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Light of wavelength 680 nm falls on two slits and produces an interference pattern in which the third-order bright red fringe is 50 mm from the central fringe on a screen 2.2 m away. What is the distance between the two slits?

1) 0.034 mm
2) 0.068 mm
3) 0.136 mm
4) 0.272 mm

User Sardoan
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1 Answer

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Final answer:

To find the distance between slits in a Young's double-slit experiment where the third-order fringe is 50 mm from the central fringe, and the screen is 2.2 m away, we used the formula d = (mλL) / x and calculated it to be approximately 0.09088 mm. None of the given options exactly match, suggesting a possible error.

Step-by-step explanation:

To determine the distance between the two slits in a double-slit experiment, we can use Young's interference equation:

The equation is d sin(θ) = mλ, where d is the distance between the slits, θ is the angle of the fringe with respect to the central maximum, m is the order of the fringe, and λ is the wavelength of the light.

First, we calculate the angle θ using the formula tan(θ) = x/L, where x is the distance of the fringe from the central maximum and L is the distance from the slits to the screen.

Assuming tan(θ) ≈ sin(θ) for small angles, θ ≈ x/L.

For the third-order (m=3) bright fringe, where λ = 680 nm (680 x 10^-9 m), x = 50 mm (50 x 10^-3 m), and L = 2.2 m, we get:

d = (mλL) / x = (3 x 680 x 10^-9 m x 2.2 m) / (50 x 10^-3 m) ≈ 0.09088 x 10^-3 m or 0.09088 mm.

Comparing this result with the given options, none match exactly, but option (2) 0.068 mm is the closest choice. However, if there is an error in the options or typing, we would refuse to answer. Assuming an option was mistyped, the correct answer might be (1) 0.09088 mm.

User Djra
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