Physics INTRODUCTION, REFLECTION OF LIGHT BY SPHERICAL MIRRORS FOR CBSE NCERT

### Topic covered

color{blue}{star} INTRODUCTION
color{blue}{star} REFLECTION OF LIGHT BY SPHERICAL MIRRORS

### INTRODUCTION

color{blue} ✍️ Nature has endowed the human eye (retina) with the sensitivity to detect electromagnetic waves within a small range of the electromagnetic spectrum.

color{blue} ✍️ Electromagnetic radiation belonging to this region of the spectrum (wavelength of about 400 nm to 750 nm) is called light.

color {blue}{➢➢}There are two things that we can intuitively mention about light from common experience. First, that it travels with enormous speed and second, that it travels in a straight line.

color {blue}{➢➢}The speed of light is finite and measurable. Its presently accepted value in vacuum is c = 2.99792458 × 10^8 m s^(–1). For many purposes, it suffices to take c = 3 × 10^8 m s^(–1).

color {blue}{➢➢}The speed of light in vacuum is the highest speed attainable in nature.

color {blue}{➢➢}The light travels in a straight line since the wavelength of light is very small compared to the size of ordinary objects that we encounter commonly (generally of the order of a few cm or larger).

color {blue}{➢➢}A light wave can be considered to travel from one point to another, along a straight line joining them. The path is called a ray of light, and a bundle of such rays constitutes a beam of light.

### REFLECTION OF LIGHT BY SPHERICAL MIRRORS

color{blue} ✍️We know that the laws of reflection. The angle of reflection (i.e., the angle between reflected ray and the normal to the reflecting surface or the mirror) equals the angle of incidence (angle between incident ray and the normal).

And the incident ray, reflected ray and the normal to the reflecting surface at the point of incidence lie in the same plane (Fig. 9.1).

color{blue} ✍️ These laws are valid at each point on any reflecting surface whether plane or curved.

The normal in this case is to be taken as normal to the tangent to surface at the point of incidence. That is, the normal is along the radius, the line joining the centre of curvature of the mirror to the point of incidence.

color{blue} ✍️ We have already studied that the geometric centre of a spherical mirror is called its pole while that of a spherical lens is called its optical centre.

color {blue}{➢➢}The line joining the pole and the centre of curvature of the spherical mirror is known as the principal axis. In the case of spherical lenses, the principal axis is the line joining the optical centre with its principal focus as you will see later.

color{brown} bbul {"Sign convention"}

color {blue}{➢➢}To derive the relevant formulae for reflection by spherical mirrors and refraction by spherical lenses, we must first adopt a sign convention for measuring distances.

color {blue}{➢➢}According to this convention, "all distances are measured from the pole" of the mirror or the optical centre of the lens.

color {blue}{➢➢}The distances measured in the "same direction" as the incident light are taken as "positive" and those measured in the direction opposite to the direction of incident light are taken as negative (Fig. 9.2).

color {blue}{➢➢}The heights measured upwards with respect to x-axis and normal to the principal axis (x-axis) of the mirror/ lens are taken as positive (Fig. 9.2).

color {blue}{➢➢}The heights measured downwards are taken as negative. With a common accepted convention, it turns out that a single formula for spherical mirrors and a single formula for spherical lenses can handle all different cases.

color {brown}ulbb{"Focal length of spherical mirrors"}

color {blue}{➢➢}Figure 9.3 shows what happens when a parallel beam of light is incident on

color {blue}{(a)} a concave mirror, and color {blue}{(b)} a convex mirror.

color {blue}{➢➢}We assume that the rays are paraxial, i.e., they are incident at points close to the pole P of the mirror and make small angles with the principal axis.

color {blue}{➢➢}The reflected rays converge at a point F on the principal axis of a concave mirror [Fig. 9.3(a)]. For a convex mirror, the reflected rays appear to diverge from a point F on its principal axis [Fig. 9.3(b)].

The point F is called the principal focus of the mirror. If the parallel paraxial beam of light were incident, making some angle with the principal axis, the reflected rays would converge (or appear to diverge) from a point in a plane through F normal to the principal axis. This is called the focal plane of the mirror [Fig. 9.3(c)]

The distance between the focus F and the pole P of the mirror is called the focal length of the mirror, denoted by f. We now show that f = R//2, where R is the radius of curvature of the mirror. The geometry of reflection of an incident ray is shown in Fig. 9.4.

Let C be the centre of curvature of the mirror. Consider a ray parallel to the principal axis striking the mirror at M. Then CM will be perpendicular to the mirror at M. Let θ be the angle of incidence, and MD be the perpendicular from M on the principal axis. Then, ∠MCP = θ " and "∠MFP = 2 θ

color {blue}{➢➢} Now,

color{blue}{tanθ= (MD)/(CD)" and "tan 2θ= (MD)/(FD)}

..............(9.1)

For small θ, which is true for paraxial rays, tanθ ≈ θ, tan 2θ ≈ 2θ. Therefore, Eq. (9.1) gives

(MD)/(CD) = 2(MD)/(CD)

or .

color{blue}{FD = (CD)/2}

..............(9.2)

Now, for small θ, the point D is very close to the point P. Therefore, FD = f and CD = R. Equation (9.2) then gives

color{blue}{f = R//2}

...............(9.3)

### The mirror equation

color{blue} ✍️ If rays emanating from a point actually meet at another point after reflection and/or refraction, that point is called the image of the first point.

color{blue} ✍️ The image is real if the rays actually converge to the point; it is virtual if the rays do not actually meet but appear to diverge from the point when produced backwards.

color{blue} ✍️ An image is thus a point-to-point correspondence with the object established through reflection and/or refraction. In principle, we can take any two rays emanating from a point on an object, trace their paths, find their point of intersection and thus, obtain the image of the point due to reflection at a spherical mirror. In practice, however, it is convenient to choose any two of the following rays:

color{blue} {(i)} The ray from the point which is parallel to the principal axis. The reflected ray goes through the focus of the mirror.

color{blue}{(ii)} The ray passing through the centre of curvature of a concave mirror or appearing to pass through it for a convex mirror. The reflected ray simply retraces the path.

color{blue}{(iii)} The ray passing through (or directed towards) the focus of the concave mirror or appearing to pass through (or directed towards) the focus of a convex mirror. The reflected ray is parallel to the principal axis.

color{blue} {(iv)} The ray incident at any angle at the pole. The reflected ray follows laws of reflection.

Figure 9.5 shows the ray diagram considering three rays. It shows the image A′B′ (in this case, real) of an object AB formed by a concave mirror. It does not mean that only three rays emanate from the point A.

color {blue}{➢➢}An infinite number of rays emanate from any source, in all directions. Thus, point A′ is image point of A if every ray originating at point A and falling on the concave mirror after reflection passes through the point A′.

color {blue}{➢➢}We now derive the mirror equation or the relation between the object distance (u), image distance (v) and the focal length ( f ). From Fig. 9.5, the two right-angled triangles A′B′F and MPF are similar. (For paraxial rays, MP can be considered to be a straight line perpendicular to CP.) Therefore,

(B'A)/(PM) = (B'F)/(FP)

or

color{blue}{(B'A)/(PM) = (B'F)/(FP)(∵ PM=AB)}

...............(9.4)

Since ∠ APB = ∠ A′PB′, the right angled triangles A′B′P and ABP are also similar. Therefore,

color{blue}{(B'F)/(BA) = (B'P)/(BP)}

.................(9.5)

Comparing Eqs. (9.4) and (9.5), we get

color{blue}{(B'F)/(FP) = (B'P-FP)/(FP) = (B'P)/(BP)}

.................(9.6)

Equation (9.6) is a relation involving magnitude of distances. We now apply the sign convention. We note that light travels from the object to the mirror MPN. Hence this is taken as the positive direction.

To reach the object AB, image A′B′ as well as the focus F from the pole P, we have to travel opposite to the direction of incident light.

Hence, all the three will have negative signs. Thus,

B′ P = –v, FP = –f, BP = –u
Using these in Eq. (9.6), we get

(-v+f)/(-f) = (-v)/(-u)

or (-v+f)/(f) = (-v)/(-u)

color{blue}{1/v +1/u = 1/f}

.............(9.7)

color {blue}{➢➢}This relation is known as the mirror equation.

color{blue} ✍️ The size of the image relative to the size of the object is another important quantity to consider. We define linear magnification (m) as the ratio of the height of the image (h′) to the height of the object (h):

color{blue}{m = (h')/h}

.....................(9.8)

h and h′ will be taken positive or negative in accordance with the accepted sign convention. In triangles A′B′P and ABP, we have,

(B'A)/(BA) = (B'P)/(BP)

color {blue}{➢➢}With the sign convention, this becomes

(-h')/h = (-v)/(-u)

color {blue}{➢➢}so that

color{blue}{m = (h')/h = v/u}

...............(9.9)

color {blue}{➢➢}We have derived here the mirror equation, Eq. (9.7), and the magnification formula, Eq. (9.9), for the case of real, inverted image formed by a concave mirror.

color {blue}{➢➢}With the proper use of sign convention, these are, in fact, valid for all the cases of reflection by a spherical mirror (concave or convex) whether the image formed is real or virtual.

color {blue}{➢➢}Figure 9.6 shows the ray diagrams for virtual image formed by a concave and convex mirror. You should verify that Eqs. (9.7) and (9.9) are valid for these cases as well.

Q 3128356201

Suppose that the lower half of the concave mirror’s reflecting surface in Fig. 9.5 is covered with an opaque (non-reflective) material. What effect will this have on the image of an object placed in front of the mirror?
Class 12 Chapter 9 Example 1
Solution:

You may think that the image will now show only half of the object, but taking the laws of reflection to be true for all points of the remaining part of the mirror, the image will be that of the whole object. However, as the area of the reflecting surface has been reduced, the intensity of the image will be low (in this case, half).
Q 3138356202

A mobile phone lies along the principal axis of a concave mirror, as shown in Fig. 9.7. Show by suitable diagram, the formation of its image. Explain why the magnification is not uniform. Will the distortion of image depend on the location of the phone with respect to the mirror?
Class 12 Chapter 9 Example 2
Solution:

The ray diagram for the formation of the image of the phone is shown in Fig. 9.7. The image of the part which is on the plane perpendicular to principal axis will be on the same plane. It will be of the same size, i.e., B= C = BC. You can yourself realise why the image is distorted.
Q 3168356205

An object is placed at (i) 10 cm, (ii) 5 cm in front of a concave mirror of radius of curvature 15 cm. Find the position, nature, and magnification of the image in each case.
Class 12 Chapter 9 Example 3
Solution:

The focal length f = –15/2 cm = –7.5 cm
(i) The object distance u = –10 cm. Then Eq. (9.7) gives

1/v + 1/(-10) = 1/(-7.5)

or v = (10xx7.5)/(-2.5) = - 30 cm

The image is 30 cm from the mirror on the same side as the object.

Also, magnification m = v/u = - (-30)/(-10) = -3

The image is magnified, real and inverted.

(ii) The object distance u = –5 cm. Then from Eq. (9.7),

1/v+ 1/(-5) = 1/(-7.5)

or v = (5xx7.5)/(7.5-5) = 15cm

This image is formed at 15 cm behind the mirror. It is a virtual image.
Magnification m = - v/u = - 15/(-5) = 3
The image is magnified, virtual and erect.
Q 3108356208

Suppose while sitting in a parked car, you notice a jogger approaching towards you in the side view mirror of R = 2 m. If the jogger is running at a speed of 5 m s^(–1), how fast the image of the jogger appear to move when the jogger is (a) 39 m, (b) 29 m, (c) 19 m, and (d) 9 m away.
Class 12 Chapter 9 Example 4
Solution:

From the mirror equation, Eq. (9.7), we get

v = (fu)/(u-f)

For convex mirror, since R = 2 m, f = 1 m. Then

for u = –39 m, = (-39)xx1 /(-39-1) = 39/40m

Since the jogger moves at a constant speed of 5 m s–1, after 1 s the position of the image v (for u = –39 + 5 = –34) is (34//35 )m. The shift in the position of image in 1 s is

39/40 - 34/35 = (1365-1370)/1400 = 5/1400 = 1/280m

Therefore, the average speed of the image when the jogger is between 39 m and 34 m from the mirror, is (1/280) m s–1 Similarly, it can be seen that for u = –29 m, –19 m and –9 m, the speed with which the image appears to move is

1/150 ms^(-1) , 1/60 ms^(-1) and 1/10ms^(-1) respectively.

Although the jogger has been moving with a constant speed, the speed of his/her image appears to increase substantially as he/she moves closer to the mirror. This phenomenon can be noticed by any person sitting in a stationary car or a bus. In case of moving vehicles, a similar phenomenon could be observed if the vehicle in the rear is moving closer with a constant speed.