Question

In a Young’s double slit experiment, the slits are placed $0.32mm$ apart. Light of wavelength $\lambda =500nm$ incident on the slits. The total number of bright fringes that are observed in the angular range - $-{{30}^{\circ }}\le \theta \le {{30}^{\circ }}$ is
A. 320
B. 641
C. 321
D. 640

Answer 1

Hint: To solve the given question, use the formula for the path difference between the two waves in the Young’s double slit experiment. Apply the condition on the path difference for which a maxima is obtained. Then find the number of maximas for the given range.

Formula used:
$\Delta x=d\sin \theta $

Complete step by step answer:
 Young’s double slit experiment is an experiment to obtain the interference pattern of two coherent sources of light.
The interference pattern is obtained in the form of fringes such a way that there are consecutive maximas and minimas, all separated by equal distances. Here maxima is a bright fringe with maximum intensity and minima is a dark fringe with minimum intensity.
The interference pattern is obtained on a screen. The intensity of the resultant wave at any point on the screen depends on the path difference between the two waves. In Young’s double slit experiment the path difference is equal to $\Delta x=d\sin \theta $ …..(i)
, where $\Delta x$ is the path difference between the two waves, d is the distance between the slits and $\theta $ is the angle that the line joining the midpoint of d and the point on the screen makes with the horizontal axis.
The ${{n}^{th}}$ maxima (i.e. of maximum intensity) is obtained when $\Delta x=n\lambda $, where $\lambda $ is the wavelength of the light.
Substitute this value in (i).
$\Rightarrow n\lambda =d\sin \theta $ …. (ii).
It is given that $d=0.32mm=32\times {{10}^{-5}}m$ and $\lambda =500nm=500\times {{10}^{-9}}m$.
Since we are suppose find the number of bright fringes from angular range $-{{30}^{\circ }}\le \theta \le {{30}^{\circ }}$, substitute $\theta ={{30}^{\circ }}$ in (ii). Also substitute the values of d and $\lambda $.
$\Rightarrow n\left( 500\times {{10}^{-9}} \right)=\left( 32\times {{10}^{-5}} \right)\sin {{30}^{\circ }}$
$\Rightarrow n={{\dfrac{\left( 32\times {{10}^{-5}} \right)\sin 30}{\left( 500\times {{10}^{-9}} \right)}}^{\circ }}=\dfrac{\left( 32\times {{10}^{-5}} \right)(0.5)}{\left( 500\times {{10}^{-9}} \right)}=320$.
This means that at $\theta ={{30}^{\circ }}$, ${{320}^{th}}$ maxima is formed.
Therefore, for the range of ${{0}^{\circ }}<\theta \le {{30}^{\circ }}$, there are 320 maximas present.
Similarly, for the range of $-{{30}^{\circ }}\le \theta <{{0}^{\circ }}$, there are 320 maximas present.
And there is a maxima at $\theta ={{0}^{\circ }}$ called central maxima.
Therefore, the total number of maximas (bright fringe\s) for the given range is $320+320+1=641$.

So, the correct answer is “Option B”.

Note:
For ${{n}^{th}}$ minima (dark fringe), the path difference between the two waves must be equal to $\Delta x=\left( 2n-1 \right)\dfrac{\lambda }{2}$, i.e. odd multiple of half of the wavelength of the light.
The distance between any two consecutive maximas or any two consecutive minimas is called fringe width.
Also note that the distance between the two slits is very much smaller compared to the distance between the slits and the screen.