MyRank: YOUNG DOUBLE SLIT EXPERIMENT

Hello MyRankers, here is the explanation of YOUNG DOUBLE SLIT EXPERIMENT

In Young’s double slit experiment, if monochromatic light is replaced by white light then central fringe will be white; all other fringes will be coloured. White light consists of colours between violet and red (VIBGYOR). Wavelength $\lambda$ is the shortest for violet light and longest for red light. At the central fringe, the path difference for all colours is zero. Hence at the central fringe, all colours superpose to give a white fringe. The first bright fringe after the central fringe will be of violet colour.
In Young’s double slit experiment, if one of the slits is covered with a transparent film or sheet of thickness t and refractive index $\mu$ , then

The path difference at the centre of the screen will not be zero, it will be equal to $\left( \mu -1 \right)t$
The entire fringe pattern will shift by an amount ${{y}_{0}}=\frac{\left( \mu -1 \right)tD}{d}$
At the centre of the screen there will be a bright fringe if $\left( \mu -1 \right)t=m\lambda ;m=1,2,3,...$
At the centre of the screen there will be a dark fringe if $\left( \mu -1 \right)t=\left( m-\frac{1}{2} \right)\lambda ;m=1,2,3,...$
The fringe width will remain the same.
The intensity of light from the covered slit will decrease due to absorption by the film or sheet. Hence intensity of bright fringes will decrease and dark fringes will have some finite intensity (because the two interfering beam do not now have equal intensity). Hence the fringe pattern will become less distinct.

If one of the slits in Young’s double slit experiment is closed (or covered with black paper), the interference pattern is replaced by single slit diffraction pattern which has a bright central fringe bordered on both sides by fringes of decreasing intensity.
If Young’s interference experiment is performed in still water rather than in air, the fringe width will decrease. Since the refractive index of water is greater than that of the air, the speed of light in water $\left( \upsilon \right)$   will be less than that in air (c).
Since the frequency of light is the same in all media ,  and
which give $\frac{{{\lambda }_{w}}}{{{\lambda }_{a}}}=\frac{\upsilon }{c}=\frac{1}{{{\mu }_{w}}}$
Now     ${{\mu }_{w}}=4/3$ .  Hence ${{\lambda }_{w}}<{{\lambda }_{a}}$ . Fringe width   $\beta \propto \lambda$ .   Hence   $\beta$ in water  < $\beta$ in air.
In Young’s interference experiment, if the beam of light has two wavelengths ${{\lambda }_{1}}$ and ${{\lambda }_{2}}$ , their maxima will coincide if , where ${{n}_{1}}$ and ${{n}_{2}}$ are integers.
In an interference experiment if the two coherent light sources have intensities in the ratio $n:1$ , i.e, ,then the ratio of the intensity of maxima and minima in the interference pattern is
$\frac{{{I}_{\max }}}{{{I}_{\min }}}=\frac{{{I}_{1}}+{{I}_{2}}+2\sqrt{{{I}_{1}}{{I}_{2}}}}{{{I}_{1}}+{{I}_{2}}-2\sqrt{{{I}_{1}}{{I}_{2}}}}$
$=\frac{{{\left( \sqrt{{{I}_{1}}}+\sqrt{{{I}_{2}}} \right)}^{2}}}{{{\left( \sqrt{{{I}_{1}}}-\sqrt{{{I}_{2}}} \right)}^{2}}}={{\left[ \frac{\sqrt{\frac{{{I}_{1}}}{{{I}_{2}}}}+1}{\sqrt{\frac{{{I}_{1}}}{{{I}_{2}}}}-1} \right]}^{2}}={{\left( \frac{\sqrt{n}+1}{\sqrt{n}-1} \right)}^{2}}$
In an interference experiment with two coherent light sources, if the ratio of the intensities of maxima and minima in the interference pattern is $n:1$ , i.e.
$\frac{{{I}_{\max }}}{{{I}_{\min }}}=n$
then the ratio of the intensities of the coherent sources is

$\frac{{{I}_{1}}}{{{I}_{2}}}={{\left( \frac{\sqrt{n}+1}{\sqrt{n}-1} \right)}^{2}}$
because

$\frac{{{I}_{\max }}}{{{L}_{\min }}}={{\left[ \frac{\sqrt{\frac{{{I}_{1}}}{{{I}_{2}}}}+1}{\sqrt{\frac{{{I}_{1}}}{{{I}_{2}}}}-1} \right]}^{2}}$

$\Rightarrow \sqrt{n}=\frac{\sqrt{\frac{{{I}_{1}}}{{{I}_{2}}}}+1}{\sqrt{\frac{{{I}_{1}}}{{{I}_{2}}}}-1}\Rightarrow \frac{{{I}_{1}}}{{{I}_{2}}}={{\left( \frac{\sqrt{n}+1}{\sqrt{n}-1} \right)}^{2}}$

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Tuesday, April 14, 2015

YOUNG DOUBLE SLIT EXPERIMENT

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