In a double slit experiment, the distance between slits is increased 10 times whereas their distance from screen is halved, then what is the fringe width?

become $\frac{1}{20}$

The correct answer is Option (3) → become $\frac{1}{20}$

The fringe width (β),

$β=\frac{λD}{d}$

where,

λ = Wavelength of light

D = Distance between the slits and screen.

d = distance between the slits

$β_{initial}=\frac{λD}{d}$

$β_{new}=\frac{λD'}{d'}=\frac{λ(D/2)}{10d}=\frac{β_{initial}}{20}$

The fringe width (β) becomes $\frac{1}{20}th$.