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In a certain double slit experimental arrangement, interference fringes of width 1.0 mm each are observed when light of wavelength 5000 Å is used. Keeping the set-up unaltered if the source is replaced by another wavelength of 6000 Å, the fringe width will be -
  • a)
    0.5 mm
  • b)
    1.00 mm
  • c)
    1.2 mm
  • d)
    1.5 mm
Correct answer is option 'C'. Can you explain this answer?
Most Upvoted Answer
In a certain double slit experimental arrangement, interference fringe...
To calculate the distance between the slits (d), we can use the formula for the fringe width (Δy):

Δy = (λ * D) / d

Where:
Δy = Fringe width (1.0 mm = 0.001 m)
λ = Wavelength of light (5000 Å = 5000 x 10^-10 m)
D = Distance from the double slit to the screen (unknown)
d = Distance between the slits (unknown)

Rearranging the formula, we have:

d = (λ * D) / Δy

Plugging in the values, we get:

d = (5000 x 10^-10 m * D) / 0.001 m

Simplifying:

d = 5 x 10^-6 D

So, the distance between the slits (d) is equal to 5 x 10^-6 times the distance from the double slit to the screen (D).
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In a certain double slit experimental arrangement, interference fringes of width 1.0 mm each are observed when light of wavelength 5000 Åis used. Keeping the set-up unaltered if the source is replaced by another wavelength of 6000 Å, the fringe width will be -a)0.5 mmb)1.00 mmc)1.2 mmd)1.5 mmCorrect answer is option 'C'. Can you explain this answer?
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