Young's Double Slit Experiment
Ram thrown a stone into water.
Then he saw water is moving in the form of waves.
Similarly, The light also shows the wave nature.
The wave nature of light was proved with the help of Young's Double Slit Experiment by using phenomenon of interference.
Let's study the Young's Double Slit Experiment.
Young took an ordinary light source
$S$
such as a light bulb.
The light from bulb made to pass through a very small slit
$S_{0}$
.
Now, the light is coming from
$S_{0}$
was made to pass through two small slits
$S_{1}$
and
$S_{2}$
.
$S_{1}$
and
$S_{2}$
were separated by a very small distance.
The source
$S_{0}$
illuminated the sources
$S_{1}$
and
$S_{2}$
, as a result, the light from
$S_{1}$
and
$S_{2}$
become coherent.
One screen was kept in front of these two sources.
Then he observed the alternative light and dark bands were formed on the screen.
Calculation of path difference.
Let
$S_{1}$
and
$S_{2}$
be two fine slits at small distance
$d$
apart.
Let them be illuminated by monochromatic light of wavelength l.
The screen
$MN$
is at a distance D from the slits
$S_{1}$
and
$S_{2}$
.
The waves from
$S_{1}$
and
$S_{2}$
superimpose upon each other and an interference pattern is obtained on the screen.
The point C is equidistant from
$S_{1}$
and
$S_{2}$
, therefore the path difference between the waves will be zero.
So, point C is of maximum intensity. It is called the central maximum.
For another point P at a distance 'x' from C, the path difference at P.
Now
$S_{1}S_{2}$
= EF = d,
$S_{1}$
E =
$S_{2}$
F = D
Then, applying Pythagorous theorem in right angle
$△S_{2}PF$
.
Similarly in
$△S_{1}PE$
.
Hence the path difference is.
On expanding binomially, the path difference will be.
For bright fringes, the path difference is an integral multiple of the wavelength
$λ$
.
Here
$X$
represents the distance of nth bright fringe from point
$C$
.
Similarly for dark fringes, the path difference
$X$
is.
Further, the separation
$β$
between the centers of two consecutive bright fringe is the width of a dark fringe and vice versa.
Therefore, the intensity of both light and dark fringes are the same.
Revision.
The formula of path difference is.
Path difference
$X$
of the light fringes.
This is the formula for the path difference
$X$
of dark fringes.
This is the separation between the two fringes.
The end.