Question

Intensity of central fringe in interference pattern is $0.01W/{{m}^{2}}$ then find the intensity at a point having the path difference $\dfrac{\lambda }{3}$ on screen from the center (in$mW/{{m}^{2}}$)
$\begin{align}
  & A.\text{ }2.5 \\
 & B.\text{ }5 \\
 & C.\text{ }7.5 \\
 & D.\text{ }10 \\
\end{align}$

Answer 1

Hint: In this question first we have to find intensity of light $\left( {{I}_{0}} \right)$ and phase difference $\left( \phi \right)$in first case central fringe intensity is given so by using intensity equation we can find intensity of light after that we can find phase difference $\phi $ using relation between phase difference and path difference by using this two values we can find the intensity at the given wavelength.
Formula used:
$I=4{{I}_{0}}{{\cos }^{2}}\dfrac{\phi }{2}$
And
$\phi =\dfrac{2\pi }{\lambda }\Delta x$

Complete answer:
It is given that intensity of central fringe on the screen
${{I}_{c}}=0.01W/{{m}^{2}}$
To find the intensity of fringe we will use below formula,
$I=4{{I}_{0}}{{\cos }^{2}}\dfrac{\phi }{2}......\left( 1 \right)$
Where, I = intensity of light after polarization
${{I}_{0}}$= original intensity of light
$\phi =$ Phase difference
To find the intensity of any path difference we will have to find the intensity of original light$\left( {{I}_{0}} \right)$.
Now in case of central fringe $I={{I}_{c}}$ and phase difference $\left( \phi \right)$will be zero now substitute this values in equation (1)
$\begin{align}
  & \Rightarrow {{I}_{c}}=4{{I}_{0}}{{\cos }^{2}}\left( 0 \right) \\
 & \Rightarrow 0.01=4{{I}_{0}}\left( \because {{\cos }^{2}}\left( 0 \right)=1 \right) \\
 & \therefore {{I}_{0}}=\dfrac{0.01}{4}....\left( 2 \right) \\
\end{align}$
Now in order to find the phase difference we will use relation between phase difference and path difference
$\phi =\dfrac{2\pi }{\lambda }\Delta x......\left( 3 \right)$
$\Delta x$ = path difference
$\lambda =$ Wavelength
Value of path difference is given
$\Delta x=\dfrac{\lambda }{3}.....\left( 4 \right)$
Now put values of $\Delta x$ in equation (3)
$\begin{align}
  & \Rightarrow \phi =\dfrac{2\pi }{\lambda }\times \dfrac{\lambda }{3} \\
 & \therefore \phi =\dfrac{2\pi }{3}.....\left( 5 \right) \\
\end{align}$
Now we will put value of equation (2) and equation (5) in equation (1)
$\begin{align}
  & \Rightarrow I=4{{I}_{0}}{{\cos }^{2}}\dfrac{\phi }{2} \\
 & \Rightarrow I=4\times \dfrac{0.01}{4}\times {{\cos }^{2}}\left( \dfrac{2\pi }{3\times 2} \right) \\
 & \Rightarrow I=0.01\times {{\cos }^{2}}\left( \dfrac{\pi }{3} \right) \\
 & \Rightarrow I=0.01\times {{\left( \dfrac{1}{2} \right)}^{2}} \\
 & \Rightarrow I=\dfrac{0.01}{4} \\
 & \therefore I=0.0025W/{{m}^{2}} \\
\end{align}$
We have to convert in $mW/{{m}^{2}}$ so we have to multiply by 1000 now
$\begin{align}
  & \Rightarrow I=0.0025\times 1000 \\
 & \therefore I=2.5mW/{{m}^{2}} \\
\end{align}$

So the correct option is (A) .

Note:
Intensity of light in a particular direction per unit solid angle can be defined as the measure of the wavelength-weighted power emitted by a light source. Additionally, when we put value of ${{\cos }^{2}}\left( \dfrac{\pi }{3} \right)$ we have to remember at square sometimes we can do mistakes by putting values of ${{\cos }^{2}}\left( \dfrac{\pi }{3} \right)=\left( \dfrac{1}{2} \right)$ instead of${{\left( \dfrac{1}{2} \right)}^{2}}$ .