The fringe width in Young's double slit experiment is $2.4 × 10^{-4} m$, when red light of wavelength 6400 Å is used. The decrease in the fringe width, if light of wavelength 4000 Å is used instead of red light, will be

$0.9 × 10^{-4} m$

The correct answer is Option (1) → $0.9 × 10^{-4} m$

Given:

Fringe width, $β_1 = 2.4 \times 10^{-4}\,m$

Wavelength of red light, $λ_1 = 6400\,Å = 6.4 \times 10^{-7}\,m$

Wavelength of new light, $λ_2 = 4000\,Å = 4.0 \times 10^{-7}\,m$

Formula:

$\frac{β_2}{β_1} = \frac{λ_2}{λ_1}$

Substitute:

$β_2 = β_1 \times \frac{λ_2}{λ_1} = 2.4 \times 10^{-4} \times \frac{4.0 \times 10^{-7}}{6.4 \times 10^{-7}}$

$β_2 = 2.4 \times 10^{-4} \times 0.625$

$β_2 = 1.5 \times 10^{-4}\,m$

Decrease in fringe width:

$Δβ = β_1 - β_2 = (2.4 - 1.5) \times 10^{-4}$

$Δβ = 0.9 \times 10^{-4}\,m$

Final Answer: Decrease in fringe width = 9 × 10⁻⁵ m