Question

(a) In Young’s double slit experiment, describe briefly how bright and dark fringes are obtained on the screen kept in front of a double slit. Hence obtain the expression for the fringe width.
(b) The ratio of the intensities at minima to the maxima in the Young’s double slit experiment is 9:25. Find the ratio of the widths of the slits.

Answer 1

Hint:The resultant intensity of the wave at a point depends on the path difference between the two waves. Due to this we get a pattern of bright and dark fringes. With help of geometry, find an expression for the path difference in terms of the position of the point. Then find the expression for fringe width. Use the formulas ${{I}_{\max }}={{\left( \sqrt{{{I}_{1}}}+\sqrt{{{I}_{2}}} \right)}^{2}}$${{I}_{\min }}={{\left( \sqrt{{{I}_{1}}}-\sqrt{{{I}_{2}}} \right)}^{2}}$
Formula used:
${{I}_{\max }}={{\left( \sqrt{{{I}_{1}}}+\sqrt{{{I}_{2}}} \right)}^{2}}$
${{I}_{\min }}={{\left( \sqrt{{{I}_{1}}}-\sqrt{{{I}_{2}}} \right)}^{2}}$

Complete answer:
(a) In Young’s double slit experiment is experiment light waves from a monochromatic light source are illuminated from two slits, which are separated by a distance d. A screen is placed parallel to the two slits at a distance D such that d<<<Due to interference of the two waves at a point on the screen a resultant wave is formed of new intensity. The intensity of the light depends on the path difference between the two waves. Therefore, we get a pattern of consecutive bright and dark fringes on the screen. Here, bright fringe is the point with maximum intensity and dark fringe is point with minimum intensity.

Now, suppose the two waves superimpose at point P that is at a distance y from O. Let this line joining P and S make an angle $\theta $.
Since D>>>>d, $\angle P{{S}_{1}}E=\angle PSO=\angle P{{S}_{2}}F=\theta $.
Considering P as the centre of a circle and ${{S}_{1}}P$ as the radius of the circle, draw an arc from point ${{S}_{1}}$. Let the arc intersect ${{S}_{2}}P$. Let the point of intersection be A. Since ${{S}_{1}}$A is of a very small length, it is almost a perpendicular line segment to ${{S}_{2}}P$ (as shown).
Here, ${{S}_{1}}P$ and ${{S}_{2}}A$ are radii of the circle. Therefore, ${{S}_{1}}P$= AP.
Path difference between the two light waves is $\Delta x={{S}_{2}}P-{{S}_{1}}P$ …. (i).
We know ${{S}_{2}}P={{S}_{2}}A+AP$ and ${{S}_{1}}P$= AP.
Substitute the value of ${{S}_{2}}P$ and ${{S}_{1}}P$ in equation (i).
Therefore,
$\Delta x={{S}_{2}}A+AP-AP={{S}_{2}}A$.
Consider the triangle ${{S}_{1}}{{S}_{2}}A$. It is right angled at A.
Hence, use the sine ratio.
Therefore, $\sin \theta =\dfrac{{{S}_{2}}A}{{{S}_{1}}{{S}_{2}}}=\dfrac{{{S}_{2}}A}{d}$
$\Rightarrow {{S}_{2}}A=d\sin \theta $
Consider the triangle PSO.
$\Delta $PSO is a right angled triangle.
Hence, $\tan \theta =\dfrac{y}{D}$.
For a small angle, $\sin \theta \approx \tan \theta $.
Therefore,
$\Rightarrow {{S}_{2}}A=d\tan \theta $.
$\Rightarrow {{S}_{2}}A=d.\dfrac{y}{D}$
Therefore, $\Delta x={{S}_{2}}A=\dfrac{yd}{D}$
$\Rightarrow y=\dfrac{D}{d}\Delta x$
For the nth bright fringe $\Delta x=n\lambda $, where $\lambda $ is the wavelength of light.
And for the nth dark fringe $\Delta x=(2n-1)\dfrac{\lambda }{2}$.
The distance between the two consecutive bright or dark fringes is called fringe width ($\beta $).
 Let ${{y}_{n-1}}$ and ${{y}_{n}}$ be the distances of two consecutive bright fringes.
Therefore,
  ${{y}_{n-1}}-{{y}_{n}}=\dfrac{D}{d}\Delta {{x}_{1}}-\dfrac{D}{d}\Delta {{x}_{2}}$
$\Rightarrow \beta =\dfrac{D}{d}\left( n\lambda -(n-1)\lambda \right)$
$\Rightarrow \beta =\dfrac{\lambda D}{d}$

(b) The maximum and minimum intensities due to the interference depend on the intensities of the light waves from the two slits (${{I}_{1}}$ and ${{I}_{2}}$) such that
${{I}_{\max }}={{\left( \sqrt{{{I}_{1}}}+\sqrt{{{I}_{2}}} \right)}^{2}}$
${{I}_{\min }}={{\left( \sqrt{{{I}_{1}}}-\sqrt{{{I}_{2}}} \right)}^{2}}$
It is given that $\dfrac{{{I}_{\max }}}{{{I}_{\min }}}=\dfrac{25}{9}$.
Therefore,
$\dfrac{{{I}_{\max }}}{{{I}_{\min }}}=\dfrac{25}{9}=\dfrac{{{\left( \sqrt{{{I}_{1}}}+\sqrt{{{I}_{2}}} \right)}^{2}}}{{{\left( \sqrt{{{I}_{1}}}-\sqrt{{{I}_{2}}} \right)}^{2}}}$
$\Rightarrow \dfrac{\left( \sqrt{{{I}_{1}}}+\sqrt{{{I}_{2}}} \right)}{\left( \sqrt{{{I}_{1}}}-\sqrt{{{I}_{2}}} \right)}=\dfrac{5}{3}$ (Intensity is always positive, so ignore the negative values).
$\Rightarrow 3\sqrt{{{I}_{1}}}+3\sqrt{{{I}_{2}}}=5\sqrt{{{I}_{1}}}-5\sqrt{{{I}_{2}}}$
$\Rightarrow 8\sqrt{{{I}_{2}}}=2\sqrt{{{I}_{1}}}$
$\Rightarrow \dfrac{\sqrt{{{I}_{1}}}}{\sqrt{{{I}_{2}}}}=4$
$\Rightarrow \sqrt{\dfrac{{{I}_{1}}}{{{I}_{2}}}}=4\Rightarrow \dfrac{{{I}_{1}}}{{{I}_{2}}}=16$.
The ratio of the widths of the slits is equal to the ratio of intensities of the light coming from the slits respectively.
Therefore, the ratio of the widths of the slits is equal to 16:1.

Note:
Remember, the formula $y=\dfrac{D}{d}\Delta x$ holds true only when the distance of separation (d) of the two slits is very much smaller to the distance (D) between the slits and the screen, i.e. d<<<The intensity at a point on the screen is given as $I={{I}_{1}}+{{I}_{2}}+2\sqrt{{{I}_{1}}}\sqrt{{{I}_{2}}}\cos \phi $, where $\phi $ is the phase difference between the waves at that point.
With the above formula we get that
 ${{I}_{\max }}={{\left( \sqrt{{{I}_{1}}}+\sqrt{{{I}_{2}}} \right)}^{2}}$
${{I}_{\min }}={{\left( \sqrt{{{I}_{1}}}-\sqrt{{{I}_{2}}} \right)}^{2}}$
The relation between path difference and phase difference is given as $\dfrac{\Delta x}{\lambda }=\dfrac{\phi }{2\pi }$.