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**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${{\left( \dfrac{1}{2} \right)}^{2}}$ .

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