(a) Let S be a narrow slit illuminated by a monochromatic source of light and
S1 and
S2 two similar parallel slits very close together and equidistant from S.
Displacement of the wave produced by
S1 is given by
y1=a cos ωt and the displacement of the wave produced by
S2 is given by
y2=a cos (ωt+ϕ) The resultant displacement is given by
y=y1+y2 y=a cos ωt+a cos(ωt+ϕ) y=2a cos ϕ/2 cos(ωt+ϕ/2) The amplitude of the resultant displacement is
2a cos ϕ/2. The intensity of light is directly proportional to the square of amplitude of the wave.
The resultant intensity is given by
I=4a2 cos2 ϕ/2 For constructive interference the intensity of light at point P is maximum, if
cos ϕ=max=+1:ϕ=0, 2π, 4π.... ϕ=2nπ where n=0, 1, 2, .....
Path difference
=λ2π×2nπ=nλ For distructive interference the intensity of light at point P is minimum, if
cosϕ=−1 ∴ϕ=π, 3π, 5π, ..... or
ϕ=(2n−1)π where n=1, 2, 3, ....
The corresponding path difference between the two waves
x=λ2πϕ=λ2π(2n−1)π=(2n−1)λ2 (b) (i) As the width of the slits is increased, the fringe width decrease. It is because,
β∝1d (ii) The different colours of white light will produce different interference patterns but the central bright fringes due to all colours are at the same positions. Therefore, the central bright fringe is white in colour. Since the wavelength of the blue light is smallest, the fringe closed on the either side of the central white fringe is blue and farthest is red. Beyond a few fringes, no clear fringe pattern will be visible.