In a modified young’s double slit experiment, a monochromatic uniform and parallel beam of light of wavelength 6000A o. and intensity 10πW/m2 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 2000A0 and refractive index 1.5 for the wavelength of 6000A o is placed in front of aperture A as shown in figure. Calculate the power (in μW) received at the focal spot F of the lens. The lens is symmetrically placed with respect to the aperture. Assume that 10% of the power received by each aperture goes in the original direction and is brought to the focal spot.

Question

In a modified young’s double slit experiment, a monochromatic uniform and parallel beam of light of wavelength 6000A      o. and intensity 10πW/m2  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 2000A0 and refractive index 1.5 for the wavelength of 6000A      o is placed in front of aperture A as shown in figure. Calculate the power (in μW) received at the focal spot F of the lens. The lens is symmetrically placed with respect to the aperture. Assume that 10% of the power received by each aperture goes in the original direction and is brought to the focal spot.

Accepted Answer

The intensities of light from the sources $S_{1}$ and $S_{2}$ are given as

$P_{1} = (\frac{10}{π}) \times π {(0.01)}^{2} = 10^{- 5} W$

$P_{2} = (\frac{10}{π}) \times π {(0.02)}^{2} = 4 \times 10^{- 5} W$

The intensities of sources after emerging from the lenses are

$P_{A} = 0.10 \times 10^{- 5} W = 10^{- 6} W$

$P_{B} = 0.10 \times 4 \times 10^{- 5} W = 4 \times 10^{- 6} W$

The path difference produced due to film

$Δ x = (μ - 1) t = (1.5 - 1) \times 2000 \times 10^{- 10} = 10^{- 7} m$

$\Rightarrow ϕ = \frac{2 π}{λ} \times Δ x = \frac{2 π}{6000 \times 10^{- 10}} \times 10^{- 7}$

$\Rightarrow ϕ = \frac{π}{3} r a d i a n$

As refraction through lens does not introduce any path difference so above will be the net phase difference in the two light beams superposing at focal point of the lens. So net intensity received at F is given as

$P = P_{A} + P_{B} + 2 \sqrt{P_{A} P_{B}} \cos ϕ$

$= 10^{- 6} + 4 \times 10^{- 6} + \sqrt{10^{- 6} \times 4 \times 10^{6}} \cos \frac{π}{3} = 7 \times 10^{- 6} W$

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