Chapter 14 SNELLS LAW According to Snells law The ratio of the sine of the angle of incidence to the sine of the angle of refraction is always constant Mathematically ID: 654170
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Slide1
Refraction Of Light & Optical Instruments
Chapter 14Slide2
SNELL’S LAW
According to Snell’s law
"The ratio of the sine of the angle of incidence to the sine of the angle of refraction is always constant. "
Mathematically,
Sine <
i
/sine <r = constant
or
sin<
i
/sine< r =
mew
where
mew
= Refractive index of the material of medium.Slide3
TOTAL INTERNAL REFLECTION
When light rays enter from one medium to the other, they are refracted. If we increase the angle of incidence, angle of refraction will also increase. At certain angle of incidence light rays are reflected back to the first medium instead of refraction. This condition or phenomenon is called Total Internal Reflection.Slide4
CRITICAL ANGLE
The angle of incidence at which the angle of refraction will become 90
o
is called Critical Angle. If angle of incidence further increased then instead of refraction, reflection will take place.Slide5
DEFECTS OF VISION
There are four common defects of vision:
1.
SHORT SIGHTEDNESS
OR MYOPIA
2.
LONG SIGHTEDNESS
OR HYPER METROPIA
3.
ASTIGMATISM
4.
PRESBYOPIASlide6
SHORT SIGHTEDNESS OR MYOPIA
SYMPTOMS
In Myopia, a person can not see distant objects clearly, but he can see clearly the objects near to him.
REASON
The reason for Myopia is either the focal length of lens of eye is too short or the eyeball is very much elongated.
WHAT HAPPENS IN
MYOPIA
In Myopia, light rays from a distant object are focused in front of the Retina.Slide7
CORRECTION OF DEFECT
This defect can be corrected by using a concave lens of suitable focal lengthSlide8
ASTIGMATISM - PRESBYOPIA
ASTIGMATISM
If the cornea or the surface of eye is not perfectly spherical. In this situation the eye has different focal points in different planes and an object is not focused clearly on the retina.
CORRECTION OF
DEFECT
ASTIGMATISM
is corrected by using asymmetrical lenses which have different radii of curvature in different planes
PRESBYOPIA
or
lack of
accommodating
At old age, the eye lens loses its natural elasticity and ability to change its shape and the
ciliary
muscles weaken resulting in a lack of accommodation. This type of long sightedness is called "
PRESBYOPIA
".
CORRECTION OF
DEFECT
This defect can be corrected by using convex lens for long sighted person and concave lens for short sighted person
.Slide9
LONG SIGHTEDNESS
OR HYPER
METROPIA
SYMPTOMS
In
HYPER METROPIA
, a person can not see objects clearly which are near to him, but he can see clearly distant objects
REASON
The reason for
HYPER METROPIA
is that either the focal length of the lens of eye is too long or the eyeball is too short.
WHAT HAPPENS
IN HYPER
METROPIA
In
HYPER METROPIA
, light rays from a near object are focused behind the Retina.Slide10
CORRECTION OF DEFECT
This defect can be corrected by using a convex lens of suitable focal lengthSlide11
POWER OF LENS
Power of lens is defined as the reciprocal of the focal length of the lens in meters.
FORMULA:
Power = 1/f
(in meter)
Unit of power of lens is
Dioptre
.
DIOPTRE
Dioptre is defined as the power of lens whose focal length is one meter if f =1 meter then the power of the lens = 1
dioptre
.Slide12
Image Formation by convex lens
POSITION OF OBJECT
When the object is placed at infinity
POSITION OF OBJECT
When the object is placed at infinity
NATURE AND POSITION OF IMAGE
1.
The image will form at the principal focus (F).
2.
The image will be real and inverted.
3.
The image will be very small in size.Slide13
POSITION
OF OBJECT
When the object is placed beyond 2F
Nature and position of image
1
.
The image will form between F and 2F.
2.
The image will be real and inverted.
3.
The image will be smaller in size.Slide14
POSITION OF OBJECT
When the object is placed at 2F
NATURE AND POSITION OF IMAGE
1.
The image will form at 2F.
2.
The image will be real and inverted.
3.
The image will be equal in the size of object.Slide15
POSITION OF OBJECT
When the object is placed between F and 2F
NATURE AND POSITION OF IMAGE
1.
The image will form beyond 2F.
2.
The image will be real and inverted.
3.
The image will be magnified.Slide16
POSITION OF OBJECT
When the object is placed at F
NATURE AND POSITION OF IMAGE
1.
The image will form at infinity.
2.
The image will be real and inverted.
3.
The image will be highly magnified.Slide17
POSITION OF OBJECT
When the object is placed between the pole (P) and
F
NATURE AND POSITION OF IMAGE
1.
The image will form on the same side of object.
2.
The image will be virtual and erect.
3.
The image will be magnified.Slide18
ASTRONOMICAL TELESCOPE
Introduction
It is an optical instrument used to view heavenly bodies such as
moon,stars
, planets and distant object.
Construction
Astronomical telescope consists of two convex lenses:
1:
Objective
2:Eye
pieceObjective
The objective is a convex lens of large focal length and large aperture. It usually made of two convex lenses in contact with each other to reduce the chromatic and spherical aberrations.Slide19
Eye piece
The eye piece is also a convex lens .Its focal length is smaller than that of objective. It is also a combination of two lenses.
The objective is mounted on a wide metallic tube while the eye piece is mounted on a small tube .The distance b/w the eye piece and the objective can be changed by moving tubes.Slide20
WORKING
The rays coming from a distant object falls on objective as parallel beam at some angle say "
a
" and these rays after refraction and passing through the objective converge at its focus and make an inverted & real image
AB. This
image acts as an object for the eye piece. The distance of the eye piece is so adjusted that the image
AB
lies within the focal length of the eye piece. The eye piece forms the final image .The final image is magnified ,virtual and inverted with respect to object. The final image is formed at infinity.Slide21
WORKINGSlide22
MAGNIFYING POWER
The magnifying power (M) of astronomical telescope is given by:
It is because the object is at infinite distance and hence the angle subtended by the object at eye may be taken as the angle subtended by the object at objective.
M =
b/a ............(1)
since
a
and
b
are small angles, therefore we can take:
a
= tan
a
...................
and
.......
..............
b
= tan
b
.............Slide23
MAGNIFYING POWER
In right angled triangles
DAOB
&
DAEB
This expression shows that in order to obtain high magnification, focal length of object must be large and that of eye piece is small.Slide24
LENGTH OF TELESCOPE
The distance b/w objective lens and the eye piece is equal to the length of the telescope.
From figure
:
OE = length of telescope =L
But
OE
= OB + BE
OB
= Fo & BE = Fe OR
L = focal length of objective + focal length of eye pieceSlide25
THIN LENS FORMULA FOR CONCAVE LENS
FOR CONCAVE
LENS
Consider an object placed in front of a concave lens of focal length "
f
" on the principle axis of the lens. Concave lens forms a virtual and erect image at a distance of "
q
" from the optical centre of the lens as shown in the diagram below.Slide26
FORMULA FOR CONCAVE LENS
Consider similar triangles
and
Similarly
in
triangles
and
Slide27
FORMULA FOR CONCAVE LENS
Comparing equation (1) and (2)
p (f - q) =
fq
pf
-
pq
=
fq
Dividing both sides by
"
pqf
"
1/f - 1/p = 1/qSlide28
COMPOUND MICROSCOPE
Compound microscope is an optical instrument which is used to obtain high magnification.
Construction
It consists of two converging lenses:
Objective
Eye piece
Objective
The lens in front of object is called
objective
. Its focal length f
1
=
f
o
is taken to be very small .The objective forms a real, inverted, and magnified image of the object placed just beyond the focus of objective.
Eye piece
The lens towards the observer's eye is called piece .Focal length of eye piece is greater than the focal length of objective. Eye piece works as a magnifying glass.Slide29
Working
The objective is so adjusted that the object is very closed to its focus. The objective forms a real,
inverted and magnified image of the abject beyond 2fo on the right hand side. The eye piece is so adjusted that it forms a virtual image at the least distance of distinct vision "
d
" .The final image is
highly magnified.Slide30
Magnifying power
In order to determine the magnifying power of a compound microscope ,we consider an object
oo
' placed
in front of objective at a distance p
1
. Objective forms an inverted image
II'
at a distance of q
1
from objective.
Magnification produced by :Mo
= size of image / size of object
M
o
= q
1
/ p
1
--------------- (1)
Eye piece works as a magnifying glass. It further magnifies the first image formed by objective.
Magnification produced by the eye piece is given by:
M
e
= size of image / size of object
M
e
= q
2
/ p
2
the
objective is given by:Slide31
Magnifying power
We know that the eye piece behaves as a magnifying glass therefore the final image will be formed at least distance of distinct vision
i.e
at 25 cm from the eye. Hence q
2
= d
M
e
= d / p
2
--------------- (2) Using thin lens formula for eye piece :Slide32
Magnifying power
1/f
2
= 1/q
2
+ 1/p
2
Here f
2
=
f
e, q2 = - d and p = p21/
f
e
= 1/-d + 1/p
2
1/
f
e
= -1/d + 1/p2
Multiplying both sides by "d"
d/
f
e
= -d/d + d/p
2
d/
f
e
= -1 + d/p
2
1 + d/
f
e
= d/p
2
d/p
2
= 1 + d/
f
e
----------------(3)
Comparing equation (2) and (3)
M
e
=
1 + d/
f
e
--------(4)Slide33
Magnifying Power
Total magnification is equal to the product of the magnification produced by the objective and the eye piece.
M =M
o
X
M
e
M = (q
1
/p
1)(1 + d/fe)
In order to get maximum magnification, we must decrease p
1
and increase q
1
.Thus maximum possible
value of p
1
is fo
i.e
p =
fo
and maximum possible value of q
1
is the length of microscope
i.e
q
1
= L
Therefore the magnification produced by a compound d microscope is given by:
M = (L/
fO
)(1 + d/
f
e
)