Question

Find the intensity at a point on a screen in Young’s double-slit experiment where interfering waves of equal intensity have a path difference of
I. $\dfrac{\lambda }{4}$
II. $\dfrac{\lambda }{3}$

Answer 1

We will first discuss the intensity. The intensity is a type of radiant energy defined as the power transferred per given unit area, where the area can be measured perpendicularly to the direction of the propagation of the radiant energy. Hence to find the intensity at a point on a screen we will use the concept of the intensity for the given wavelength$\lambda$.

Formula used:
Intensity formula
$I = 4{I_0}{\cos ^2}\dfrac{\phi }{2}$
where $\phi$ is the phase difference.

Complete step by step answer:
We will consider the intensity at a point on a screen in Young’s double-slit experiment for the interfering waves of equal intensity which can be given as
$I = 4{I_0}{\cos ^2}\dfrac{\phi }{2}$ …… $(1)$
where $\phi$ is the phase difference.
As we know that the phase difference can be defined as the differences when two or more alternating quantities when reached at their zero or maximum values and the phase difference can be given by the formula as
$\Delta \phi = \dfrac{{2\pi }}{\lambda } \times \Delta x$ ……. $(2)$
Where $\Delta x$ is the path difference.

I. For $\dfrac{\lambda }{4}$
Here for the given, first case, the path difference is given as $\Delta x = \dfrac{\lambda }{4}$, and substitute it in the equation $(2)$. Hence
$\Delta \phi = \dfrac{{2\pi }}{\lambda } \times \dfrac{\lambda }{4}$
$\Rightarrow \Delta \phi = \dfrac{\pi }{2}$
Now substituting the value of $\Delta \phi = \dfrac{\pi }{2}$ in the equation $(1)$, which provides us the intensity given as $I = 4{I_0}{\cos ^2}\dfrac{\phi }{2}$
$\Rightarrow I = 4{I_0}{\cos ^2}\left( {\dfrac{1}{2} \times \dfrac{\pi }{2}} \right)$ …… $(3)$
As we know that the the value of $\cos \left( {\dfrac{\pi }{4}} \right) = \dfrac{1}{{\sqrt 2 }}$, thus putting this value in the equation $(3)$ gives us
$I = 4{I_0}{\cos ^2}\left( {\dfrac{\pi }{4}} \right)$
$\Rightarrow I = 4{I_0} \times \left( {\dfrac{1}{2}} \right)$
Hence this gives us the final answer of the intensity as
$I = 4{I_0} \times \dfrac{1}{2}$
$\Rightarrow I = 2{I_0}$
Therefore the intensity at a point on a screen in Young’s double-slit experiment where interfering waves of equal intensity have a path difference of $\dfrac{\lambda }{4}$ is given as $2{I_0}$.

II. For $\dfrac{\lambda }{3}$
Here for the given, second case, the path difference is given as $\Delta x = \dfrac{\lambda }{3}$, and substitute it in the equation $(2)$. Hence
$\Delta \phi = \dfrac{{2\pi }}{\lambda } \times \dfrac{\lambda }{3}$
$\Rightarrow \Delta \phi = \dfrac{{2\pi }}{3}$
Now substituting the value of $\Delta \phi = \dfrac{{2\pi }}{3}$ in the equation $(1)$, which provides us the intensity given as $I = 4{I_0}{\cos ^2}\dfrac{\phi }{2}$
$\Rightarrow I = 4{I_0}{\cos ^2}\left( {\dfrac{1}{2} \times \dfrac{{2\pi }}{3}} \right)$ …… $(4)$
As we know that the the value of $\cos \left( {\dfrac{\pi }{3}} \right) = \dfrac{1}{2}$, putting this value in the equation $(4)$ gives us
$I = 4{I_0}{\cos ^2}\left( {\dfrac{\pi }{3}} \right)$
$\Rightarrow I = 4{I_0} \times {\left( {\dfrac{1}{2}} \right)^2}$
Hence this gives us the final answer of the intensity as
$I = 4{I_0} \times \dfrac{1}{4}$
$\therefore I = {I_0}$

Therefore the intensity at a point on a screen in Young’s double-slit experiment where interfering waves of equal intensity have a path difference of $\dfrac{\lambda }{3}$ is given as ${I_0}$.

Note: While dealing with such questions one should ensure that the proper methods are used and while dealing with trigonometric functions like $\sin$ and $\cos$ it should be ensured the proper conversion should be made from degree to radian or vice versa.