# 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.

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

$P=IS$ , 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 .
$\Delta x =(\mu-1) t$
$\qquad=(1.5-1) \times 2000=1000 A^{\circ}$