Question

In YDSE, $d = 2\;{\rm{mm}}$ and $\lambda = 500\;{\rm{mm}}$. If the intensities of two slits are ${{\rm{I}}_0}$ and ${\rm{9}}\;{{\rm{I}}_0}$, then find intensity at $y = \dfrac{1}{6}\;{\rm{mm}}$.
A.${\rm{7}}{{\rm{I}}_0}$
B.${\rm{10}}{{\rm{I}}_0}$
C.${\rm{16}}{{\rm{I}}_0}$
D.${\rm{4}}{{\rm{I}}_0}$

Answer 1

Hint: This question is based on the Young’s double slit experiment. We have to know the Young’s double slit experiment. This experiment is used to get the interference pattern to be visible, coherent sources are required. If the sources are coherent, only then would there be constructive interference.

Complete step by step answer:
Given: The distance between the two slits is $d = 2\;{\rm{mm}}$, the fringe width is $D = 2\;{\rm{m}}$ the wavelength of the wavelength is $\lambda = 500\;{\rm{mm}}$ and the intensity of two slits are ${{\rm{I}}_0}$ and ${\rm{9}}\;{{\rm{I}}_0}$.
We have to find the intensity at $y = \dfrac{1}{6}\;{\rm{mm}}$.

Formula used:
First, we use the formula to find the path difference is given as,
$\Delta x = \dfrac{{yd}}{D}$
Here, $y$ is the displacement.
Second, we use the formula to find the phase difference is given as,
$\phi = \dfrac{{2\pi \Delta x}}{\lambda }$
And last, we use the resultant intensity formula we get,
$I = {I_1} + {I_2} + 2\sqrt {{I_1}{I_2}\cos \phi } $

Young's double-slit experiment is used to explain the interference pattern of dark and bright fringes. Young diffracted the light at two nearby slits, hence generating interference patterns.

So, first we have to calculate the path difference.
Therefore,
$\Delta x = \dfrac{{yd}}{D}$
Now, substitute the values in above equation we get,
$\begin{array}{l}
\Delta x = \dfrac{{\dfrac{1}{6}\; \times {{10}^{ - 3}}\;{\rm{m}} \times 2 \times {{10}^{ - 3}}\;{\rm{m}}}}{{2\;{\rm{m}}}}\\
\Rightarrow \Delta x = \dfrac{{{{10}^{ - 6}}}}{6}\;{\rm{m}}
\end{array}$

Now, we have to find the phase difference because in the resultant intensity formula there is a use of phase difference. So, we use the relation between the phase difference and path difference is given as,
\[\begin{array}{l}
\phi = \dfrac{{2\pi \times \dfrac{{{{10}^{ - 6}}}}{6}}}{{\left( {5 \times {{10}^{ - 7}}} \right)}}\\
\Rightarrow \phi = \dfrac{{2\pi }}{3}\\
\Rightarrow \phi = 120^\circ
\end{array}\]

We have to find the intensity at $y = \dfrac{1}{6}\;{\rm{mm}}$. Therefore, the expression of the resultant intensity is given as,
$I = {I_1} + {I_2} + 2\sqrt {{I_2}{I_1}\cos 120^\circ } $
Substitute the values in above equation we get,
$\begin{array}{l}
I = {I_0} + 9{I_0} + 2\sqrt {{I_0} \times 9{I_0} \times \cos 120^\circ } \\
\Rightarrow I = 10{I_0} - 3{I_0}\\
\Rightarrow I = 7{I_0}
\end{array}$

Thus, the correct option is (A).

Note:In this question, students have the knowledge about the Young’s double slit experiment. The formula of phase difference and intensity should be known. We have to use the step by step procedure to get resultant intensity.