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A pair of narrow, parallel slits separated by 0.275 mm is illuminated by the green component from a mercury vapor lamp (λ = 546.1 nm). The interference pattern is observed on a screen 1.25 m from the plane of the parallel slit. Calculate the distance from the central maximum to the first bright region on either side of the central maximum.

User Vidul
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1 Answer

2 votes

Answer:

2.481 mm

Step-by-step explanation:

Given that

Distance between the slits, d = 0.275 mm = 2.75*10^-4 m

Wavelength of the lamp, λ = 546.1 nm

Distance between slit and the screen, L = 1.25 m

To solve this question, we would be applying the formula for Young's double slit fringes, from the path difference equal to a whole number of wavelengths.

mλ = dy/L

The question states that we should find the distance from the centre to the 1st order bright fringe, thus, m = 1

Now, we apply all these values into the question

1 * 546.1*10^-9 = (2.75*10^-4 * y) / 1.25

1.25 * 1 * 546.1*10^-9 = 2.75*10^-4 * y

6.825*10^-7 = 2.75*10^-4 * y

y = 6.825*10^-7 / 2.75*10^-4

y = 0.002481 m or 2.481 mm

User Dramasea
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