Question

Light of wavelength 530 nm is incident on two slits that are spaced 1.0 mm apart. If each of the slits has a width of 0.10 mm, how many interference maxima lie within the central diffraction peak?

Answer 1

Solution:

Answer : 21

Fringe width = y = D/d where = wavelength = 650 x 10^-9 m

Distance to the screen = L

Slit width = w = 0.10 mm

Slit separation = d = 1mm = 10 * w

y_D = L/w and interference maxima y₁ = m L / d

To see how many bright fringes are in the central diffraction envelope, we have to equate y_D to y₁ .

m L / d =   L / w =>m / 10 w = 1/w

m = 10 .

This implies that 10th interference maximum will coincide with the 1st diffraction minimum.

Taking the number on either side of the central maximum, m= 10 x 2 =20

Within the first Diffraction envelope there are 20 +1 = 21 interference maxima visible .