Question

In Young’s double slit experiment $\beta $ is the fringe width and ${{I}_{0}}$ is the intensity of the central bright fringe. What will be the intensity at a distance (x) from the central bright fringe?
$\begin{align}
  & (A){{I}_{0}}\cos \left( \dfrac{x}{\beta } \right) \\
 & (B){{I}_{0}}{{\cos }^{2}}\left( \dfrac{x}{\beta } \right) \\
 & (C){{I}_{0}}{{\cos }^{2}}\left( \dfrac{\pi x}{\beta } \right) \\
 & (D)\dfrac{{{I}_{0}}}{4}{{\cos }^{2}}\left( \dfrac{\pi x}{\beta } \right) \\
\end{align}$

Answer 1

Hint: We know that for a given electromagnetic radiation, be it white light or anything else. The intensity of radiation is directly proportional to the square of its amplitude. We shall use this fact to calculate intensity of radiation at the central bright fringe and then at any point on the screen having a specific phase difference, with respect to the central bright fringe.

Complete answer:
It has been given in the problem that the intensity of the central bright fringe is equal to ${{I}_{0}}$ .
Then, this can be written as:
$\Rightarrow {{I}_{0}}=k{{(2A)}^{2}}$ [Let this expression be equation number (1)]
Where, A is the amplitude.
Now, let us move to an arbitrary point on screen. Then,
Let the path difference of this point be given as: $\vartriangle x$
And having a phase difference of: $\phi $
Now, the amplitude of radiation at a point whose phase difference is $\phi $ is given by:
$\Rightarrow {{A}_{1}}=2A\cos (\dfrac{\phi }{2})$
Therefore, the new intensity at that point will be given as:
$\begin{align}
  & \Rightarrow {{I}_{1}}=k{{\left( 2A\cos (\dfrac{\phi }{2}) \right)}^{2}} \\
 & \Rightarrow {{I}_{1}}=k{{(2A)}^{2}}{{\cos }^{2}}(\dfrac{\phi }{2})\text{ } \\
\end{align}$
$\Rightarrow {{I}_{1}}={{I}_{0}}{{\cos }^{2}}(\dfrac{\phi }{2})\text{ }$ [Let this expression be equation number (2)]
Let the above expression
Now, putting the value of phase difference as:
$\begin{align}
  & \Rightarrow \phi =\dfrac{2\pi }{\lambda }(\vartriangle x) \\
 & \Rightarrow \phi =\dfrac{2\pi }{\lambda }\left( \dfrac{dx}{D} \right) \\
\end{align}$
Using, the formula of fringe width as:
$\Rightarrow \beta =\dfrac{\lambda D}{d}$
We get:
$\begin{align}
  & \Rightarrow \phi =\dfrac{2\pi x}{\beta } \\
 & \Rightarrow \dfrac{\phi }{2}=\dfrac{\pi x}{\beta } \\
\end{align}$
Using this value of phase difference in equation number (2), we get:
$\begin{align}
  & \Rightarrow {{I}_{1}}={{I}_{0}}{{\cos }^{2}}(\dfrac{\phi }{2})\text{ } \\
 & \therefore {{I}_{1}}={{I}_{0}}{{\cos }^{2}}(\dfrac{\pi x}{\beta })\text{ } \\
\end{align}$
Hence, the intensity at a point distance (x) from the central bright fringe is given by ${{I}_{0}}{{\cos }^{2}}(\dfrac{\pi x}{\beta })$.

Hence, option (C) is the correct option.

Note:
In problems like these, we used very simple and basic formulas which were all under their definition. But the execution of all these formulas simultaneously made the problem a bit hard and lengthy to solve. So, one should be very careful while moving ahead in such types of problems as it might take some time to solve and get to the final answer. Also, the intensity of light above and below the central maxima at some distance (x) will be the same.