SATHEE: Optics 6 Question 56

Optics 6 Question 56

58. In a modified Young’s double slit experiment, a monochromatic uniform and parallel beam of light of wavelength $6000 \AA$ and intensity $(10 / π) W m^{- 2}$ is incident normally on two apertures $A$ and $B$ of radii $0.001 m$ and $0.002 m$ respectively. A perfectly transparent film of thickness $2000 \AA$ and refractive index 1.5 for the wavelength of $6000 \AA$ is placed in front of aperture $A$ (see figure). Calculate the power (in W) received at the focal spot $F$ of the lens. The lens is symmetrically placed with respect to the apertures. Assume that $10$ of the power received by each aperture goes in the original direction and is brought to the focal spot. $(1989, 8 M)$

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Answer:

Correct Answer: 58. $7 \times 10^{- 6} W$

$\begin{array}{lll} 59. (a) 1.17 m m & (b) 1.56 m m & 60 . 5892 \AA \end{array}$

Solution:

Power received by aperture $A$ ,

$P_{A} = I (π r_{A}^{2}) = \frac{10}{π} (π) (0.001)^{2} = 10^{- 5} W$

Power received by aperture $B$ ,

$\begin{aligned} P_{B} = I (π r_{B}^{2}) & = \frac{10}{π} (π) (0.002)^{2} \\ = 4 \times 10^{- 5} W \end{aligned}$

Only $10$ of $P_{A}$ and $P_{B}$ goes to the original direction. Hence, $10$ of $P_{A} = 10^{- 6} = P_{1}$ (say)

and $10$ of $P_{B} = 4 \times 10^{- 6} = P_{2}$ (say)

Path difference created by slab

$\begin{aligned} Δ x & = (μ - 1) t \\ = (1.5 - 1) (2000) = 1000 \AA \end{aligned}$

Corresponding phase difference,

$φ = \frac{2 π}{λ} \cdot Δ x = \frac{2 π}{6000} \times 1000 = \frac{π}{3}$

Now, resultant power at the focal point

$\begin{aligned} P & = P_{1} + P_{2} + 2 \sqrt{P_{1} P_{2}} \cos φ \\ = 10^{- 6} + 4 \times 10^{- 6} + 2 \sqrt{(10^{- 6}) (4 \times 10^{- 6})} \cos \frac{π}{3} \\ = 7 \times 10^{- 6} W \end{aligned}$

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Optics 6 Question 56

Answer:

इंजीनियरिंग की तैयारी

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Optics 6 Question 56

Answer:

इंजीनियरिंग की तैयारी

चिकित्सा तैयारी

एनसीईआरटी पुस्तकें और समाधान

महत्वपूर्ण संसाधन

अन्य संसाधन

पाठ्यक्रम

परीक्षा मंच

नमूना मॉक टेस्ट

सदस्यता

एनसीईआरटी व्यायाम वीडियो समाधान

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