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?
Answer & explanation
Correct answer: option 3
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 $\frac{1}{20}th$.