Notes for 12th Physics - Interference and Diffraction by ednexa.com

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Diffraction of Light
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1. What is diffraction? Explain diffraction due to single
slit.
Ans:
Diffraction : Light travels in straight line direction.
When an opaque obstacle is placed in front of light
source, the shadow of obstacle is formed. It is
observed that, the light rays which incident on the
edges of the obstacle, these light rays cannot travel
in straight line. These light rays are deviates.
Encroachment of light in the geometric shadow of
obstacle, takes place. This phenomenon is called
as diffraction of light.
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i.
observed by Garibaldi in 1665.
ii.
studies by Newton and Hooks.
iii.
Thomas Young was the first scientist who
applied a wave theory to this phenomenon.
“The phenomenon of bending of light around
the
corners
of
opaque
obstacle
and
encroachment of light in the geometries
shadow of opaque obstacle is called as
diffraction of light.”
ii. Diffraction of light is possible only when the size
of obstacle comparable to the wave length of
light.
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Diffraction
due
to
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single
slit
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2. Formation of first minima & first maxima on the
basis of Huygen’s wave theory.
According to Huygen’s theory, every point, on the
slit acts as secondary source of light. All the
secondary waves emitted, from all points O. These
all waves are in same phase, when reaches at
point O. Due to this maxima is formed at centre of
the screen as show in diagram.
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Formation of first minima :
i.
According to Huygen’s wave theory, secondary
waves are travel in all possible direction. When
secondary waves makes an angle with the
direction of incident beam then few waves
reaches at point N, starting from point A and
point C. These waves are not in same phase.
There is a finite phase difference between
these waves starting from point A and C when
reaches at point NI as shown in the diagram. If
this phase difference is A/2 then point N, is a
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dark point, i.e. a dark band is formed.
Secondary waves are also emitting from point B
& C produces a dark point at N1.
ii. The path difference between the wave emitting
from B & C at point N1, is also
2
From the diagram, path difference between
point A & B is AD = AB sinθ = λ , because path
difference between AC
2
2
is and between CB is
also AB sin θ = λ i.e. d sin θ = λ (AB = d)
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In the same way dark bands are produced at
points N2, N3, .............
iii. For first dark band we have the relation
d sin θ = λ .
For Second dark band dsin θ = 2 λ. Similarly
for nth dark band dsin λ = n θ. Values of θ is
different for different dark bands.
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Formation of first maxima
i.
To explain the formation of first maxima (or first
secondary maxima), let us divide the slit in to
their equal sections as shown in diagram.
For specific angle, the secondary waves
emitting form two outer section cancels the
effect of each other i.e. wave emitting from
outer sections are in out of phase with each
other. The inner section produces a maxima at
M1 with less intensity.
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ii. In the same way we can explain the formation
of second, third...... maxima by dividing the slit
in to 5,7,9.... equal parts. From the diagram,
path difference for first maxima is AD = 3 λ / 2
i.e. path different between the end points of
each section is λ / 2 , There are three sections
therefore the path difference between point A
and B is 3 λ / 2 . From the diagram, path
different AD = d sin θ
d sin θ = 3 λ /2
This is the relation for first maxima.
For second maxima, path difference is
5 λ /2
d sin θ = 5 λ /2
For nth maxima, path difference is,
dsin θ = ( 2n + 1) λ / 2 where n = 1, 2, 3 .........
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Rayleijh’s criteria for resolution of image.
i.
“Separation of image of two objects, which
are very close to each other, using an
instrument is called as resolution of image.”
ii. When the image of point objects are formed
using optical instruments, these images are not
clear because of diffraction of light. Diffraction
pattern of two images which are separated by
limit distance is as show in diagram (a)
If the two objects are very close to each other,
diffraction pattern produced by two images is
overlapped and we cannot distinguish two
separate images as shown in diagram (b)
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iii. If such closed objects are seen through optical
instruments then these images are resolved.
The diffraction pattern after resolution is as
show in diagram (c)
These two images are appears to be a single
object.
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In this diffraction pattern the central maxima
due to first object i-overlapped on the first
minima due to second object and vice versa.
Due i this there are two peaks with small dip in
the middle. In this case we can say that the
objects are just resolved.
This is the limiting separation. According to
Lord Ray light if is called as limit of resolution of
two images.
“Two point sources of equal intensity are said to
be just resolves by an optical instrument if the
central maxima of diffraction pattern of one falls
on the first minima of the diffraction pattern due
to other.” OR
“The smallest separation between the two point
object at which they appear just separated is
called as limit of resolution of optical instrument
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and the reciprocal of limit of resolution is called
as resolving power of instrument.”
4. Explain
resolving
power
of
microscope
and
telescope.
Ans:
Microscope is an optical instrument used to
magnify the object which is close to the instrument.
Telescope is an optical instrument used to magnify
the objects which are far away from the installment.
Resolving power of microscope :
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i.
“Resolving power of microscope is defined as it
is the reciprocal of least separation between
two close objects, so that they appears just
separate, when observed through microscope.”
iii. According to Abbe the least distance of
separation so that the object gets separated
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using microscope is given as, According to
definition of resolving power of microscope,
Resolving power =
1
2
sin
d
From this relation it is clear that, to get
maximum resolution
1. Refractive index p-must be large.
2. To increase the valve of q, aperture of
objective should be large.
In the above relation the term 2μ sin θ is called
as numerical aperture microscope.
Resolving power of telescope
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i.
“Resolving power of telescope is defined as the
reciprocal
of
smallest
angular
separation
between two distant object, so that they
appears
ii. According to Airy, the angular separation is
given by instrument but the resolving power of
instrument
is
depends
up
of
diffraction
phenomenon.
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