Question

In Young's double-slit interference experience separation is made 3 fold. The fringe width becomes

• A. $sixfold$times
• B. $threefold$times
• C. $3/6-fold$ times
• D. $1/3-fold$times

Answer 1

The correct option is D $1/3-fold$times

Given,

The distance between slits is $3$

Let the new fringes width be ${\beta }^{\prime }$

Then,

${\beta }^{\prime }=\frac{\lambda d}{3d}=\frac{1}{3}\beta$

Therefore, fringe width also becomes $\frac{1}{3}$ times.