asked 74.2k views
3 votes
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.

asked
User Vidul
by
8.3k points

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

answered
User Dramasea
by
8.5k points
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