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In a modified YDSE a monochromatic uniform and parallel beam of light of wavelength 6000 A and intensity $\bigg(\large\frac{10}{\pi}\bigg) \large\frac{w}{m^2}$ is incident normally on two circular operates A and B of radii $0.001\;m $ and $0.002\;m$ respectively. A perfectly transparent film of thickness $2000 A^{\circ}$ and refractive index $1.5 $ for wavelength $6000\;A^{\circ}$ is placed in front of aperture A. Calculate the pour (in waters) received at Focal spot F of lens. The lens is symmetrically placed with respect to aperture. Assume $10 \%$ of power received by each aperture goes in original direction and is bought to focal speet.

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

1 Answer

 $ 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.
answered Feb 13, 2014 by meena.p
edited Jul 28, 2014 by meena.p

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