Question

In a modified YDSE a monochromatic uniform and parallel beam of light of wavelength $6000\mathring{A}$ and intensity $(10/\pi )W/m^2$ is incident normally on two circular apertures $A$ and $B$ of radii $0.001m$ and $0.002m$, respectively. A perfectly transparent film of thickness $2000\mathring{A}$ and refractive index $1.5$ for the wavelength of $6000\mathring{A}$ is placed in front of aperture $A$ (see figure). Calculate the power in watts received at the focal spot $F$ of the lens. The lens is symmetrically placed w.r.t. the apertures. Assume that $10\%$ of the power received by each aperture goes in the original direction and is brought to the focal spot.

• A. $3\times 10^{-6}W$
• B. $1\times 10^{-6}W$
• C. $5\times 10^{-6}W$
• D. $7\times 10^{-6}W$

Answer 1

Answer: (D)

The power transmitted by apertures $A$ and $B$ are
$P_A=\left(\frac{10}{\pi }\right)\left[\pi (0.001{)}^2\right]=10^{-5}W$
$P_B=\left(\frac{10}{\pi }\right)\left[\pi \times (0.002{)}^2\right]=4\times 10^{-5}W$
Only $10\%$ of transmitted power reaches the focus.
$P_A^{\prime }=10^{-5}\times \frac{10}{100}=10^{-6}W$
$P_B^{\prime }=4\times 10^{-5}\times \frac{10}{100}=4\times 10^{-6}W$
The resultant power at the focus after superposition of two waves is
$P=P_A^{\prime }+P_B^{\prime }+2\sqrt{P_A^{\prime }{P}_B^{\prime }}\cos \varphi$
where $\varphi$ is phase difference
The introduction of mica sheet in the path of $A$ creates a path difference $(\mu -1)t$.
$(\mu -1)t=(1.5-1)\times 2000\mathring{A}=1000\mathring{A}$
Phase difference
$\Delta \varphi =\frac{2\pi }{\lambda }(\mu -1)t=\frac{2\pi }{6000}\times 1000=\frac{\pi }{3}$
Therefore
$P=10^{-6}+4\times 10^{-6}+2\sqrt{\left(10^{-6}\right)\left(4\times 10^{-6}\right)}\cos \frac{\pi }{3}$
$=7\times 10^{-6}W$