Question

A beam of coherent light of wavelength 550 nm is incident normally to the plane of a pair of two slits S₁ and S₂ each of width 1.2 × 10⁻⁶ m separated by 1.1 mm. Dark and bright fringes are observed on a screen 2.2 m away from the plane of the slits.

Calculate:

1. Fringe width.
2. Distance of second dark fringe from central maximum.
3. What will happen when the entire apparatus is immersed in water?

Answer 1

Given: Wavelength of light (λ) = 550 nm

= 550 × 10⁻⁹ m

Slit separation (d) = 1.1 × 10⁻³ m

Distance of screen (D) = 2.2 m

Slit width (a) = 1.2 × 10⁻⁶ m

(I) Fringe Width (β) = `(λD)/d`

= `(550 xx 10^(−9) xx 2.2)/(1.1 xx 10^(-3))`

= `(1210.0 xx 10^(−9))/(1.1 xx 10^(-3))`

= `(1.21 xx 10^(−6))/(1.1 xx 10^(-3))`

= 1.1 × 10⁻³ m

= 1.1 nm

(II) Position of nth dark fringe (yn) = `(2n - 1)/2 β`

For the second dark fringe (n = 2).

= `((2 xx 2 - 1))/2 β`

= `((4 - 1))/2 β`

= `3/2 β`

= 1.5 × 1.1

= 1.65 nm

(III) When immersed in water, the speed of light decreases, which reduces the wavelength:

`λ_"water" = λ_"air"/μ`   ...(where μ ≈ 1.33 for water)

Since fringe width (β) is directly proportional to wavelength, the fringe width decreases.

Thus, the new fringe width:

β' = `β/μ`

= `1.1/1.33`

= 0.827 nm