f) for this particular case. State clearly the assumptions and the sign convention used.

Answer: For an object placed between F and C, the image is real, inverted and enlarged, and formed beyond C. The formula is \(\dfrac1v+\dfrac1u=\dfrac1f\).

(The ray diagram is not reproduced here: a ray from the top of the object parallel to the principal axis is reflected through F, a ray through C is reflected back along itself, and the two meet beyond C to give the top of the inverted image.)

The derivation is the same as for a real image given in the previous answer, using similar triangles ABP ~ A′B′P and EPF ~ A′B′F.

Assumptions: the mirror has a small aperture; the object lies close to the principal axis; the rays make small angles with the principal axis. Sign convention: the New Cartesian sign convention described above (distances measured from the pole; distances in the direction of the incident light are positive).

Express the magnification produced by a spherical mirror in terms of (i) u and f, (ii) v and f.

Answer: (i) \(m=\dfrac{f}{f-u}\); (ii) \(m=\dfrac{f-v}{f}\).

(i) From the mirror formula, \(\dfrac1v=\dfrac1f-\dfrac1u=\dfrac{u-f}{uf}\), so \(v=\dfrac{uf}{u-f}\). Then \(m=-\dfrac vu=-\dfrac{f}{u-f}=\dfrac f{f-u}\).

(ii) \(\dfrac1u=\dfrac1f-\dfrac1v=\dfrac{v-f}{vf}\), so \(u=\dfrac{vf}{v-f}\). Then \(m=-\dfrac vu=-\dfrac{v-f}{f}=\dfrac{f-v}f\).

EXERCISE 2: SINGLE CORRECT ANSWER TYPE QUESTIONS

A child walks towards a fixed plane mirror at a speed of 5 km h⁻¹. The velocity of the image with respect to the mirror is:

1. 5 km h⁻¹
2. −5 km h⁻¹
3. 10 km h⁻¹
4. −10 km h⁻¹

Answer: (B) −5 km h⁻¹.

The image moves towards the mirror with the same speed as the child, in the opposite direction.

The letter that does not show lateral inversion is:

1. Z
2. M
3. O
4. W

Answer: As given in the key: (A). However, Z is a letter that does show lateral inversion; M, O and W look the same in a mirror.

(The question as printed seems faulty: M, O and W are symmetrical about a vertical line and look unchanged in a plane mirror, while Z does not. The key’s answer (A) is kept but flagged.)

In a plane mirror, an object is 0.5 m in front of the mirror. The distance between the object and the image is:

1. 0.5 m
2. 1 m
3. 0.25 m
4. 0.75 m

Answer: (B) 1 m.

The image is 0.5 m behind the mirror, so the distance between object and image is 0.5 + 0.5 = 1 m.

An object 0.5 m tall is in front of a plane mirror at a distance of 0.2 m. The size of the image formed is:

1. 0.2 m
2. 0.5 m
3. 0.1 m
4. 1 m

Answer: (B) 0.5 m.

A plane mirror forms an image of the same size as the object.

A plane mirror is approaching you at 10 cm s⁻¹. Your image shall approach you with a speed of:

1. +10 cm s⁻¹
2. −10 cm s⁻¹
3. +20 cm s⁻¹
4. −20 cm s⁻¹

Answer: (C) +20 cm s⁻¹.

When the mirror moves towards you with speed v, the image moves towards you with speed 2v = 20 cm/s.

The path along which light travels in a homogeneous medium is called a:

1. beam of light
2. ray of light
3. pencil of light
4. none of these

Answer: (B) ray of light.

A thin layer of water is transparent but a very thick layer of water is:

1. translucent
2. opaque
3. most transparent
4. none of these

Answer: (A) translucent, as given in the key.

(A very thick layer of water allows only some light to pass and does not let us see clearly through it, so it is translucent.)

Air is not visible because it:

1. is nearly perfectly transparent
2. neither absorbs nor reflects light
3. transmits the whole of light
4. all of the above are correct

Answer: (D) all of the above are correct.

According to the laws of reflection of light:

1. the angle of incidence is equal to the angle of reflection
2. the angle of incidence is less than the angle of reflection
3. the angle of incidence is greater than the angle of reflection
4. none of these

Answer: (A) the angle of incidence is equal to the angle of reflection.

Which of the following correctly represents the graphical relation between the angle of incidence (i) and the angle of reflection (r)? (A) a curve rising upwards with increasing slope; (B) a straight line sloping downwards; (C) a curve rising and flattening; (D) a straight line rising from the origin.

Answer: (D). Since r = i, the graph is a straight line through the origin.

A light ray falls on a mirror and deviates by 60°. Then the angle of reflection will be:

1. 30°
2. 90°
3. 60°
4. 180°

Answer: (C) 60°.

Deviation = 180° − 2i = 60°, so i = 60° and r = i = 60°.

A ray of light is incident on a plane mirror at an angle θ. If the angle between the incident and reflected rays is 80°, what is the value of θ?

1. 40°
2. 50°
3. 45°
4. 55°

Answer: (B) 50°, as given in the key.

The angle between the incident and reflected rays is 2i = 80°, so the angle of incidence is 40°. The key’s 50° is the angle that the incident ray makes with the mirror surface (90° − 40°), so θ must be meant as the glancing angle. If θ meant the angle of incidence, the answer would be 40°; this is flagged.

Light shows:

1. random propagation
2. curvilinear propagation
3. rectilinear propagation
4. none of these

Answer: (C) rectilinear propagation.

The image of the moon is formed by a concave mirror whose radius of curvature is 4.8 m at a time when the distance of the moon is 2.4 × 10⁸ m. If the diameter of the image is 2.2 cm, the diameter of the moon is:

1. 1.1 × 10⁶ m
2. 2.2 × 10⁶ m
3. 2.2 × 10⁸ m
4. 2.2 × 10¹⁰ m

Answer: (B) 2.2 × 10⁶ m.

The moon is very far away, so the image forms at the focus: f = R/2 = 2.4 m. Magnification = image size / object size = v/u = 2.4/(2.4 × 10⁸) = 10⁻⁸. Diameter of the moon = 2.2 cm / 10⁻⁸ = 2.2 × 10⁸ cm = 2.2 × 10⁶ m.

The focal length of a concave mirror is f and the distance of the object from the principal focus is a. The magnitude of the magnification obtained will be:

1. (f + a)/f
2. f/a
3. √(f/a)
4. f²/a²

Answer: (B) f/a.

The object distance is u = f + a, so \(|m|=\dfrac{f}{|f-u|}=\dfrac fa\).

The magnification of an object placed 10 cm from a convex mirror of radius of curvature 20 cm will be:

1. 0.2
2. 0.5
3. 1
4. infinity

Answer: (B) 0.5.

f = +10 cm, u = −10 cm: \(\dfrac1v=\dfrac1{10}+\dfrac1{10}=\dfrac15\), so v = 5 cm and \(m=-\dfrac vu=\dfrac5{10}=0.5\).

The image formed by a concave mirror is observed to be virtual, erect and larger than the object. Then the position of the object should be:

1. between the focus and the centre of curvature
2. at the centre of curvature
3. beyond the centre of curvature
4. between the pole of the mirror and the focus

Answer: (D) between the pole of the mirror and the focus.

The magnification produced by a concave mirror:

1. is always more than one
2. is always less than one
3. is always equal to one
4. may be less than or greater than one

Answer: (D) may be less than or greater than one.

It depends on the position of the object.

Choose the correct relation between u, v and R:

1. R = 2uv/(u + v)
2. R = 2/(u + v)
3. R = 2(u + v)/(uv)
4. none of these

Answer: (A) R = 2uv/(u + v).

From \(\dfrac1v+\dfrac1u=\dfrac1f=\dfrac2R\), we get \(R=\dfrac{2uv}{u+v}\).

The image formed by a concave mirror is real, inverted and of the same size as the object. The position of the object should be:

1. beyond C
2. between C and F
3. at C
4. at F

Answer: (C) at C.

A boy is standing in front of a plane mirror at a distance of 3 m from it. What is the distance between the boy and his image?

1. 3 m
2. 4.5 m
3. 6 m
4. none of these

Answer: (C) 6 m.

The image is 3 m behind the mirror, so the distance between the boy and his image is 3 + 3 = 6 m.

ANSWER KEY (as given in the PDF)

EXERCISE-1

• 1. π − 2i
• 2. virtual
• 5. 3
• 6. 1
• 7. plane mirror
• 8. 1/v + 1/u = 1/f ⇒ linear
• 9. as I ∝ 1/r²
• 10. They reflect light parallel
• 15. No change
• 26. Image becomes brighter
• 27. Yes

EXERCISE-2

Irregular reflection or Diffused reflection

The phenomenon due to which a parallel beam of light, travelling through some medium, gets reflected in various possible directions, on striking some rough surface is called irregular reflection or diffused reflection.

Please click on below link to download pdf file for Class 10 Science Light Reflection Exam Notes.