Question

The maximum intensity in Young’s double slit experiment is ${I_0}$. Distance between the slits is $d = 5\lambda $, where $\lambda $ is the wavelength of monochromatic light used in the experiment. What will be the intensity of light in front of one of the slits on a screen at a distance$D = 10d$?
A. ${I_0}$
B. ${I_0}/4$
C. $\dfrac{3}{4}{I_0}$
D. ${I_0}/2$

Answer 1

Hint:The intensity of the light in the Young’s double slit experiment varies with the distance between the slits, path difference and distance of the screen from the source of the light.

Complete Step by Step Answer:
The distance between the slits is $d = 5\lambda $, wavelength of monochromatic light is $\lambda $and distance of one of the slits from the screen is $D = 10d$.
Write the equation to calculate the distance of the source from the screen.
$y = \dfrac{d}{2}$
Substitute $d$ as $5\lambda $ in the above equation.
$y = \dfrac{{5\lambda }}{2}$
Write the equation to calculate the path difference.
$\Delta x = \dfrac{{yd}}{D}$
Substitute $y$ as $\dfrac{{5\lambda }}{2}$, $D$ as $10d$ and $d$ as $5\lambda $ in the above equation.
$\begin{array}{l}
\Delta x = \left( {\dfrac{{5\lambda }}{2}} \right)\left( {\dfrac{{5\lambda }}{{10\left( {5\lambda } \right)}}} \right)\\
 = \dfrac{\lambda }{4}
\end{array}$
Write the equation to calculate the path difference.
$\phi = \dfrac{{2\pi }}{\lambda }\left( {\Delta x} \right)$
Substitute $\Delta x$ as $\dfrac{\lambda }{4}$in the above equation.
\[\begin{array}{l}
\phi = \dfrac{{2\pi }}{\lambda }\left( {\dfrac{\lambda }{4}} \right)\\
 = \dfrac{\pi }{2}
\end{array}\]
Write the equation to calculate the intensity of light in front of one of the slit.
$I = {I_0}{\cos ^2}\left( {\dfrac{\phi }{2}} \right)$
Substitute $\phi $ as \[\dfrac{\pi }{2}\] in the above equation.
$\begin{array}{l}
I = {I_0}{\cos ^2}{\left( {\dfrac{{\left( {\dfrac{\pi }{2}} \right)}}{2}} \right)^2}\\
 = {I_0}{\cos ^2}\left( {\dfrac{\pi }{4}} \right)\\
 = {I_0}{\left( {\dfrac{1}{{\sqrt 2 }}} \right)^2}\\
 = \dfrac{{{I_0}}}{2}
\end{array}$
Therefore, the intensity of light in front of one of the slits is $\dfrac{{{I_0}}}{2}$ and the option (D) is correct.

Note:Make sure to calculate the path difference, phase difference of the light and distinguish between the distance between the slit and distance of the source from the slit.