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
Answer & explanation
Correct answer: option 1
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