Question

If the distance between the first maxima and the central fringe in Young's double-slit experiment is 2 mm, what is the distance between the third minima and the central fringe?

A) 4 mm
B) 6 mm
C) 8 mm
D) 10 mm

Answer 1

The distance between the third minima and the central fringe in Young's double-slit experiment is three times the distance of the first maxima to the central fringe, which results in 6 mm.

Step-by-step explanation:

The distance between the first maxima and the central fringe in Young's double-slit experiment correlates with the position of other maxima and minima according to the interference pattern. The position of a minima in the interference pattern is given by d × sin(θ) = m × λ, where d is the slit separation, λ is the wavelength, θ is the angle of incidence, and m is the order number of the minima (m is a half-integer for minima). Since the third minima would be one and a half orders higher than the first maxima, and assuming the distance to the screen and the wavelength of light remain consistent, the distance between the third minima and the central fringe would be 3 times the distance of the first maxima from the central fringe. Therefore, the correct answer is C) 6 mm.