$(a)\;7 \times 10^{-4}\;watt \\ (b)\;7 \times 10^{-2} watt \\ (c)\;7 \times 10^{-6} watt \\ (d)\;7 \times 10^{-8} watt $

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$ P=IS$ , where A is the surface area

So Power received at A and B is respectively.

$P_A=\large\frac{10}{\pi}$$ \times \pi (0.001)^2 =10^{-5}w$

and $P_B=\large\frac{10}{\pi} $$\times \pi (0.002)^2= 4 \times 10^{-5}w$

and as only 10 % of incident power passes,

$P_A'= \large\frac{10}{100} $$\times 10^{-5}=10^{-6} w$

and $P_B'= \large\frac{10}{100} $$ 4 \times 10^{5}= 4 \times 10^{-6}w$

Now as due to introduction of film the path difference produced .

$\Delta x =(\mu-1) t $

$\qquad=(1.5-1) \times 2000=1000 A^{\circ}$

So, $ \phi =\large\frac{2 \pi}{\lambda} (\Delta x)= \large\frac{2\pi}{6000} $$ \times 1000 =\large\frac{n}{3}$

But as in interference,

$I=I_1+I_2+ 2 \sqrt {I_1 I_2} \cos \phi$

and if s is area of focal spot,

$P= IS= I_AS+I_BS+ 2S(\sqrt {I_A+I_B}) \cos \phi$

ie $P= PA'+PB'+2 \sqrt {PA'P_B'} \cos (\large\frac{\pi}{3})$

$P= 10^{-6} [1+4+2 \sqrt {1 \times 4}) \times (\frac{1}{2})]$

$\qquad= 7 \times 10^{-6}\;watt$

Hence c is the correct answer.

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