RAY OPTICSMODULE 2

ATOMIC ENERGY EDUCATION SOCIETY

MUMBAI

RAY OPTICS

REFLECTION REFRACTION OPTICAL INSTRUMENTS

Spherical mirrors

Mirror formula

Total internal reflection and its

applications, Optical fibres

Refraction at spherical surfaces

Lenses

Magnification, Power of a lensLens maker's formula Combination of thin lenses in contact

Refraction of light through a prism

SCATTERING OF LIGHT

Microscopes Astronomical telescopes

MODULE 1 MODULE 2 MODULE 3

REFRACTION

When a beam of light travels from one transparent medium to another transparent medium, a part of light gets reflected back

into the first medium while the rest enters the other. The direction of propagation of an obliquely incident ray of light that enters

the other medium, changes at the interface of the two media. This phenomenon is called refraction of light.

The 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.

r

in

sin

sin21

𝑛21 =𝑛2𝑛1

=sin𝑖

sin𝑟=

𝑣1𝑣2

=𝑓𝑟𝑒𝑞𝑢𝑒𝑛𝑐𝑦 × 𝜆1𝑓𝑟𝑒𝑞𝑢𝑒𝑛𝑐𝑦 × 𝜆2

=𝜆1𝜆2

𝑛21 =𝑛2𝑛1

=𝜆1𝜆2

𝑛 𝛼1

𝜆

𝑛21 = relative refractive index

𝑛1 and 𝑛2 are called absolute refractive index

Principle of reversibilityAccording to principle of reversibility of light, if the path of the light is reversed after suffering a number of reflections and

refractions, then it retraces its path. This means that if a light ray travels from medium 1 to medium 2 and has angle of incidence

and angle of refraction as i and r respectively, then if the light is incident from medium 2 at an angle r, then the angle of

refraction in medium 1 will be i.

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 𝑛21 =1

𝑛12

is true according to the principle of reversibility 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

For a rectangular slab, refraction takes place at two interfaces (air-glass and glass-air). It is evident from Fig. that r2 = i1.

The angle of deviation produced by a glass slab is 0∘

( incident ray is parallel to emergent ray).The emergent ray does suffer

lateral displacement/shift with respect to the incident ray.

The formula for lateral shift(y) is given by where t is the thickness of the glass slab.

In daily life we may experience the bottom of a tank or an object at the bottom of a tank filled with water appears to be raised. The

refractive index of the medium of water is the ratio of real depth (h2) to the apparent depth, (h1)

𝑦 =𝑡. sin 𝑖1 − 𝑟1

cos𝑟1

The refraction of light through the atmosphere is responsible advance sun rise and delayed sunset. The sun is visible

a little before the actual sunrise and until a little after the actual sunset due to refraction of light through the

atmosphere (Fig). By actual sunrise means the actual crossing of the horizon by the sun. The actual and apparent

positions of the sun with respect to the horizon shown in the figure. There is a variation in the refractive index of air

with respect to vacuum. The refractive index of air is 1.00029 and the refractive index of vacuum is 1.Sun rays bend

due to this variation as they enter the earth’s atmosphere Due to this, there is apparent shift in the direction of the

sun is by about half a degree and the apparent sunrise is advanced about 2 minutes and the apparent sunset is also

delayed about 2 minutes. The apparent flattening of the sun like oval shape at sunset and sunrise is also due to the

same phenomenon.

Atmospheric Refraction

Advance sunrise and delayed sunset due to atmospheric refraction

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, As the angle of incidence increases, the angle of

refraction also increases. It increases till for the ray AO3, upto an angle ic called critical angle. At this critical angle , the angle of refraction is 900.

Critical angle for the given pair of media is defined as the angle of incidence corresponding to an angle of refraction 900.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 in Fig. If the angle of

incidence is increased beyond ic (e.g., the ray AO4), refraction in the rarer medium is not possible, and the incident ray is totally reflected in the same

denser medium. This is called total internal reflection. When light gets reflected by a surface, normally some fraction of it gets transmitted. The reflected

ray is always less intense than the incident ray, howsoever smooth the reflecting surface may be. In total internal reflection no transmission of light takes

place. From Snell’s law , we shall write , when i= ic and r= 900 ,

For values of i larger than ic , Snell’s law of refraction cannot be satisfied, and hence no refraction is possible

𝑛12 =sin𝑖𝑐sin𝑟

=sin𝑖𝑐sin90

= sin𝑖𝑐 =1

𝑛21

Total internal reflection in nature and its technological applications

Mirage: The refractive index of air increases with its density.

The density of hotter air is less than that of cooler air and hence the refractive index of hotter air is less than that of cooler air. On hot summer days, the

air near the ground becomes hotter than the air at higher levels.Due to this the refractive index of air decreases towards the ground ( if the air is still).

When Light from a tall object (tree), passes through such medium whose refractive index decreases towards the ground and hence a ray of light from

the tree successively bends away from the normal.

If the angle of incidence for the air near the ground exceeds the critical angle, the ray of light undergoes total internal reflection. To a distant observer,

the light appears to be coming from below the ground and assumes that light is being reflected from water body near the tall object. Such inverted

images of distant tall objects cause an optical illusion to the observer. This phenomenon is called mirage. This type of mirage is common in hot deserts.

We might have noticed that while moving in a vehicle during a hot summer day, a distant patch of road on a highway, appears to be having water on the

surface. But when we reach the place we do not find any water.

(ii) Diamond: The sparkling brilliance of diamond is mainly due to the total internal reflection of light inside them. The critical angle

for diamond-air interface( 24.4°) is very small. The light entering a diamond undergoes total internal reflection inside it.

Diamonds found in nature rarely exhibit the brilliance for which they are known.. By cutting the diamond suitably with some

technical skill, multiple total internal reflections can be made to occur and hence it sparkles brilliantly.

(iii) Prism: Prisms can be designed to bend light by 90º or by 180º using of total internal reflection principle. This concept is also

used to invert images without changing their size. In the first two cases, the critical angle ic for the material of the prism must be

less than 45º.

(iv) Optical fibersPrinciple: Optical fibers too make use of the phenomenon of total internal reflection.

Construction: Optical fibers are fabricated with high quality composite glass/quartz fibers having a radius of approximately 10-6 m. A

bundle of thin optical fibres forms an optical pipe or light pipe. Each fiber consists of a core and cladding. The refractive index of the

material of the core is higher than that of the cladding.

Working: Optical fibers are fabricated such that total internal reflection occurs along the length of the fiber when a light signal incident

at angle of incidence is greater than the critical. Since light undergoes total internal reflection at each stage, there is no appreciable

loss in the intensity of the light signal.. Even if the fiber is bent, light can easily travel along its length.

Uses: Optical fibers can also be used for transmission of optical signals.

They are used for transmitting audio and video signals through long distances.These signals can be converted to light by suitable

transducers.

They are used as a ‘light pipe’ to facilitate visual examination of patients internal organs like esophagus, stomach and intestines

Light undergoes successive total internal reflections as it moves through an optical fibre.

Refraction at a spherical surface

Let us consider a spherical surface with center of curvature C, and radius of

curvature R. Let O be the point object placed on the principal axis and I be the

image formed inside the spherical surface on the principal axis Assume that the

rays are incident from a medium of refractive index n1, to another of refractive

index n2.Consider a ray ON meeting the spherical surface.Draw the normal at N

and let us take i and r be angle of incidence and angle of reflection respectively.

tan NOM = ; tan NCM = ; tan NIM =OM

MN

MC

MN

MI

MN

Now, for NOC, i is the exterior angle. =, i = NOM + NCM= +OM

MN

MC

MN

, r = NCM – NIM = -MC

MN

MI

MN

Now, by Snell’s law n1 sin i = n2 sin r and for small angles n1i = n2r = n1 ( + )= n2( - ) OM

MN

MC

MN

MC

MN

MI

MN

MC

nn

MI

n

OM

n 1221 Applying the Cartesian sign convention,OM = –u, MI = +v, MC =+R

R

nn

u

n

v

n 1212 We will get,

From the geometry of the figure shown we shall write

Similarly, for INC, NCM is the exterior angle =

Refraction by a lens-Lens maker’s formula

(a) The position of object, and the imageformed by a double convex lens

b) Refraction at the first spherical surface (c) Refraction at the second spherical surface

(i) The first refracting surface forms the image I1 of the object O [Fig.(b)]

R

nn

u

n

v

n 1212

At the first interface ABC, putting u=u and v=v1 ,

we get from eqn 1 that

2 1 2 1

1 1

n n n n

BI OB BC

At the second interface* ADC ,

since the ray travels from denser to

rarer medium , interchanging n1 and n2

And putting u=v1 and v=v in eqn 1

,

1 2 1 2

1 2

n n n n

DI DI DC

For a thin lens,

We shall use BI1 = DI1

Adding Eqn 2 and Eqn 3, we get ---

…..Eqn 2

……Eqn 3

1 12 1

1 2

1 1( )

n nn n

v u R R

Eqn 4

(ii)The image I1 acts as a virtual object for the second surface that forms the image at I [Fig(c)

Eqn 1

2 1 2 1

1 1

n n n n

v u R

1 2 1 2

1 2

n n n n

v v R

=

=

Suppose the object is at infinity, then u= and v = f, the above equation will become

21

1 2

1 1 1( 1)n

f R R

Is called lens maker’s formula

21 1

1 1 2

1 1 1 11

nn n

v u n R R

21

1 2

1 1 1 1( 1)n

v u R R

Taking n1 outsideOn both LHS and RHS

gives Where 𝑛21 =𝑛2𝑛1

Comparing the LHS of eqn 5 and eqn 6,we get

…….Eqn 5

…..Eqn 6

1 1 1

f v u

Is called thin lens formula

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.

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 falling at unit distant from the optical center

1tan

hP

f f For h= 1 m and for small angles tan=

fP

1

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.

𝜃

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. The first lens produces an image at I1. Image I1 is real, it serves as a virtual

object for the second lens B, producing the final image at I. Since the lenses are thin, we assume the optical centers of

the lenses to be coincident at a central point P

For the image formed by the first lens A, we get

11

111

fuv

For the image formed by the second lens B, we get

………Eqn 1

21

111

fvv

………..Eqn 2

Adding Eqn 1 and eqn 2

1 2

1 1 1 1

v u f f

Also we shall write effective power as P=P1+P2

and effective magnification m=m1xm2

1 1 1

f v u

Using thin lens formula

1 2

1 1 1

f f f Equivalent focal length

of combination is thus

Bibliography

I Acknowledge that

The content and diagrams are taken from NCERT SYLLABUS 2020-21 AND NCERT TEXT BOOK

CLASS XII- PHYSICS

THANK YOU

Dr.A.JEBIN JOEL

PGT(SS)

AECS 4,MUMBAI