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In a Young’s double slit experiment, 12 fringes are observed to be formed in a certain segment of the screen when light of wavelength 600 nm is used. If the wavelength of light is changed to 400 nm, number of fringes observed in the same segment of the screen is given by
  • a)
    12
  • b)
    18
  • c)
    24
  • d)
    30
Correct answer is option 'B'. Can you explain this answer?
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Explanation:

Given data:
- Wavelength of light used in the first case, λ1 = 600 nm
- Number of fringes observed in the first case, n1 = 12
- Wavelength of light used in the second case, λ2 = 400 nm
- Number of fringes observed in the second case, n2 = ?

Formula:
The number of fringes formed in a double-slit interference pattern is given by the formula:
n = (D * λ) / d
where:
- n = number of fringes
- D = distance between the slits and the screen
- λ = wavelength of light
- d = distance between the two slits

Calculations:
Let's assume D and d remain constant for both cases.
For the first case:
n1 = (D * λ1) / d
12 = (D * 600) / d
For the second case:
n2 = (D * λ2) / d
We know that D and d are constant, so n2/n1 = λ2/λ1
n2 = n1 * (λ2/λ1)
n2 = 12 * (400/600)
n2 = 8 * 2
n2 = 16
Therefore, the number of fringes observed in the second case when the wavelength of light is changed to 400 nm is 16, which corresponds to option B.
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