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Fresnel Biprism
Augustin-Jean Fresnel was a French
physicist who contributed significantly to
the establishment of the wave theory of
light and optics.
He gave a simple
arrangement
for
the production of
interference
pattern.
Prism
A prism is a wedge-shaped transparent body
which causes incident light to be separated by
color. The separation by color occurs since
different colors corresponding to different
wavelengths.
It is a device used to refract light, reflect it or
break it up (to disperse it) into its constituent
spectral colours.
C
B
D
E
A
  (   1)
Biprism
It consists of two thin acute angled prisms
joined at the bases. It is constructed as a
single prism of obtuse angle of 179º. The
acute angle  on both side is about 30´. A
portion of the incident light is refracted
downward and a portion upward.
179º


a
179º


b
c
S
c
a
b
The prism is placed with the refracting edge parallel to the line
Source S such that Sa is normal to the face bc of prism.
c
A
E
Fringes of equal width
S
C
d
a
B
Z1
Z2
b
D
F
Fringes of large width
When light incident from S on the lower portion of prism
It bents upwards and appears to come from virtual source
B. Similarly light from S incident on the upper portion of
Prism bents downwards appear to come from A.
So A and B are two virtual coherent sources.
AB = d
C is the screen
Distance between source and eyepiece = D
Interference fringes of equal width will be occur between EF
portion of the screen. Beyond EF portion fringes of large
width will be produced.
Since C is equidistant from A and B so at C maximum fringe
intensity will occur. On both sides of C alternate bright and
dark fringes will appear.
According to the previous theory
the fringe width
 
D
d
n D
So position of bright fringes from C =
d
( 2n  1)D
Position of dark fringes from C =
2d
n = 0,1,2,3….
So the wavelength of light will be

d
D
Determination of the distance between the two sources (d)
M
L1
A convex lens (L1) is placed between the prism and eyepiece (M), such
that the image of the virtual sources A and B are seen in the field of
view of the eyepiece.
Suppose the distance between the images of A and B as
seen by the eyepiece is d1.
So ,
d1 v n
 
d u m
……..(1)
Eyepiece is moved horizontally to determine the fringe width.
Suppose for crossing 20 bright fringes from the field of view, the
Eyepiece has moved through a distance l.
So the fringe width be,
l

20
Now move the lens towards eyepiece and bring it to other positon L2
So that again images of A and B are seen clearly in the the field of view
Of eyepiece. Again if the distance between the two images be d2
L2
d2
v
m


d
u
n
……..(2)
Multiplying (1) and (2) we get
d1 d 2

1
2
d
 d  d1 d 2
Substituting the values of , d, D we calculate the
value of wavelength () of given monochromatic light.
Fringes with white light
When white light is used the center fringe at C is white since
all waves will constructively interfere here while the fringes
on the both side of C are colored because the fringe
width () depends on wavelength of light.
The fringe pattern in fresnel’s biprism is totally different
From that of fresnel’s mirrors.
In biprism it depends on refraction
In mirror it depends on reflection
For white light the two coherent virtual sources are
produced by Refraction and the distance between the
two sources depends upon the refractive index which
intern depends upon the wavelength.
So, for green light the distance between the two
sources is different to that with red light.
The distance of the nth bright fringe from the centre
with monochromatic light
nD
y
d
Where
d  2(   1)z1
nD
y 
2(   1)z1
For green light,
yg 
ng D
2(  g  1)z1
For red light,
nr D
yr 
2( r  1)z1