LIGHT

Introduction to Optics:

The branch of physics which deals with light and other electromagnetic waves is called Optics.

Electromagnetic radiation with wavelength in the range about 4,000 to 7,000 to which human eye is sensitive is called visible region.

Geometrical optics: A light beam can be treated as a ray whose propagation is governed by simple geometrical rules. The part of optics that deals with such phenomena is known as geometrical optics.

Physical optics: The treatment of light by considering the wave property of light is known as physical optics.

The object distance is shown by u and image distance by v There are two types of images, real and virtual. An image which can be captured on a

screen is called real image whereas the later cant.

Theories to explain Light

There are two fundamental theories which explain the nature and properties of light.

Newtons Corpuscular Theory:

The nature of light was first explained by Sir Isaac Newton by considering the corpuscular nature of light.

In this theory, light was considered as composition of small and tiny particles called corpuscular.

The colour of light varies with size of the particle. And it was considered that light will have highest speed in solids and lowest speed in

gaseous media. Corpuscular theory explained reflection and refraction of light successfully and failed to

explain other phenomena.

Huygens Wave Theory

Huygens explained different properties of light like reflection, refraction, interference and diffraction by considering the Wave nature of light.

He introduced a concept of wavefrontwhich is a source of secondary waves. Later the electromagnetic nature of light was introduced which confirms that light is an

electromagnetic radiation. The velocity of light (EM waves) is maximum in vacuum wiz 3X108 m/s.

The intensity of light is measured in Candela(Cd) The product of wavelength and frequency of light gives its velocity.

c =

where - frequency,

- wavelength.

Light exhibits a property called Polarization which concludes that light is a transverse wave.

Other than visible region, we have different electromagnetic waves like -rays, X-rays, Ultra Violet rays, infrared rays, micro and radio waves which have different applications in the modern world.

Reflection of Light

When waves of any type strike the interface between two optical materials, the waves will bounce back. This phenomenon is known as reflection.

Angle of incidence: The angle between the normal drawn to the surface and incident ray is called as angle of incidence.

Angle of reflection: The angle between the normal drawn to the surface and reflected ray is called as angle of incidence.

The angle of incidence, normal and reflect rays.

Laws of reflection: 1. The incident angle and reflected angle are always equal. 2. Incident ray, reflected ray and normal lie on the same plane.

Reflection from a plane surface:

If the object is real, the image formed by a plane mirror is virtual, erect, of same size and at the same distance from the mirror. The diagram is shown below.

Fig. Image formation by a plane mirror

Q1. A point source of light S placed at a distance L in front of the centre of a mirror of width d, hangs vertically on a wall. A man walks in front of the mirror along a line parallel to the mirror at a distance of 2L from it as shown. What is the greatest distance over which he can see image of the light source in the mirror?

Solution: The geometrical representation of the problem is as follows.

Here HI = AB = d

DS = CD =

AH = 2AD GH = 2CD = 2

= d

Similarly IJ = d

Hence GJ = GH+HI+IJ

= d+d+d = 3d Hence the man can see the image of light source for a range of distance 3d.

Two mirrors separated by an angle :

If there is a geometrical arrangement of two mirrors, as shown below,

Then the number of images that can be observed are *

360 1

if 360

is even

360

if 360

is odd

*the angle must be lesser than 1800.

Q2. There is an arrangement of two mirrors separated by 600. Find the number of images observed.

Solution: From the formula.

Number of images observed is: 5

Q3. Two plane mirrors M1 and M2 are inclined at angle as shown. A ray of light 1, which is parallel to M1 strikes M2 and after two reflections, the ray 2 becomes parallel to M2.

Find the angle .

Solution:

Let the angle of incidence of ray 1 is . From the laws of reflection, the angle of reflection also .

Hence = 900 .

Therefore angle between M2 and reflected ray is . Similarly at M1 also makes an angle .

Hence we have a triangle in which 180o

60o

Spherical Mirrors

A curved mirror is a mirror with a curved reflecting surface. The surface may be either convex (bulging outward) or concave(bulging inward). Most curved mirrors have surfaces that are shaped like part of a sphere

Concave Spherical Mirror Convex Spherical Mirror

Image formation by a concave mirror:

The object location determines the position of image, and its type. The following table gives the information about images formed by a concave mirror.

These are shown below geometrically,

Image formation by a convex mirror:

The type and nature of images formed by the convex mirror are given by the following table.

These are shown below geometrically.

Sign convention:

To solve the problems related to image formation, correct distance measurements with proper sign will give correct solutions. Usually,

Object will be placed left side of the mirror, which implies light moves from left to right. The distances parallel to the principal axis are measured from the pole of the mirror. All the distances measured to the right of the origin (along + x-axis) are taken as positive

while those measured to the left of the origin (along x-axis) are taken as negative. Distances measured perpendicular to and above the principal axis (along + y-axis) are

taken as positive while below the principal axis (along y-axis) are taken as negative.

These points are shown below.

Reflection from a spherical surface:

A spherical mirror is simply a piece cut out of a reflective sphere. It has a center of curvature, C, which corresponds to the center of the sphere it was cut

from; a radius of curvature, R which corresponds to the radius of the sphere. A focal point (the point where parallel light rays are focused to) which is located half the

distance from the mirror to the center of curvature. The focal length, f, is therefore:

Focal length of a spherical mirror: f = R / 2

The ray diagrams of refection from spherical surface are shown below.

Rules of image formation by a mirror. 1. A ray which is parallel to principal axis will meet the axis at focal point.

2. A ray which is passing through the focal point, will pass parallel to axis, after the

reflection.

3. A ray which is passing through the centre of curvature of a mirror, will bounce back

into the same direction.

4. A ray incident obliquely to the principal axis, towards a point P (pole of the mirror), on the concave mirror or a convex mirror is reflected obliquely. The incident and reflected rays follow the laws of reflection.

Mirror Formula:

The image formation by a spherical mirror is governed by the mirror formula given by. 1 = 1

+ 1

Where u is the object distance from the optical element v is the image distance from the optical element andf is the focal length of the mirror.

Q4. A convex mirror used for rear-view on an automobile has a radius of curvature of 3.00 m. If a bus is located at 5.00 m from this mirror, find the position of the image.

Solution: Given

Radius of curvature, R= + 3.00 m; Object-distance, u= 5.00 m;

Image-distance, v=?

Focal length of the mirror f =3.00 1.5

2 2R m

From mirror formula, 1 1 1v f u

1v

1 1

1.5 5.00

1 1

1.5 5.00

6.507.50

7.50 1.156.50

v m

Magnification:

Magnification is a numerical quantity which gives the size of the image relative to the object. It is expressed as the ratio of height of the image to the height of the object.

i.e., m is defined as

m =

and mathematically, m =

=

=

Q5. Find the magnification in the above problem.

Solution: We have, object distance u = 5.00m

image distance v=1.15m.

We know that: m =

=

=

1.15 0.235.00

m

Q6. A mirror forms the image of an object with unity magnification. The mirror has a focal length of 20 cm. Find the image distance. ( Take the concept of real image)

Solution: The magnification is unity provided both object and image are at same distances from the mirror.

Let u=v= x and f = 20cm.

From the mirror formula. 1 = 1

+ 1

120 = 1 + 1

On solving x = 40 cm.

Note: The image will have unit magnification, if the object is placed at a distance of 2f.

Q7. Find the distance of objective from a concave mirror of focal length 10 cm so that image size is four times the size of the object.

Solution: Concave mirror can form real as well as virtual image. Here nature of image is not given in the question. So we will consider two possible cases.

Case1: When image is real.

Real image is formed on the same side of the object, i.e., u,vand f all are negative.

Let u = -x

Then v= - 4x and f = - 10 cm.

Substituting in mirror formula 1 = 1

+ 1

110 = 14 + 1

110 = 54 Hence x = 12.5cm

Case 2. When the image virtual. In case of a mirror image is virtual when it is formed behind the mirror. i.e., u and f are negative while v is positive.

Let u = -y then v = +4y and f = -10cm.

From mirror formula: 1 = 1

+ 1

We have 110 = 14 + 1

110 = 34 Or y= 7.5 cm.

Q8.A ray of light is incident on a plane mirror along a vector i j k . The normal on incidence

point is along i j . Find a unit vector along the reflected ray.

Solution: Reflection of a ray is just like an elastic collision of a ball with a horizontal ground. Component of incident rat along the inside normal gets reversed while the component perpendicular to it remains unchanged. Thus the component of incident ray vector A i j k

parallel to normal. i.e., i j gets reversed while perpendicular to it, i.e., k remains unchanged. Thus the reflected ray can be written as.

R i j k

Therefore a unit vector along the reflected ray will be,

Refraction of Light:

When the light travels from one medium to another medium, a part of it, reflects back and the other part will enter into the second medium. When it passes into the second medium, it will bent slightly from its actual path. This bending of light when the medium is changed is called refraction of light.

The angle made by the incident ray with the normal drawn to the boundary of two media is called angle of incidence (i).

The angle made by the light ray with normal after the refraction is called angle of refraction(r).

This is illustrated in the following figure.

31 3

R i j krR

r i j k

Refraction of light when medium is changed.

Refractive Index:

For two particular media, the ratio of the sine of the angle of incidence and the sine of the angle of refraction is a constant.

i.e.,

= This constant is called as refractive index of the medium and this principle is called Snells law.

The refractive index of a medium is also defined as the ratio of velocity of light in vacuum to the velocity of light in that medium.

i.e., refractive index () =

As this is the ratio between same physical parameters, there is no units for refractive index.

As the velocity of light in vacuum is always greater than the velocity of light in medium, the refractive index is always greater than 1.

Velocity of light in vacuum = 3 X 108 m/sec. For example, the velocity of light in glass is 2 X 108 m/sec. Hence, the refractive index

of glass is 1.5. During refraction, the medium with greater refractive index is called as Denser

medium while the medium with lower refractive index is called as Rarer medium

Laws of refraction:

1. Snells law: The ratio of sine of the angle of incidence to the sine of the angle of refraction is a constant and is equal to the ratio of refractive indices of second medium to the first.

= 2/1

2. The incident ray, reflected ray and refracted ray all will lie in the same plane.

Facts during Refraction:

From the Snells law, it is clear that a) A light ray passing from rarer medium to denser medium will move towards the

normal. b) A light ray passing from denser medium to rarer medium will move away from the

normal.

These two are shown in below.

At point O the ray move towards normal and at O the ray move away from the normal.

If the wavelengths of light are 1 and 2 in the media 1 and 2 and if the speed of light is v1 and v2 respectively. Then

1 1 1

2 2 2

vv

If the light is travelling from rarer medium to denser medium then

2 1

2 1

2 1

i rv v

If the light is travelling from denser medium to rarer medium then

2 1

2 1

2 1

i rv v

Refraction at a single surface:

When the refraction, occurs at single surface, the observed length/height of an object will vary according to the formula

actual

apparentdd

Shift due to a glass slab:

When the light passes through a glass slab, there will be small shift from its original path due to refraction.

The shift is given by the formula

11shift t

Where t thickness of the glass slab.

- the refractive index of the slab.

The phenomenon is shown below.

Q9. Refractive index of glass with respect to water is 9/8. Refractive index of glass with respect to air is 3/2. Find the refractive index of water with respect to air.

Solution:

Given: 9 / 8w g and 3 / 2a g

We have 1a g g w w a .

Substituting the given data, 4 / 3a w

Q10. Find the speed of light of wavelength =780nm (in air) in a medium of refractive index 1.55.

Solution:

From the definition of refractive index, cv

cv

= 8

83 10 1.94 10 /1.5

m s

Q11. What is the wavelength of the light given in the above question, in the given medium.

Solution:

By definition.

780 5031.55

airmedium nm

Q12. An observer measures the depth of a water tank is 12m., filled with water. What is the deviation from the true depth of the tank?

Solution: The original depth and the apparent depth are related as:

actualapparent

dd

Here refractive index of the liquid (water) is 1.33

Hence the original depth of tank

1.33 12 15.96original apparent

original

d dd metres

The deviation from the true value is given by

100

12 15.96 10015.96

24.81%

measured original

original

d dd

Here the - sign indicates that there is a lesser measurement from the original.

Total Internal reflection:

Consider the refraction of light from a denser medium to rarer medium. In this case, the light moves away from the normal.

If the angle of incidence increased constantly, at a particular angle of incidence (called as critical angle), the light will pass through the boundary of the two media.

Further, if the angle of incidence increased (i.e., i>critical angle), then the light will reflect back completely to the same medium. This phenomenon is called as Total Internal Reflection.

Conditions for total internal reflection:

1. The light must travel from denser medium to rarer medium 2. The angle of incidence must be larger than the critical angle.

The critical angle for a set of two media is calculated as follow.

sin sinD Ri r

For critical angle, r=900

sin sin 90

sin

D c R

Rc

D

Hence

The applications of Total internal reflection can be observed in 1. Total reflecting prisms 2. Optical fiber.

Q13. Find the critical angle for air and glass system.

Solution:

Given media Glass: Refractive index 1.5

Air: Refractive index 1

Critical angle exists when the light travels from Denser medium (glass) to rarer medium(air)

Hence sin RcD

1sin1.5c

1sin 0.6667c

Refraction through spherical surface:

There are two types of spherical surfaces mainly, convex and concave. Both are used in many applications to obtain images of different size, shape and

nature. A lens, either a convex lens or a concave lens, has two spherical surfaces. Each of

these surfaces forms a part of a sphere. The centres of these spheres are called centres of curvature of the lens.

The centre of curvature of a lens is usually represented by the letter C. Central point of a lens is its optical centre. It is usually represented by the letter O. A ray of light through the optical centre of a lens passes without suffering any

deviation. The effective diameter of the circular outline of a spherical lens is called its aperture.

Refraction by spherical lenses: A transparent material bound by two surfaces, of which one or both surfaces are spherical, forms a lens. This means that a lens is bound by at least one spherical surface.In such lenses, the other surface would be plane. A lens may have two spherical surfaces, bulging outwards. Such a lens is called a double convex lens. It is simply called a convex lens. It is thicker at the middle as comparedto the edges. Convex lens converges light rays as shown in Fig.

Hence convex lenses are called converging lenses.

Image Formation in Lenses Using Ray Diagrams Similarly, a double concave lens is bounded by two spherical surfaces, curved inwards. It is thicker at the edges than at the middle. Such lenses diverge light rays as shown in Fig.

Such lenses are called diverging lenses. A double concave lens is simply called a concave lens. To form an image, one must follow the following principles.

A ray of light from the object, parallel to the principal axis, after refraction from a convex lens, passes through the principal focus on the other side of the lens, as shown in figure. In case of a concave lens, the ray appears to diverge from the principal focus located on the same side of the lens.

A ray of light passing through a principal focus, after refraction from a convex lens,

will emerge parallel to the principal axis. A ray of light appearing to meet at the principal focus of a concave lens, after refraction, will emerge parallel to the principal axis.

A ray of light passing through the optical centre of a lens will emerge without any

deviation.

Image formation by convex lens: The nature, size and position of the image produced by a convex lens will depend on the position of the object from the lens. The following table will give the different relations.

These are shown below.

Image formation by a concave lens: The image formation by a concave lens is shown below.

Sign convention for Spherical Lenses: The sign convention is similar to spherical mirrors concepts. Lens formula and magnification: The relation between focal length of a lens f, object distance u and image distance v is called as

lens formula. This is given by 1 1 1v u f

This formula can be used for all spherical lenses with proper sign convention. Magnification: The magnification produced by a spherical mirror is defined as the ratio of height of the image to the height of the object. =

The object and image distance can also be related with magnification given by

'h vm

h u

Q14. A concave lens has focal length of 15 cm. At what distance should the object from the lens be placed so that it forms an image at 10 cm from the lens? Also, find the magnification produced by the lens.

Solution: A concave lens always forms a virtual, erect image on the same side of the object.

Image distance v = - 10 cm Focal length f = - 15 cm Object distance u=?

Since 1 1 1v u f or

1 1 1u v f

1 1 110 15

1 1 110 15

1 3 2 130 30

u

u

u

Or u = -30 cm. Thus object should be placed at 30cm from the lens. Magnification m=v/u

Therefore 10 0.3330

m

The positive sign shows that the image is erect and virtual. The image is one-third of the size of the object. Q15. A 2.0 cm tall object is placed perpendicular to the principal axis of a convex lens of focal length 10 cm. The distance of the object from the lens is 15 cm. Find the nature, position and size of the image. Solution:

The positive sign of v shows that the image is formed at a distance of 30 cm on the other side of the optical center. The image is real and inverted.

Power of a Lens: Power of a lens gives how effectively the lens can bent the light rays. A lens with high power will bend the rays more than a lens with lesser power. Power of a lens is defined as the reciprocal of its focal length taken in meters.

= 1()

= 100( )

The SI unit of power of a lens is diopter. It is denoted by the letter D.If f is expressed in meters, then, power is expressed in diopters. Thus,1diopter is the power of a lens whose focal length is 1 metre. 1D = 1m1.

If there is a combination of lenses with powers P1, P2, P3.. then, The resultant power is given by 1 2 3 ......P P P P

The power of a convex lens is positive and that of a concave lens is negative.

Doublet lens system: The combination of a convex and concave lens is called as Doublet. This combination is used to reduce chromatic aberration in the optical instruments. The focal length of the combination is given by

1 2

1 1 1F f f

Q16. A convex lens has focal length of 25 cm. Find its power. Solution: Given f = 25cm = 0.25m. Power of the lens is given by P = 1/f

P=1/0.25=4 Diopters.

INTERFERENCE OF LIGHT

Interference: Interference is a phenomenon observed when two or more waves superimposed in a medium. It is defined as the re-distribution of the intensity in the region of superposition.

The interference of light is explained by principle of super position. According to this principle, when two or more waves are acting on a particle, the resultant amplitude of the particle will be the algebraic sum of the individual amplitudes.

If 1, 2y y are the two waves acting on a medium, the resultant amplitude is the algebraic sum of two

components. i.e,

1 2y y y for waves in same direction

and 1 2y y y for waves in opposite direction.

The same principle can be extended for n waves acting simultaneously on a particle.

We use physical optics to explain the phenomenon of Interference. Physical optics deals directly with light as a wave and takes into account phase differences

when light waves combine. (Note that by light we really mean any kind of electromagnetic radiation. Once again, however, we will mainly be concerned with visible light.)

Coherence: When two waves are having same amplitude, same frequency and zero or constant phase difference, then the two waves are said to be coherent waves and the phenomenonis said to be coherence.

Coherence is a necessary phenomenon to observe the interference. Also the sources must be monochromatic. When two waves interfere, there is a combination of bright field and dark fields. The

bright field is due to constructive interference and the dark field is due to destructive interference.

Constructive interference occurs when the path difference between the two interfering waves is either zero of n or for they must be in phase.

Destructive interference occurs when the path difference between the two interfering

waves is 2 12

n or for the waves out of phase.