Chapter 9 Ray opticals and opticals instruments

Chapter Nine
RAY OPTICS
AND OPTICAL
INSTRUMENTS
9.1 INTRODUCTION
Nature has endowed the human eye (retina) with the sensitivity to detect
electromagnetic waves within a small range of the electromagnetic
spectrum. Electromagnetic radiation belonging to this region of the
spectrum (wavelength of about 400 nm to 750 nm) is called light. It is
mainly through light and the sense of vision that we know and interpret
the world around us.
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. It took some time for people to realise that
the speed of light is finite and measurable. Its presently accepted value
in vacuum is c = 2.99792458 × 108
m s–1
. For many purposes, it suffices
to take c = 3 × 108
m s–1
. The speed of light in vacuum is the highest
speed attainable in nature.
The intuitive notion that light travels in a straight line seems to
contradict what we have learnt in Chapter 8, that light is an
electromagnetic wave of wavelength belonging to the visible part of the
spectrum. How to reconcile the two facts? The answer is that 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).
In this situation, as you will learn in Chapter 10, a light wave can be
considered to travel from one point to another, along a straight line joining
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FIGURE 9.1 The incident ray, reflected ray
and the normal to the reflecting surface lie
in the same plane.
FIGURE 9.2 The Cartesian Sign Convention.
them. The path is called a ray of light, and a bundle of such rays
constitutes a beam of light.
In this chapter, we consider the phenomena of reflection, refraction
and dispersion of light, using the ray picture of light. Using the basic
laws of reflection and refraction, we shall study the image formation by
plane and spherical reflecting and refracting surfaces. We then go on to
describe the construction and working of some important optical
instruments, including the human eye.
9.2 REFLECTION OF LIGHT BY SPHERICAL MIRRORS
We are familiar with 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). Also that
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). These laws are valid
at each point on any reflecting surface whether
plane or curved. However, we shall restrict our
discussion to the special case of curved surfaces,
that is, spherical surfaces. 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.
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.
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.
9.2.1 Sign convention
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. In this book, we
shall follow the Cartesian sign
convention. According to this
convention, all distances are measured
from the pole of the mirror or the optical
centre of the lens. 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).
The heights measured upwards with respect to x-axis and normal to the
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principal axis (x-axis) of the mirror/lens are taken as positive (Fig. 9.2).
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.
9.2.2 Focal length of spherical mirrors
Figure 9.3 shows what happens when a parallel beam of light is incident
on (a) a concave mirror, and (b) a convex mirror. 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. 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)].
FIGURE 9.3 Focus of a concave and convex mirror.
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 q be the angle of incidence, and MD
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be the perpendicular from M on the principal axis. Then,
ÐMCP = q and ÐMFP = 2q
Now,
tanq =
MD
CD
and tan 2q =
MD
FD
(9.1)
For small q, which is true for paraxial rays, tanq » q,
tan 2q » 2q. Therefore, Eq. (9.1) gives
MD
FD
= 2
MD
CD
or, FD =
CD
2
(9.2)
Now, for small q, the point D is very close to the point P.
Therefore, FD = f and CD = R. Equation (9.2) then gives
f = R/2 (9.3)
9.2.3 The mirror equation
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. 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. 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:
(i) The ray from the point which is parallel to the
principal axis. The reflected ray goes through
the focus of the mirror.
(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.
(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.
(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.
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¢.
FIGURE 9.4 Geometry of
reflection of an incident ray on
(a) concave spherical mirror,
and (b) convex spherical mirror.
FIGURE 9.5 Ray diagram for image
formation by a concave mirror.
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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 B F
PM FP
′ ′ ′
=
or
B A B F
BA FP
′ ′ ′
= (∵PM = AB) (9.4)
Since Ð APB = Ð A¢PB¢, the right angled triangles A¢B¢P and ABP are
also similar. Therefore,
B A B P
B A B P
′ ′ ′
= (9.5)
Comparing Eqs. (9.4) and (9.5), we get
B P – FP
B F B P
FP FP 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 v
f u
+
=
–
or
–
v f v
f u
=
v
f
v
u
= +
1
Dividing it by v, we get
1 1 1
v u f
+ =
(9.7)
This relation is known as the mirror equation.
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):
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 B P
BA BP
′ ′ ′
=
With the sign convention, this becomes
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– –
h v
h u
′
=
–
so that
m = –
h v
h u
′
= (9.9)
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. 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. 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.
FIGURE 9.6 Image formation by (a) a concave mirror with object between
P and F, and (b) a convex mirror.
E
XAMPLE
9.1
Example 9.1 Suppose that the lower half of the concave mirror’s
reflecting surface in Fig. 9.6 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?
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).
Example 9.2 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?
FIGURE 9.7
E
XAMPLE
9.2
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E
XAMPLE
9.3
E
XAMPLE
9.4
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.
Example 9.3 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.
Solution
The focal length f = –15/2 cm = –7.5 cm
(i) The object distance u = –10 cm. Then Eq. (9.7) gives
– – .
1 1 1
10 7 5
v
+ =
or
.
.
10 7 5
2 5
v
×
−
= = – 30 cm
The image is 30 cm from the mirror on the same side as the object.
Also, magnification m =
( 30)
– – – 3
( 10)
v
u
−
= =
−
The image is magnified, real and inverted.
(ii) The object distance u = –5 cm. Then from Eq. (9.7),
1 1 1
5 7.5
v
+ =
− −
or
( )
.
. –
5 7 5
15 cm
7 5 5
v
×
=
=
This image is formed at 15 cm behind the mirror. It is a virtual image.
Magnification m =
15
– – 3
( 5)
v
u
= =
−
The image is magnified, virtual and erect.
Example 9.4 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.
Solution
From the mirror equation, Eq. (9.7), we get
fu
v
u f
=
−
For convex mirror, since R = 2 m, f = 1 m. Then
for u = –39 m,
( 39) 1 39
m
39 1 40
v
− ×
= =
− −
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.
E
XAMPLE
9.2
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E
XAMPLE
9.4
FIGURE 9.8 Refraction and reflection of light.
The shift in the position of image in 1 s is
1365 1360
39 34 5 1
m
40 35 1400 1400 280
−
− = = =
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 –1 –1
1 1 1
m s , m s and m s ,
150 60 10
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.
9.3 REFRACTION
When a beam of light encounters another transparent medium, a part of
light gets reflected back into the first medium while the rest enters the
other. A ray of light represents a beam. The direction of propagation of an
obliquely incident (0°< i < 90°) ray of light that enters the other medium,
changes at the interface of the two media. This phenomenon is called
refraction of light. Snell experimentally obtained the following laws
of refraction:
(i) The incident ray, the refracted ray and the
normal to the interface at the point of
incidence, all lie in the same plane.
(ii) The ratio of the sine of the angle of incidence
to the sine of angle of refraction is constant.
Remember that the angles of incidence (i ) and
refraction (r ) are the angles that the incident
and its refracted ray make with the normal,
respectively. We have
21
sin
sin
i
n
r
= (9.10)
where n21 is a constant, called the refractive
index of the second medium with respect to the
first medium. Equation (9.10) is the well-known
Snell’s law of refraction. We note that n21
is a
characteristic of the pair of media (and also depends on the wavelength
of light), but is independent of the angle of incidence.
From Eq. (9.10), if n21
> 1, r < i, i.e., the refracted ray bends towards
the normal. In such a case medium 2 is said to be optically denser (or
denser, in short) than medium 1. On the other hand, if n21 <1, r > i, the
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refracted ray bends away from the normal. This
is the case when incident ray in a denser
medium refracts into a rarer medium.
Note: Optical density should not be
confused with mass density, which is mass
per unit volume. It is possible that mass
density of an optically denser medium may
be less than that of an optically rarer
medium (optical density is the ratio of the
speed of light in two media). For example,
turpentine and water. Mass density of
turpentine is less than that of water but
its optical density is higher.
If n21
is the refractive index of medium 2 with
respect to medium 1 and n12 the refractive index
of medium 1 with respect to medium 2, then it
should be clear that
12
21
1
n
n
= (9.11)
It also follows that if n32 is the refractive index
of medium 3 with respect to medium 2 then n32
=
n31 × n12, where n31 is the refractive index of
medium 3 with respect to medium 1.
Some elementary results based on the laws of
refraction follow immediately. For a rectangular
slab, refraction takes place at two interfaces (air-
glass and glass-air). It is easily seen from Fig. 9.9
that r2 = i1, i.e., the emergent ray is parallel to the
incident ray—there is no deviation, but it does
suffer lateral displacement/shift with respect to the
incident ray. Another familiar observation is that
the bottom of a tank filled with water appears to be
raised (Fig. 9.10). For viewing near the normal direction, it can be shown
that the apparent depth (h1) is real depth (h2) divided by the refractive
index of the medium (water).
9.4 TOTAL INTERNAL REFLECTION
When light travels from an optically denser medium to a rarer medium
at the interface, it is partly reflected back into the same medium and
partly refracted to the second medium. This reflection is called the internal
reflection.
When a ray of light enters from a denser medium to a rarer medium,
it bends away from the normal, for example, the ray AO1
B in Fig. 9.11.
The incident ray AO1 is partially reflected (O1C) and partially transmitted
(O1
B) or refracted, the angle of refraction (r) being larger than the angle of
incidence (i). As the angle of incidence increases, so does the angle of
FIGURE 9.10 Apparent depth for
(a) normal, and (b) oblique viewing.
FIGURE 9.9 Lateral shift of a ray refracted
through a parallel-sided slab.
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refraction, till for the ray AO3, the angle of
refraction is p/2. The refracted ray is bent
so much away from the normal that it
grazes the surface at the interface between
the two media. This is shown by the ray
AO3
D in Fig. 9.11. If the angle of incidence
is increased still further (e.g., the ray AO4),
refraction is not possible, and the incident
ray is totally reflected. This is called total
internal reflection. When light gets
reflected by a surface, normally some
fraction of it gets transmitted. The
reflected ray, therefore, is always less
intense than the incident ray, howsoever
smooth the reflecting surface may be. In
total internal reflection, on the other hand,
no transmission of light takes place.
The angle of incidence corresponding to an angle of refraction 90°,
say ÐAO3N, is called the critical angle (ic ) for the given pair of media. We
see from Snell’s law [Eq. (9.10)] that if the relative refractive index of the
refracting medium is less than one then, since the maximum value of sin
r is unity, there is an upper limit to the value of sin i for which the law
can be satisfied, that is, i = ic such that
sin ic = n21 (9.12)
For values of i larger than ic
, Snell’s law of refraction cannot be
satisfied, and hence no refraction is possible.
The refractive index of denser medium 1 with respect to rarer medium
2 will be n12
= 1/sinic
. Some typical critical angles are listed in Table 9.1.
FIGURE 9.11 Refraction and internal reflection
of rays from a point A in the denser medium
(water) incident at different angles at the interface
with a rarer medium (air).
A demonstration for total internal reflection
All optical phenomena can be demonstrated very easily with the use of a
laser torch or pointer, which is easily available nowadays. Take a glass
beaker with clear water in it. Add a few drops of milk or any other
suspension to water and stir so that water becomes a little turbid. Take
a laser pointer and shine its beam through the turbid water. You will
find that the path of the beam inside the water shines brightly.
TABLE 9.1 CRITICAL ANGLE OF SOME TRANSPARENT MEDIA WITH RESPECT TO AIR
Substance medium Refractive index Critical angle
Water 1.33 48.75
Crown glass 1.52 41.14
Dense flint glass 1.62 37.31
Diamond 2.42 24.41
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Shine the beam from below the beaker such that it strikes at the
upper water surface at the other end. Do you find that it undergoes partial
reflection (which is seen as a spot on the table below) and partial refraction
[which comes out in the air and is seen as a spot on the roof; Fig. 9.12(a)]?
Now direct the laser beam from one side of the beaker such that it strikes
the upper surface of water more obliquely [Fig. 9.12(b)]. Adjust the
direction of laser beam until you find the angle for which the refraction
above the water surface is totally absent and the beam is totally reflected
back to water. This is total internal reflection at its simplest.
Pour this water in a long test tube and shine the laser light from top,
as shown in Fig. 9.12(c). Adjust the direction of the laser beam such that
it is totally internally reflected every time it strikes the walls of the tube.
This is similar to what happens in optical fibres.
Take care not to look into the laser beam directly and not to point it
at anybody’s face.
9.4.1
9.4.1
9.4.1
9.4.1
9.4.1 Total internal reflection in nature and
Total internal reflection in nature and
its technological applications
(i) Prism: Prisms designed to bend light by 90° or by 180° make use of
total internal reflection [Fig. 9.13(a) and (b)]. Such a prism is also
used to invert images without changing their size [Fig. 9.13(c)].
In the first two cases, the critical angle ic
for the material of the prism
must be less than 45°. We see from Table 9.1 that this is true for both
crown glass and dense flint glass.
(ii) Optical fibres: Nowadays optical fibres are extensively used for
transmitting audio and video signals through long distances. Optical
fibres too make use of the phenomenon of total internal reflection.
Optical fibres are fabricated with high quality composite glass/quartz
fibres. Each fibre consists of a core and
cladding. The refractive index of the
material of the core is higher than that
of the cladding.
When a signal in the form of light is
directed at one end of the fibre at a suitable
angle, it undergoes repeated total internal
reflections along the length of the fibre and
finally comes out at the other end (Fig.
9.14). Since light undergoes total internal
reflection at each stage, there is no
appreciable loss in the intensity of the light
signal. Optical fibres are fabricated such
that light reflected at one side of inner
surface strikes the other at an angle larger
than the critical angle. Even if the fibre is
bent, light can easily travel along its length.
Thus, an optical fibre can be used to act as
an optical pipe.
A bundle of optical fibres can be put to
several uses. Optical fibres are extensively
used for transmitting and receiving
FIGURE
FIGURE
FIGURE
FIGURE
FIGURE 9.12
9.12
9.12
9.12
9.12
Observing total internal
reflection in water with
a laser beam (refraction
due to glass of beaker
neglected being very
thin).
FIGURE
FIGURE
FIGURE
FIGURE
FIGURE 9.13
9.13
9.13
9.13
9.13 Prisms designed to bend rays by
90° and 180° or to invert image without changing
its size make use of total internal reflection.
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electrical signals which are converted to light
by suitable transducers. Obviously, optical
fibres can also be used for transmission of
optical signals. For example, these are used
as a ‘light pipe’ to facilitate visual examination
of internal organs like esophagus, stomach
and intestines. You might have seen a
commonly available decorative lamp with fine
plastic fibres with their free ends forming a
fountain like structure. The other end of the
fibres is fixed over an electric lamp. When the
lamp is switched on, the light travels from the bottom of each fibre and
appears at the tip of its free end as a dot of light. The fibres in such
decorative lamps are optical fibres.
The main requirement in fabricating optical fibres is that there should
be very little absorption of light as it travels for long distances inside
them. This has been achieved by purification and special preparation of
materials such as quartz. In silica glass fibres, it is possible to transmit
more than 95% of the light over a fibre length of 1 km. (Compare with
what you expect for a block of ordinary window glass 1 km thick.)
9.5 REFRACTION AT SPHERICAL SURFACES
AND BY LENSES
We have so far considered refraction at a plane interface. We shall now
consider refraction at a spherical interface between two transparent media.
An infinitesimal part of a spherical surface can be regarded as planar
and the same laws of refraction can be applied at every point on the
surface. Just as for reflection by a spherical mirror, the normal at the
point of incidence is perpendicular to the tangent plane to the spherical
surface at that point and, therefore, passes through its centre of
curvature. We first consider refraction by a single spherical surface and
follow it by thin lenses. A thin lens is a transparent optical medium
bounded by two surfaces; at least one of which should be spherical.
Applying the formula for image formation by a single spherical surface
successively at the two surfaces of a lens, we shall obtain the lens maker’s
formula and then the lens formula.
9.5.1 Refraction at a spherical surface
Figure 9.15 shows the geometry of formation of image I of an object O on
the principal axis of a spherical surface with centre of curvature C, and
radius of curvature R. The rays are incident from a medium of refractive
index n1, to another of refractive index n2. As before, we take the aperture
(or the lateral size) of the surface to be small compared to other distances
involved, so that small angle approximation can be made. In particular,
NM will be taken to be nearly equal to the length of the perpendicular
from the point N on the principal axis. We have, for small angles,
tan ÐNOM =
MN
OM
FIGURE 9.14 Light undergoes successive total
internal reflections as it moves through an
optical fibre.
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E
XAMPLE
9.5
tan ÐNCM =
MN
MC
tan ÐNIM =
MN
MI
Now, for DNOC, i is the exterior angle. Therefore, i
= ÐNOM + ÐNCM
i =
MN MN
OM MC
+ (9.13)
Similarly,
r = ÐNCM – ÐNIM
i.e., r =
MN MN
MC MI
− (9.14)
Now, by Snell’s law
n1 sin i = n2 sin r
or for small angles
n1
i = n2
r
Substituting i and r from Eqs. (9.13) and (9.14), we get
1 2 2 1
OM MI MC
n n n n
−
+ = (9.15)
Here, OM, MI and MC represent magnitudes of distances. Applying the
Cartesian sign convention,
OM = –u, MI = +v, MC = +R
Substituting these in Eq. (9.15), we get
2 1 2 1
n n n n
v u R
−
− = (9.16)
Equation (9.16) gives us a relation between object and image distance
in terms of refractive index of the medium and the radius of
curvature of the curved spherical surface. It holds for any curved
spherical surface.
Example 9.5 Light from a point source in air falls on a spherical
glass surface (n = 1.5 and radius of curvature = 20 cm). The distance
of the light source from the glass surface is 100 cm. At what position
the image is formed?
Solution
We use the relation given by Eq. (9.16). Here
u = – 100 cm, v = ?, R = + 20 cm, n1
= 1, and n2
= 1.5.
We then have
1.5 1 0.5
100 20
v
+ =
or v = +100 cm
The image is formed at a distance of 100 cm from the glass surface,
in the direction of incident light.
FIGURE 9.15 Refraction at a spherical
surface separating two media.
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9.5.2 Refraction by a lens
Figure 9.16(a) shows the geometry of image formation by a double convex
lens. The image formation can be seen in terms of two steps:
(i) The first refracting surface forms the image I1
of the object O
[Fig. 9.16(b)]. The image I1
acts as a virtual object for the second surface
that forms the image at I [Fig. 9.16(c)]. Applying Eq. (9.15) to the first
interface ABC, we get
1 2 2 1
1 1
OB BI BC
n n n n

  (9.17)
A similar procedure applied to the second
interface* ADC gives,
2 1 2 1
1 2
DI DI DC
n n n n

   (9.18)
For a thin lens, BI1
= DI1
. Adding
Eqs. (9.17) and (9.18), we get
n n
n n
1 1
2 1
1 1
OB DI BC DC
1 2
+ = − +






( ) (9.19)
Suppose the object is at infinity, i.e.,
OB ® ¥ and DI = f, Eq. (9.19) gives
n
f
n n
1
2 1
1 1
= − +






( )
BC DC
1 2
(9.20)
The point where image of an object
placed at infinity is formed is called the
focus F, of the lens and the distance f gives
its focal length. A lens has two foci, F and
F¢, on either side of it (Fig. 9.17). By the
sign convention,
BC1
= + R1
,
DC2
= –R2
So Eq. (9.20) can be written as
(9.21)
Equation (9.21) is known as the lens
maker’s formula. It is useful to design
lenses of desired focal length using surfaces
of suitable radii of curvature. Note that the
formula is true for a concave lens also. In
that case R1
is negative, R2
positive and
therefore, f is negative.
FIGURE 9.16 (a) The position of object, and the
image formed by a double convex lens,
(b) Refraction at the first spherical surface and
(c) Refraction at the second spherical surface.
* Note that now the refractive index of the medium on the right side of ADC is n1
while on its left it is n2
. Further DI1
is negative as the distance is measured
against the direction of incident light.
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From Eqs. (9.19) and (9.20), we get
1 1 1
OB DI
n n n
f
+ = (9.22)
Again, in the thin lens approximation, B and D are both close to the
optical centre of the lens. Applying the sign convention,
BO = – u, DI = +v, we get
1 1 1
v u f
− = (9.23)
Equation (9.23) is the familiar thin lens formula. Though we derived
it for a real image formed by a convex lens, the formula is valid for both
convex as well as concave lenses and for both real and virtual images.
It is worth mentioning that the two foci, F and F¢, of a double convex
or concave lens are equidistant from the optical centre. The focus on the
side of the (original) source of light is called the first focal point, whereas
the other is called the second focal point.
To find the image of an object by a lens, we can, in principle, take any
two rays emanating from a point on an object; trace their paths using the
laws of refraction and find the point where the refracted rays meet (or
appear to meet). In practice, however, it is convenient to choose any two
of the following rays:
(i) A ray emanating from the object parallel to the principal axis of the
lens after refraction passes through the second principal focus F¢ (in
a convex lens) or appears to diverge (in a concave lens) from the first
principal focus F.
(ii) A ray of light, passing through the optical
centre of the lens, emerges without any
deviation after refraction.
(iii) (a) A ray of light passing through the first
principal focus of a convex lens [Fig. 9.17(a)]
emerges parallel to the principal axis after
refraction.
(b) A ray of light incident on a concave lens
appearing to meet the principal axis at
second focus point emerges parallel to the
principal axis after refraction [Fig. 9.17(b)].
Figures 9.17(a) and (b) illustrate these rules
for a convex and a concave lens, respectively.
You should practice drawing similar ray
diagrams for different positions of the object with
respect to the lens and also verify that the lens
formula, Eq. (9.23), holds good for all cases.
Here again it must be remembered that each
point on an object gives out infinite number of
rays. All these rays will pass through the same
image point after refraction at the lens.
Magnification (m) produced by a lens is
defined, like that for a mirror, as the ratio of the
size of the image to that of the object. Proceeding
FIGURE 9.17 Tracing rays through (a)
convex lens (b) concave lens.
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E
XAMPLE
9.6
in the same way as for spherical mirrors, it is easily seen that
for a lens
m =
h
h
′
=
v
u
(9.24)
When we apply the sign convention, we see that, for erect (and virtual)
image formed by a convex or concave lens, m is positive, while for an
inverted (and real) image, m is negative.
Example 9.6 A magician during a show makes a glass lens with
n = 1.47 disappear in a trough of liquid. What is the refractive index
of the liquid? Could the liquid be water?
Solution
The refractive index of the liquid must be equal to 1.47 in order to
make the lens disappear. This means n1
= n2
. This gives 1/f =0 or
f ® ¥. The lens in the liquid will act like a plane sheet of glass. No,
the liquid is not water. It could be glycerine.
FIGURE 9.18 Power of a lens.
9.5.3 Power of a lens
Power of a lens is a measure of the convergence or
divergence, which a lens introduces in the light falling on
it. Clearly, a lens of shorter focal length bends the incident
light more, while converging it in case of a convex lens
and diverging it in case of a concave lens. The power P of
a lens is defined as the tangent of the angle by which it
converges or diverges a beam of light parallel to the
principal axis falling at unit distance from the optical
centre (Fig. 9.18).
tan ; , tan
δ δ
= = =
h
f
h
f
if 1
1
or
1
f
δ = for small
value of d. Thus,
P =
1
f
(9.25)
The SI unit for power of a lens is dioptre (D): 1D = 1m–1
. The power of
a lens of focal length of 1 metre is one dioptre. Power of a lens is positive
for a converging lens and negative for a diverging lens. Thus, when an
optician prescribes a corrective lens of power + 2.5 D, the required lens is
a convex lens of focal length + 40 cm. A lens of power of – 4.0 D means a
concave lens of focal length – 25 cm.
Example 9.7 (i) If f = 0.5 m for a glass lens, what is the power of the
lens? (ii) The radii of curvature of the faces of a double convex lens
are 10 cm and 15 cm. Its focal length is 12 cm. What is the refractive
index of glass? (iii) A convex lens has 20 cm focal length in air. What
is focal length in water? (Refractive index of air-water = 1.33, refractive
index for air-glass = 1.5.)
E
XAMPLE
9.7
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Solution
(i) Power = +2 dioptre.
(ii) Here, we have f = +12 cm, R1
= +10 cm, R2
= –15 cm.
Refractive index of air is taken as unity.
We use the lens formula of Eq. (9.22). The sign convention has to
be applied for f, R1
and R2
.
Substituting the values, we have
1
12
1
1
10
1
15
= − −
−






( )
n
This gives n = 1.5.
(iii) For a glass lens in air, n2
= 1.5, n1
= 1, f = +20 cm. Hence, the lens
formula gives
1
20
0 5
1 1
1 2
= −






.
R R
For the same glass lens in water, n2
= 1.5, n1
= 1.33. Therefore,
1 33
1 5 1 33
1 1
1 2
.
( . . )
f R R
= − −





 (9.26)
Combining these two equations, we find f = + 78.2 cm.
9.5.4 Combination of thin lenses in contact
Consider two lenses A and B of focal length f1 and
f2
placed in contact with each other. Let the object
be placed at a point O beyond the focus of the first
lens A (Fig. 9.19). The first lens produces an image
at I1
. Since image I1
is real, it serves as a virtual
object for the second lens B, producing the final
image at I. It must, however, be borne in mind that
formation of image by the first lens is presumed
only to facilitate determination of the position of the
final image. In fact, the direction of rays emerging
from the first lens gets modified in accordance with
the angle at which they strike the second lens. Since the lenses are thin,
we assume the optical centres of the lenses to be coincident. Let this
central point be denoted by P.
For the image formed by the first lens A, we get
1 1
1 1 1
v u f
− = (9.27)
For the image formed by the second lens B, we get
1 2
1 1 1
v v f
− = (9.28)
Adding Eqs. (9.27) and (9.28), we get
1 2
1 1 1 1
v u f f
− = + (9.29)
If the two lens-system is regarded as equivalent to a single lens of
focal length f, we have
FIGURE 9.19 Image formation by a
combination of two thin lenses in contact.
E
XAMPLE
9.7
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E
XAMPLE
9.8
1 1 1
v u f
− =
so that we get
1 2
1 1 1
f f f
= + (9.30)
The derivation is valid for any number of thin lenses in contact. If
several thin lenses of focal length f1
, f2
, f3
,... are in contact, the effective
focal length of their combination is given by
1 2 3
1 1 1 1
f f f f
= + + + … (9.31)
In terms of power, Eq. (9.31) can be written as
P = P1
+ P2
+ P3
+ … (9.32)
where P is the net power of the lens combination. Note that the sum in
Eq. (9.32) is an algebraic sum of individual powers, so some of the terms
on the right side may be positive (for convex lenses) and some negative
(for concave lenses). Combination of lenses helps to obtain diverging or
converging lenses of desired magnification. It also enhances sharpness
of the image. Since the image formed by the first lens becomes the object
for the second, Eq. (9.25) implies that the total magnification m of the
combination is a product of magnification (m1
, m2
, m3
,...) of individual
lenses
m = m1
m2
m3
... (9.33)
Such a system of combination of lenses is commonly used in designing
lenses for cameras, microscopes, telescopes and other optical instruments.
Example 9.8 Find the position of the image formed by the lens
combination given in the Fig. 9.20.
FIGURE 9.20
Solution Image formed by the first lens
1 1 1
1 1 1
v u f
− =
1
1 1 1
30 10
v
− =
−
or v1
= 15 cm
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FIGURE 9.21 A ray of light passing through
a triangular glass prism.
E
XAMPLE
9.8
The image formed by the first lens serves as the object for the second.
This is at a distance of (15 – 5) cm = 10 cm to the right of the second
lens. Though the image is real, it serves as a virtual object for the
second lens, which means that the rays appear to come from it for
the second lens.
2
1 1 1
10 10
v
− =
−
or v2
= ¥
The virtual image is formed at an infinite distance to the left of the
second lens. This acts as an object for the third lens.
3 3 3
1 1 1
v u f
− =
or
3
1 1 1
30
v
= +
∞
or v3
= 30 cm
The final image is formed 30 cm to the right of the third lens.
9.6 REFRACTION THROUGH A PRISM
Figure 9.21 shows the passage of light through
a triangular prism ABC. The angles of incidence
and refraction at the first face AB are i and r1,
while the angle of incidence (from glass to air) at
the second face AC is r2
and the angle of refraction
or emergence e. The angle between the emergent
ray RS and the direction of the incident ray PQ
is called the angle of deviation, d.
In the quadrilateral AQNR, two of the angles
(at the vertices Q and R) are right angles.
Therefore, the sum of the other angles of the
quadrilateral is 180°.
ÐA + ÐQNR = 180°
From the triangle QNR,
r1
+ r2
+ ÐQNR = 180°
Comparing these two equations, we get
r1 + r2 = A (9.34)
The total deviation d is the sum of deviations at the two faces,
d = (i – r1 ) + (e – r2 )
that is,
d = i + e – A (9.35)
Thus, the angle of deviation depends on the angle of incidence. A plot
between the angle of deviation and angle of incidence is shown in
Fig. 9.22. You can see that, in general, any given value of d, except for
i = e, corresponds to two values i and hence of e. This, in fact, is expected
from the symmetry of i and e in Eq. (9.35), i.e., d remains the same if i
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and e are interchanged. Physically, this is related
to the fact that the path of ray in Fig. 9.21 can be
traced back, resulting in the same angle of
deviation. At the minimum deviation Dm, the
refracted ray inside the prism becomes parallel
to its base. We have
d = Dm
, i = e which implies r1
= r2
.
Equation (9.34) gives
2r = A or r =
2
A
(9.36)
In the same way, Eq. (9.35) gives
Dm
= 2i – A, or i = (A + Dm
)/2 (9.37)
The refractive index of the prism is
2
21
1
sin[( )/2]
sin[ /2]
m
A D
n
n
n A
+
= = (9.38)
The angles A and Dm
can be measured experimentally. Equation
(9.38) thus provides a method of determining refractive index of the
material of the prism.
For a small angle prism, i.e., a thin prism, Dm
is also very small, and
we get
( )
21
/2
sin[( )/2]
sin[ /2] /2
m
m
A D
A D
n
A A
+
+
= ≃
Dm
= (n21
–1)A
It implies that, thin prisms do not deviate light much.
9.7 OPTICAL INSTRUMENTS
A number of optical devices and instruments have been designed utilising
reflecting and refracting properties of mirrors, lenses and prisms.
Periscope, kaleidoscope, binoculars, telescopes, microscopes are some
examples of optical devices and instruments that are in common use.
Our eye is, of course, one of the most important optical device the nature
has endowed us with. We have already studied about the human eye in
Class X. We now go on to describe the principles of working of the
microscope and the telescope.
9.7.1 The microscope
A simple magnifier or microscope is a converging lens of small focal length
(Fig. 9.23). In order to use such a lens as a microscope, the lens is held
near the object, one focal length away or less, and the eye is positioned
close to the lens on the other side. The idea is to get an erect, magnified
and virtual image of the object at a distance so that it can be viewed
comfortably, i.e., at 25 cm or more. If the object is at a distance f, the
image is at infinity. However, if the object is at a distance slightly less
FIGURE 9.22 Plot of angle of deviation (d)
versus angle of incidence (i) for a
triangular prism.
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than the focal length of the lens, the image is virtual and closer than
infinity. Although the closest comfortable distance for viewing the image
is when it is at the near point (distance D @ 25 cm), it causes some strain
on the eye. Therefore, the image formed at infinity is often considered
most suitable for viewing by the relaxed eye. We show both cases, the
first in Fig. 9.23(a), and the second in Fig. 9.23(b) and (c).
The linear magnification m, for the image formed at the near point D,
by a simple microscope can be obtained by using the relation
FIGURE 9.23 A simple microscope; (a) the magnifying lens is located
such that the image is at the near point, (b) the angle subtanded by the
object, is the same as that at the near point, and (c) the object near the
focal point of the lens; the image is far off but closer than infinity.
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m
v
u
v
v f
v
f
= =





 =






1 1
1
– –
Now according to our sign convention, v is negative, and is equal in
magnitude to D. Thus, the magnification is
m
D
f
= +






1 (9.39)
Since D is about 25 cm, to have a magnification of six, one needs a convex
lens of focal length, f = 5 cm.
Note that m = h¢/h where h is the size of the object and h¢ the size of
the image. This is also the ratio of the angle subtended by the image
to that subtended by the object, if placed at D for comfortable viewing.
(Note that this is not the angle actually subtended by the object at the
eye, which is h/u.) What a single-lens simple magnifier achieves is that it
allows the object to be brought closer to the eye than D.
We will now find the magnification when the image is at infinity. In
this case we will have to obtained the angular magnification. Suppose
the object has a height h. The maximum angle it can subtend, and be
clearly visible (without a lens), is when it is at the near point, i.e., a distance
D. The angle subtended is then given by
tan θo
h
D
=





 » qo
(9.40)
We now find the angle subtended at the eye by the image when the
object is at u. From the relations
h v
m
h u
′
= =
we have the angle subtended by the image
tan i
h h v h
v v u u
θ
′
= = ⋅ =
− − −
»q. The angle subtended by the object, when it
is at u = –f.
θi
h
f
=





 (9.41)
as is clear from Fig. 9.23(c). The angular magnification is, therefore
m
D
f
i
o
=





 =
θ
θ
(9.42)
This is one less than the magnification when the image is at the near
point, Eq. (9.39), but the viewing is more comfortable and the difference
in magnification is usually small. In subsequent discussions of optical
instruments (microscope and telescope) we shall assume the image to be
at infinity.
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A simple microscope has a limited maximum magnification (£ 9) for
realistic focal lengths. For much larger magnifications, one uses two lenses,
one compounding the effect of the other. This is known as a compound
microscope. A schematic diagram of a compound microscope is shown
in Fig. 9.24. The lens nearest the object, called the objective, forms a
real, inverted, magnified image of the object. This serves as the object for
the second lens, the eyepiece, which functions essentially like a simple
microscope or magnifier, produces the final image, which is enlarged
and virtual. The first inverted image is thus near (at or within) the focal
plane of the eyepiece, at a distance appropriate for final image formation
at infinity, or a little closer for image formation at the near point. Clearly,
the final image is inverted with respect to the original object.
We now obtain the magnification due to a compound microscope.
The ray diagram of Fig. 9.24 shows that the (linear) magnification due to
the objective, namely h¢/h, equals
O
o
h L
m
h f
′
= = (9.43)
where we have used the result
tanβ =





 =
′






h
f
h
L
o
Here h¢ is the size of the first image, the object size being h and fo
being the focal length of the objective. The first image is formed near the
focal point of the eyepiece. The distance L, i.e., the distance between the
second focal point of the objective and the first focal point of the eyepiece
(focal length fe) is called the tube length of the compound microscope.
The
world’s
largest
optical
telescopes
http://astro.nineplanets.org/bigeyes.html
FIGURE 9.24 Ray diagram for the formation of image by a
compound microscope.
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As the first inverted image is near the focal point of the eyepiece, we
use the result from the discussion above for the simple microscope to
obtain the (angular) magnification me due to it [Eq. (9.39)], when the
final image is formed at the near point, is
m
D
f
e
e
= +
1





 [9.44(a)]
When the final image is formed at infinity, the angular magnification
due to the eyepiece [Eq. (9.42)] is
me = (D/fe ) [9.44(b)]
Thus, the total magnification [(according to Eq. (9.33)], when the
image is formed at infinity, is
m m m
L
f
D
f
o e
o e
= =











 (9.45)
Clearly, to achieve a large magnification of a small object (hence the
name microscope), the objective and eyepiece should have small focal
lengths. In practice, it is difficult to make the focal length much smaller
than 1 cm. Also large lenses are required to make L large.
For example, with an objective with fo = 1.0 cm, and an eyepiece with
focal length fe
= 2.0 cm, and a tube length of 20 cm, the magnification is
m m m
L
f
D
f
o e
o e
= =












20 25
250
1 2
Various other factors such as illumination of the object, contribute to
the quality and visibility of the image. In modern microscopes, multi-
component lenses are used for both the objective and the eyepiece to
improve image quality by minimising various optical aberrations (defects)
in lenses.
9.7.2 Telescope
The telescope is used to provide angular magnification of distant objects
(Fig. 9.25). It also has an objective and an eyepiece. But here, the objective
has a large focal length and a much larger aperture than the eyepiece.
Light from a distant object enters the objective and a real image is formed
in the tube at its second focal point. The eyepiece magnifies this image
producing a final inverted image. The magnifying power m is the ratio of
the angle b subtended at the eye by the final image to the angle a which
the object subtends at the lens or the eye. Hence
. o o
e e
f f
h
m
f h f
(9.46)
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FIGURE 9.25 A refracting telescope.
In this case, the length of the telescope tube is fo + fe.
Terrestrial telescopes have, in addition, a pair of inverting lenses to
make the final image erect. Refracting telescopes can be used both for
terrestrial and astronomical observations. For example, consider
a telescope whose objective has a focal length of 100 cm and the eyepiece
a focal length of 1 cm. The magnifying power of this telescope is
m = 100/1 = 100.
Let us consider a pair of stars of actual separation 1¢ (one minute of
arc). The stars appear as though they are separated by an angle of 100 ×
1¢ = 100¢ =1.67°.
The main considerations with an astronomical telescope are its light
gathering power and its resolution or resolving power. The former clearly
depends on the area of the objective. With larger diameters, fainter objects
can be observed. The resolving power, or the ability to observe two objects
distinctly, which are in very nearly the same direction, also depends on
the diameter of the objective. So, the desirable aim in optical telescopes is
to make them with objective of large diameter. The largest lens objective
in use has a diameter of 40 inch (~1.02 m). It is at the Yerkes Observatory
in Wisconsin, USA. Such big lenses tend to be very heavy and therefore,
difficult to make and support by their edges. Further, it is rather difficult
and expensive to make such large sized lenses which form images that
are free from any kind of chromatic aberration and distortions.
For these reasons, modern telescopes use a concave mirror rather
than a lens for the objective. Telescopes with mirror objectives are called
reflecting telescopes. There is no chromatic aberration in a mirror.
Mechanical support is much less of a problem since a mirror weighs
much less than a lens of equivalent optical quality, and can be supported
over its entire back surface, not just over its rim. One obvious problem
with a reflecting telescope is that the objective mirror focusses light inside
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SUMMARY
1. Reflection is governed by the equation Ði = Ðr¢ and refraction by the
Snell’s law, sini/sinr = n, where the incident ray, reflected ray, refracted
ray and normal lie in the same plane. Angles of incidence, reflection
and refraction are i, r ¢ and r, respectively.
2. The critical angle of incidence ic
for a ray incident from a denser to rarer
medium, is that angle for which the angle of refraction is 90°. For
i > ic
, total internal reflection occurs. Multiple internal reflections in
diamond (ic
@ 24.4°), totally reflecting prisms and mirage, are some
examples of total internal reflection. Optical fibres consist of glass
fibres coated with a thin layer of material of lower refractive index.
Light incident at an angle at one end comes out at the other, after
multiple internal reflections, even if the fibre is bent.
FIGURE 9.26 Schematic diagram of a reflecting telescope (Cassegrain).
the telescope tube. One must have an eyepiece and the observer right
there, obstructing some light (depending on the size of the observer cage).
This is what is done in the very large 200 inch (~5.08 m) diameters, Mt.
Palomar telescope, California. The viewer sits near the focal point of the
mirror, in a small cage. Another solution to the problem is to deflect the
light being focussed by another mirror. One such arrangement using a
convex secondary mirror to focus the incident light, which now passes
through a hole in the objective primary mirror, is shown in Fig. 9.26.
This is known as a Cassegrain telescope, after its inventor. It has the
advantages of a large focal length in a short telescope. The largest telescope
in India is in Kavalur, Tamil Nadu. It is a 2.34 m diameter reflecting
telescope (Cassegrain). It was ground, polished, set up, and is being used
by the Indian Institute of Astrophysics, Bangalore. The largest reflecting
telescopes in the world are the pair of Keck telescopes in Hawaii, USA,
with a reflector of 10 metre in diameter.
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3. Cartesian sign convention: Distances measured in the same direction
as the incident light are positive; those measured in the opposite
direction are negative. All distances are measured from the pole/optic
centre of the mirror/lens on the principal axis. The heights measured
upwards above x-axis and normal to the principal axis of the mirror/
lens are taken as positive. The heights measured downwards are taken
as negative.
4. Mirror equation:
1 1 1
v u f
+ =
where u and v are object and image distances, respectively and f is the
focal length of the mirror. f is (approximately) half the radius of
curvature R. f is negative for concave mirror; f is positive for a convex
mirror.
5. For a prism of the angle A, of refractive index n2
placed in a medium
of refractive index n1
,
n
n
n
A D
A
m
21
2
1
2
2
= =
+
( )

 

( )
sin /
sin /
where Dm
is the angle of minimum deviation.
6. For refraction through a spherical interface (from medium 1 to 2 of
refractive index n1
and n2
, respectively)
2 1 2 1
n n n n
v u R
−
− =
Thin lens formula
1 1 1
v u f
− =
Lens maker’s formula
1 1 1
2 1
1 1 2
f
n n
n R R
=
−
( ) −






R1 and R2 are the radii of curvature of the lens surfaces. f is positive
for a converging lens; f is negative for a diverging lens. The power of a
lens P = 1/f.
The SI unit for power of a lens is dioptre (D): 1 D = 1 m–1
.
If several thin lenses of focal length f1
, f2
, f3
,.. are in contact, the
effective focal length of their combination, is given by
1 2 3
1 1 1 1
f f f f
= + + + …
The total power of a combination of several lenses is
P = P1 + P2 + P3 + …
7. Dispersion is the splitting of light into its constituent colour.
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POINTS TO PONDER
1. The laws of reflection and refraction are true for all surfaces and
pairs of media at the point of the incidence.
2. The real image of an object placed between f and 2f from a convex lens
can be seen on a screen placed at the image location. If the screen is
removed, is the image still there? This question puzzles many, because
it is difficult to reconcile ourselves with an image suspended in air
without a screen. But the image does exist. Rays from a given point
on the object are converging to an image point in space and diverging
away. The screen simply diffuses these rays, some of which reach our
eye and we see the image. This can be seen by the images formed in
air during a laser show.
3. Image formation needs regular reflection/refraction. In principle, all
rays from a given point should reach the same image point. This is
why you do not see your image by an irregular reflecting object, say
the page of a book.
4. Thick lenses give coloured images due to dispersion. The variety in
colour of objects we see around us is due to the constituent colours
of the light incident on them. A monochromatic light may produce an
entirely different perception about the colours on an object as seen in
white light.
5. For a simple microscope, the angular size of the object equals the
angular size of the image. Yet it offers magnification because we can
keep the small object much closer to the eye than 25 cm and hence
have it subtend a large angle. The image is at 25 cm which we can see.
Without the microscope, you would need to keep the small object at
25 cm which would subtend a very small angle.
8. Magnifying power m of a simple microscope is given by m = 1 + (D/f),
where D = 25 cm is the least distance of distinct vision and f is the
focal length of the convex lens. If the image is at infinity, m = D/f. For
a compound microscope, the magnifying power is given by m = me
× m0
where me
= 1 + (D/fe
), is the magnification due to the eyepiece and mo
is the magnification produced by the objective. Approximately,
o e
L D
m
f f
= ×
where fo
and fe
are the focal lengths of the objective and eyepiece,
respectively, and L is the distance between their focal points.
9. Magnifying power m of a telescope is the ratio of the angle b subtended
at the eye by the image to the angle a subtended at the eye by the
object.
o
e
f
m
f
β
α
= =
where f0 and fe are the focal lengths of the objective and eyepiece,
respectively.
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EXERCISES
9.1 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?
9.2 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.
9.3 A tank is filled with water to a height of 12.5 cm. The apparent
depth of a needle lying at the bottom of the tank is measured by a
microscope to be 9.4 cm. What is the refractive index of water? If
water is replaced by a liquid of refractive index 1.63 up to the same
height, by what distance would the microscope have to be moved to
focus on the needle again?
9.4 Figures 9.27(a) and (b) show refraction of a ray in air incident at 60°
with the normal to a glass-air and water-air interface, respectively.
Predict the angle of refraction in glass when the angle of incidence
in water is 45° with the normal to a water-glass interface [Fig. 9.27(c)].
FIGURE 9.27
9.5 A small bulb is placed at the bottom of a tank containing water to a
depth of 80cm. What is the area of the surface of water through
which light from the bulb can emerge out? Refractive index of water
is 1.33. (Consider the bulb to be a point source.)
9.6 A prism is made of glass of unknown refractive index. A parallel
beam of light is incident on a face of the prism. The angle of minimum
deviation is measured to be 40°. What is the refractive index of the
material of the prism? The refracting angle of the prism is 60°. If the
prism is placed in water (refractive index 1.33), predict the new
angle of minimum deviation of a parallel beam of light.
9.7 Double-convex lenses are to be manufactured from a glass of
refractive index 1.55, with both faces of the same radius of
curvature. What is the radius of curvature required if the focal length
is to be 20cm?
9.8 A beam of light converges at a point P. Now a lens is placed in the
path of the convergent beam 12cm from P. At what point does the
beam converge if the lens is (a) a convex lens of focal length 20cm,
and (b) a concave lens of focal length 16cm?
9.9 An object of size 3.0cm is placed 14cm in front of a concave lens of
focal length 21cm. Describe the image produced by the lens. What
happens if the object is moved further away from the lens?
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9.10 What is the focal length of a convex lens of focal length 30cm in
contact with a concave lens of focal length 20cm? Is the system a
converging or a diverging lens? Ignore thickness of the lenses.
9.11 A compound microscope consists of an objective lens of focal length
2.0 cm and an eyepiece of focal length 6.25 cm separated by a
distance of 15cm. How far from the objective should an object be
placed in order to obtain the final image at (a) the least distance of
distinct vision (25cm), and (b) at infinity? What is the magnifying
power of the microscope in each case?
9.12 A person with a normal near point (25 cm) using a compound
microscope with objective of focal length 8.0 mm and an eyepiece of
focal length 2.5cm can bring an object placed at 9.0mm from the
objective in sharp focus. What is the separation between the two
lenses? Calculate the magnifying power of the microscope,
9.13 A small telescope has an objective lens of focal length 144cm and
an eyepiece of focal length 6.0cm. What is the magnifying power of
the telescope? What is the separation between the objective and
the eyepiece?
9.14 (a) A giant refracting telescope at an observatory has an objective
lens of focal length 15m. If an eyepiece of focal length 1.0cm is
used, what is the angular magnification of the telescope?
(b) If this telescope is used to view the moon, what is the diameter
of the image of the moon formed by the objective lens? The
diameter of the moon is 3.48 × 106
m, and the radius of lunar
orbit is 3.8 × 108
m.
9.15 Use the mirror equation to deduce that:
(a) an object placed between f and 2f of a concave mirror produces
a real image beyond 2f.
(b) a convex mirror always produces a virtual image independent
of the location of the object.
(c) the virtual image produced by a convex mirror is always
diminished in size and is located between the focus and
the pole.
(d) an object placed between the pole and focus of a concave mirror
produces a virtual and enlarged image.
[Note: This exercise helps you deduce algebraically properties of
images that one obtains from explicit ray diagrams.]
9.16 A small pin fixed on a table top is viewed from above from a distance
of 50cm. By what distance would the pin appear to be raised if it is
viewed from the same point through a 15cm thick glass slab held
parallel to the table? Refractive index of glass = 1.5. Does the answer
depend on the location of the slab?
9.17 (a) Figure 9.28 shows a cross-section of a ‘light pipe’ made of a
glass fibre of refractive index 1.68. The outer covering of the
pipe is made of a material of refractive index 1.44. What is the
range of the angles of the incident rays with the axis of the pipe
for which total reflections inside the pipe take place, as shown
in the figure.
FIGURE 9.28
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(b) What is the answer if there is no outer covering of the pipe?
9.18 The image of a small electric bulb fixed on the wall of a room is to be
obtained on the opposite wall 3m away by means of a large convex
lens. What is the maximum possible focal length of the lens required
for the purpose?
9.19 A screen is placed 90cm from an object. The image of the object on
the screen is formed by a convex lens at two different locations
separated by 20cm. Determine the focal length of the lens.
9.20 (a) Determine the ‘effective focal length’ of the combination of
the two lenses in Exercise 9.10, if they are placed 8.0cm apart
with their principal axes coincident. Does the answer depend
on which side of the combination a beam of parallel light is
incident? Is the notion of effective focal length of this system
useful at all?
(b) An object 1.5 cm in size is placed on the side of the convex lens
in the arrangement (a) above. The distance between the object
and the convex lens is 40 cm. Determine the magnification
produced by the two-lens system, and the size of the image.
9.21 At what angle should a ray of light be incident on the face of a prism
of refracting angle 60° so that it just suffers total internal reflection
at the other face? The refractive index of the material of the prism is
1.524.
9.22 A card sheet divided into squares each of size 1 mm2
is being viewed
at a distance of 9 cm through a magnifying glass (a converging lens
of focal length 9 cm) held close to the eye.
(a) What is the magnification produced by the lens? How much is
the area of each square in the virtual image?
(b) What is the angular magnification (magnifying power) of the
lens?
(c) Is the magnification in (a) equal to the magnifying power in (b)?
Explain.
9.23 (a) At what distance should the lens be held from the card sheet in
Exercise 9.22 in order to view the squares distinctly with the
maximum possible magnifying power?
(b) What is the magnification in this case?
(c) Is the magnification equal to the magnifying power in this case?
Explain.
9.24 What should be the distance between the object in Exercise 9.23
and the magnifying glass if the virtual image of each square in
the figure is to have an area of 6.25 mm2
. Would you be able to
see the squares distinctly with your eyes very close to the
magnifier?
[Note: Exercises 9.22 to 9.24 will help you clearly understand the
difference between magnification in absolute size and the angular
magnification (or magnifying power) of an instrument.]
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9.25 Answer the following questions:
(a) The angle subtended at the eye by an object is equal to the
angle subtended at the eye by the virtual image produced by a
magnifying glass. In what sense then does a magnifying glass
provide angular magnification?
(b) In viewing through a magnifying glass, one usually positions
one’s eyes very close to the lens. Does angular magnification
change if the eye is moved back?
(c) Magnifying power of a simple microscope is inversely proportional
to the focal length of the lens. What then stops us from using a
convex lens of smaller and smaller focal length and achieving
greater and greater magnifying power?
(d) Why must both the objective and the eyepiece of a compound
microscope have short focal lengths?
(e) When viewing through a compound microscope, our eyes should
be positioned not on the eyepiece but a short distance away
from it for best viewing. Why? How much should be that short
distance between the eye and eyepiece?
9.26 An angular magnification (magnifying power) of 30X is desired using
an objective of focal length 1.25cm and an eyepiece of focal length
5cm. How will you set up the compound microscope?
9.27 A small telescope has an objective lens of focal length 140cm and
an eyepiece of focal length 5.0cm. What is the magnifying power of
the telescope for viewing distant objects when
(a) the telescope is in normal adjustment (i.e., when the final image
is at infinity)?
(b) the final image is formed at the least distance of distinct vision
(25cm)?
9.28 (a) For the telescope described in Exercise 9.27 (a), what is the
separation between the objective lens and the eyepiece?
(b) If this telescope is used to view a 100 m tall tower 3 km away,
what is the height of the image of the tower formed by the objective
lens?
(c) What is the height of the final image of the tower if it is formed at
25cm?
9.29 A Cassegrain telescope uses two mirrors as shown in Fig. 9.26. Such
a telescope is built with the mirrors 20mm apart. If the radius of
curvature of the large mirror is 220mm and the small mirror is
140mm, where will the final image of an object at infinity be?
9.30 Light incident normally on a plane mirror attached to a galvanometer
coil retraces backwards as shown in Fig. 9.29. A current in the coil
produces a deflection of 3.5o
of the mirror. What is the displacement
of the reflected spot of light on a screen placed 1.5 m away?
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FIGURE 9.29
9.31 Figure 9.30 shows an equiconvex lens (of refractive index 1.50) in
contact with a liquid layer on top of a plane mirror. A small needle
with its tip on the principal axis is moved along the axis until its
inverted image is found at the position of the needle. The distance of
the needle from the lens is measured to be 45.0cm. The liquid is
removed and the experiment is repeated. The new distance is
measured to be 30.0cm. What is the refractive index of the liquid?
FIGURE 9.30
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Notes
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