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In a double-slit interference pattern, t...

In a double-slit interference pattern, the first maxima for infrared light would be

A

At the same place as the first maxima for green light

B

Closer to the center than the first maxima for green light

C

Farther from the center than the first maxima for green light

D

Infrared light does not produce an interference pattern

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To solve the problem of finding the position of the first maxima in a double-slit interference pattern for infrared light, we can follow these steps: ### Step-by-Step Solution: 1. **Understanding the Formula**: The position of the maxima in a double-slit interference pattern is given by the formula: \[ y = \frac{n \lambda D}{d} \] where: - \( y \) = distance from the central maximum to the nth maximum, - \( n \) = order of the maximum (for the first maximum, \( n = 1 \)), - \( \lambda \) = wavelength of the light, - \( D \) = distance from the slits to the screen, - \( d \) = distance between the slits. 2. **Identifying the Wavelength Range**: The problem states that the wavelength of infrared light varies from 700 nm to 1 mm. We need to convert these values into meters for consistency in units: - \( 700 \, \text{nm} = 700 \times 10^{-9} \, \text{m} \) - \( 1 \, \text{mm} = 1 \times 10^{-3} \, \text{m} \) 3. **Calculating the Position for the First Maximum**: For the first maximum (\( n = 1 \)): \[ y_1 = \frac{1 \cdot \lambda D}{d} = \frac{\lambda D}{d} \] We will calculate \( y_1 \) for both extremes of the infrared wavelength. 4. **Calculating for \( \lambda = 700 \, \text{nm} \)**: \[ y_1 = \frac{700 \times 10^{-9} \cdot D}{d} \] 5. **Calculating for \( \lambda = 1 \, \text{mm} \)**: \[ y_1 = \frac{1 \times 10^{-3} \cdot D}{d} \] 6. **Comparing the Results**: Since \( y_1 \) is directly proportional to \( \lambda \), as the wavelength increases, the distance to the first maximum also increases. Therefore, the first maximum for infrared light will be farther from the center than that for visible light (like green light). 7. **Conclusion**: The first maxima for infrared light (700 nm to 1 mm) will be farther from the center compared to the first maxima for green light (which has a wavelength of approximately 560 nm to 570 nm). ### Final Answer: The first maxima for infrared light would be farther from the center than the first maxima for green light. ---
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RESNICK AND HALLIDAY-INTERFERENCE AND DIFFRACTION -PRACTICE QUESTIONS (Single Correct Choice Type)
  1. In Young's double-slit interference experiment a first screen with a ...

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  2. Two sources, in phase and a distance d apart, each emit a wave of wav...

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  3. In a double-slit interference pattern, the first maxima for infrared ...

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  4. Light shining through two very narrow slits produces an interference ...

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  5. In Young's double-slit experiment, the slit separation is 0.5 mm and t...

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  6. Two different color beams (yellow and blue) from a point source are i...

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  7. Two wavelength of light lambda(1) and lambda(2) are sent through Youn...

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  8. In a Young's double-slit experiment, if the incident light consists o...

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  9. Two plane monochromatic coherent waves produce inter- ference pattern...

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  10. Path followed by two rays through a thin lens in air is shown in the ...

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  11. An interference pattern is formed on a screen by shining a planar wave...

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  12. A glass slab of thickness 4 cm contains the same number of waves as 5 ...

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  13. On introducing a thin sheet of mica (thickness 12 xx 10^(-7) cm) in p...

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  14. A Young's double-slit experiment is conducted in water (n(1)) as show...

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  15. In a Young's double-slit experiment, green light is incident on the t...

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  16. Why isn't the interference pattern like the one from Young's double-s...

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  17. Coherent light is incident on two fine parallel slits S(1) and S(2) as...

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  18. Three coherent, equal intensity light rays arrive at a point Pon a sc...

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  19. In a Young's double slit experiment, if the slits are of unequal widt...

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  20. Intensities of light due to the two slits of Young's double-slit exper...

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