
Find the intensity at a point on a screen in Young’s double slit experiment where the interfering waves of equal intensity have a path difference of (i) $\dfrac{\lambda }{4}$ and (ii) $\dfrac{\lambda }{3}$
Answer
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Hint: Use the value of path difference given to find out the value of phase difference. To figure out the intensity at a point in Young’s double slit experiment, use the formula to find the resultant intensity at a point.
Complete step by step answer:
Given, the intensities of interfering waves are equals that is,
${I_1} = {I_2} = {I_o}$ (1)
Phase difference can be written as,
$\phi = \dfrac{{2\pi }}{\lambda }\Delta x$
Where $\lambda $ is the wavelength and $\Delta x$ is the phase difference.
The resultant intensity at a point is given as
$I = {I_1} + {I_2} + 2\sqrt {{I_1}{I_2}} \cos \phi $(2)
(i) For path difference $\Delta x = \dfrac{\lambda }{4}$
Phase difference, $\phi $ = $\dfrac{{2\pi }}{\lambda }\Delta x$ $= \dfrac{{2\pi }}{\lambda } \times $ $\dfrac{\lambda }{4}$ = $\dfrac{\pi }{2}$
The resultant intensity at a point is
$I = {I_1} + {I_2} + 2\sqrt {{I_1}{I_2}} \cos \phi $
Putting the values of ${I_1}$, ${I_2}$ and $\phi $, we have
$I = {I_0} + {I_0} + 2\sqrt {{I_0}{I_0}} \cos$ $\dfrac{\pi }{2} \\$
$\Rightarrow I $= $2{I_0}$
(ii) For path difference $\Delta x = \dfrac{\lambda }{3}$
Phase difference, $\phi = \dfrac{{2\pi }}{\lambda }\Delta x = \dfrac{{2\pi }}{\lambda } \times \dfrac{\lambda }{3} = \dfrac{{2\pi }}{3}$
The resultant intensity at a point is
$I = {I_1} + {I_2} + 2\sqrt {{I_1}{I_2}} \cos \phi $
Putting the values of ${I_1}$, ${I_2}$ and $\phi $, we have
$I = {I_0} + {I_0} + 2\sqrt {{I_0}{I_0}} \cos \dfrac{{2\pi }}{3} \\$
$ \Rightarrow I = 2{I_0} + 2{I_0}\left( { - \dfrac{1}{2}} \right) \\$
$ \Rightarrow I = {I_0} \\$
Therefore, for a path difference of $\dfrac{\lambda }{4}$ and $\dfrac{\lambda }{3}$, the resultant intensities at a point are $2{I_0}$ and ${I_0}$ respectively.
Note:
Here, the intensities of the interfering waves are the same but in some cases the intensities might be different. So, before proceeding we should always check whether the interfering waves have same intensities or different. The double-slit experiment is a demonstration that light and matter can display characteristics of both classically defined waves and particles.
Complete step by step answer:
Given, the intensities of interfering waves are equals that is,
${I_1} = {I_2} = {I_o}$ (1)
Phase difference can be written as,
$\phi = \dfrac{{2\pi }}{\lambda }\Delta x$
Where $\lambda $ is the wavelength and $\Delta x$ is the phase difference.
The resultant intensity at a point is given as
$I = {I_1} + {I_2} + 2\sqrt {{I_1}{I_2}} \cos \phi $(2)
(i) For path difference $\Delta x = \dfrac{\lambda }{4}$
Phase difference, $\phi $ = $\dfrac{{2\pi }}{\lambda }\Delta x$ $= \dfrac{{2\pi }}{\lambda } \times $ $\dfrac{\lambda }{4}$ = $\dfrac{\pi }{2}$
The resultant intensity at a point is
$I = {I_1} + {I_2} + 2\sqrt {{I_1}{I_2}} \cos \phi $
Putting the values of ${I_1}$, ${I_2}$ and $\phi $, we have
$I = {I_0} + {I_0} + 2\sqrt {{I_0}{I_0}} \cos$ $\dfrac{\pi }{2} \\$
$\Rightarrow I $= $2{I_0}$
(ii) For path difference $\Delta x = \dfrac{\lambda }{3}$
Phase difference, $\phi = \dfrac{{2\pi }}{\lambda }\Delta x = \dfrac{{2\pi }}{\lambda } \times \dfrac{\lambda }{3} = \dfrac{{2\pi }}{3}$
The resultant intensity at a point is
$I = {I_1} + {I_2} + 2\sqrt {{I_1}{I_2}} \cos \phi $
Putting the values of ${I_1}$, ${I_2}$ and $\phi $, we have
$I = {I_0} + {I_0} + 2\sqrt {{I_0}{I_0}} \cos \dfrac{{2\pi }}{3} \\$
$ \Rightarrow I = 2{I_0} + 2{I_0}\left( { - \dfrac{1}{2}} \right) \\$
$ \Rightarrow I = {I_0} \\$
Therefore, for a path difference of $\dfrac{\lambda }{4}$ and $\dfrac{\lambda }{3}$, the resultant intensities at a point are $2{I_0}$ and ${I_0}$ respectively.
Note:
Here, the intensities of the interfering waves are the same but in some cases the intensities might be different. So, before proceeding we should always check whether the interfering waves have same intensities or different. The double-slit experiment is a demonstration that light and matter can display characteristics of both classically defined waves and particles.
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