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In Young's double-slit experiment, a student observes 8 fringes in a certain segment of the screen when a monochromatic light of 600 nm wavelength is used. If the wavelength of light is changed to 400 nm, then the number of fringes he would observe in the same region of the screen is:

A

12

B

6

C

8

D

9

Step-by-Step Solution

  1. Identify Given Values:
  • Initial number of fringes (n1n_1) = 8
  • Initial wavelength (λ1\lambda_1) = 600 nm600 \text{ nm}
  • Final wavelength (λ2\lambda_2) = 400 nm400 \text{ nm}
  • The segment of the screen (width LL) remains constant.
  1. Formula: The width of the region occupied by nn fringes is given by L=nβL = n \beta, where β=λDd\beta = \frac{\lambda D}{d} is the fringe width. Thus, L=nλDdL = \frac{n \lambda D}{d}.
  2. Relation: Since L,D,L, D, and dd are constant, the product of the number of fringes and the wavelength is constant: n1λ1=n2λ2n_1 \lambda_1 = n_2 \lambda_2
  3. Calculation: 8×600 nm=n2×400 nm8 \times 600 \text{ nm} = n_2 \times 400 \text{ nm} 4800=n2×4004800 = n_2 \times 400 n2=4800400=12n_2 = \frac{4800}{400} = 12
  4. Conclusion: The number of fringes observed with the new wavelength is 12.
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