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 $0.001 \,m$ and $0.002\, m$, 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
$P=P_{A}^{\prime}+P_{B}^{\prime}+2 \sqrt{P_{A}^{\prime} P_{B}^{\prime}} \cos \phi$
where $\phi$ 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
$\Delta \phi=\frac{2 \pi}{\lambda}(\mu-1) t=\frac{2 \pi}{6000} \times 1000=\frac{\pi}{3}$
Therefore
$P=10^{-6}+4 \times 10^{-6}+2 \sqrt{\left(10^{-6}\right)\left(4 \times 10^{-6}\right)} \cos \frac{\pi}{3}$
$=7 \times 10^{-6} \,W$