Question

In a young’s double slit experiment, the intensity at the central maximum is\[{I_0}\] . The intensity at a distance $\dfrac{\beta }{4}$ from the central maximum is ($\beta $ is fringe width)
A) \[{I_0}\]
B) $\dfrac{{{I_0}}}{2}$
C) $\dfrac{{{I_0}}}{{\sqrt 2 }}$
D) $\dfrac{{{I_0}}}{4}$

Answer 1

Hint:In this question, we can calculate the path difference using the fringe width formula. Path difference will be used in calculating the phase difference. After that we can calculate the resultant intensities using the formula $I = {I_1} + {I_2} + 2\sqrt {{I_1}{I_2}\cos \phi } $ for $\phi = 0$ and $\cos \phi = \dfrac{\pi }{2}$.

Complete step by step solution: -
According to the question, in a double slit experiment, the central maximum intensity$ = {I_0}$
We know that if a light of wavelength $\lambda $ is incident on the slits. Let \[d\] is the distance between the fringes and \[D\] is the distance between the slits and screen then the fringe width$\beta $ is given as-
$\beta = \dfrac{{\lambda D}}{d}$
But according to the question, if the distance is $y = \dfrac{\beta }{4}$ then we get the value of \[y\] after putting the value of $\beta $ .
 $
  y = \dfrac{\beta }{4} \\
   \Rightarrow y = \dfrac{{\lambda D}}{{4d}} \\
$
Now, in Young’s double slit experiment, the path difference$\Delta x$ is given by-
$\Delta x = \dfrac{{yd}}{D}$
Now putting the value of \[y\] in the above equation, we get-
$ \Rightarrow \Delta x = \left( {\dfrac{{\lambda D}}{{4d}}} \right)\dfrac{d}{D}$
On solving the above equation, we get-
$ \Rightarrow \Delta x = \dfrac{\lambda }{4}$
If $\phi $ is the phase difference corresponding to path difference $\dfrac{\lambda }{4}$, then the phase difference will be-
$\phi = \left( {\dfrac{{2\pi }}{\lambda }} \right)\Delta x$
Putting the value of $\Delta x$ in the above equation, we get-
$
   \Rightarrow \phi = \left( {\dfrac{{2\pi }}{\lambda }} \right)\dfrac{\lambda }{4} \\
   \Rightarrow \phi = \dfrac{\pi }{2} \\
$
Now, we know that the intensity due to interference is given by-
$I = {I_1} + {I_2} + 2\sqrt {{I_1}{I_2}\cos \phi } $
Where \[I\] is the resultant intensity of the two waves having intensities ${I_1}$ and ${I_2}$.
We know that at central maximum $\phi = 0$ and $I = {I_0}$ . Then let the intensity from each slit ${I_1} = {I_2} = {I_s}$
$ \Rightarrow {I_0} = {I_s} + {I_s} + 2\sqrt {{I_s}{I_s}} $
(As $\cos 0 = 1$ )
On simplifying it, we get-
$ \Rightarrow {I_0} = 4{I_s}$ …………………………..(i)
Now intensity at a distance $\dfrac{\beta }{4}$ is given by $\cos \phi = \dfrac{\pi }{2}$ , then putting this value, we get-
$I = {I_s} + {I_s} + 2\sqrt {{I_s}{I_s}\cos \dfrac{\pi }{2}} $
$ \Rightarrow I = 2{I_s}$ ………………………….(ii)
(As $\cos \dfrac{\pi }{2} = 0$ )
Now comparing equation (i) and (ii), we get-
$I = \dfrac{{{I_0}}}{2}$
Hence, the intensity at a distance $\dfrac{\beta }{4}$ from the central maximum is $I = \dfrac{{{I_0}}}{2}$.

Therefore, option B is correct.

Note: - In this question, we have to keep in mind that two resultant intensities should be found for $\phi = 0$ and $\cos \phi = \dfrac{\pi }{2}$. These two intensities can be compared to find the required relation between the intensity of central maximum and the intensity at $y = \dfrac{\beta }{4}$. As it is a double slit experiment, the light source is monochromatic and the intensity is constant throughout the experiment whether it is incident on two slits or more.