Interference fringes from sodium light (1 = 5890 Å) in a double slit experiment have an angular width 0.20°. To increase the fringe width by 10%, wavelength of light used should be
Answer & explanation
Correct answer: option 4
$β=\frac{λD}{d}$ and angular fringe width, $θ=\frac{β}{D}=\frac{λ}{d}$
$θ_1=λ_1/d,\,θ_2=λ_2/d$
$∴\frac{θ_1}{θ_2}=\frac{λ_1}{λ_2}$ or $λ_2=λ_1.\frac{θ_2}{θ_1}$
$∴5890×\frac{0.22}{0.20}=6479 Å$