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.

2.481 mm

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

the entire statement is true and correct.  

there's no diagram, nothing is illustrated, and no question is asked.

i am pretty sure it is c because when people are closer together, they cannot move as freely. picture being in a mostly full elevator, for instance. the people are right next to you, and you can barely move.