Question

Light consisting of two wavelengths 600 nm and 480 nm is used to obtain interference fringes in a double slit experiment. The screen is placed 1.0 m away from slits which are 1.0 nm apart.

1. Calculate the distance of the third bright fringe on the screen from the central maximum for wavelength 600 nm.
2. Find the least distance from the central maximum where the bright fringes due to both the wavelengths coincide.

Answer 1

Given: Wavelengths: λ₁ = 600 nm = 600 × 10⁻⁹ m and 

λ₂ = 480 nm = 480 × 10⁻⁹ m

Distance between slits (d) = 1.0 mm = 1.0 × 10⁻³ m

Distance from slits to screen (L) = 1.0 m

1. Distance of the third bright fringe for λ₁ = 600 nm

The fringe width (y) = `(m lambda L)/d`

For the third bright fringe (m = 3):

y₃ = `(3 xx (600 xx 10^-9) xx 1)/(1.0 xx 10^-3)`

= `(3 xx 600 xx 10^-9)/10^-3`

= `(1800 xx 10^-9)/10^-3`

= 1.8 mm

2. Least distance from the central maximum where bright fringes coincide:

Bright fringes from both wavelengths will coincide at a position where the order numbers m₁ and m₂ satisfy:

m₁λ₁ = m₂λ₂

The LCM of the wavelengths gives the least position where both fringes coincide, i.e., LCM (600, 480) = 2400 nm = 2.4 × 10⁻⁶ m.

Now, the fringe distance for this position:

y = `(2.4 xx 10^-6 xx 1)/10^-3`

= `(2.4 xx 10^-6 xx 1)/10^-3`

= 2.4 mm