Question

In a YDSE setup, the distance between the plane of the slits and the screen is increased by $20\%$ and the distance between the slits is increased by $50\%$. Then the percentage change in fringe width is

• A. $80\%$
• B. $20\%$
• C. $66\%$
• D. $44\%$

Answer 1

The correct option is B $20\%$

$D^{\prime }D+\frac{20}{100}D=1.2D;d^{\prime }=d+\frac{50}{100}d=1.5d$
$\therefore {\beta }^{\prime }\frac{D^{\prime }\lambda }{d^{\prime }}=\frac{D\left(1.2\right)\lambda }{d\left(1.5\right)}0.8\lambda =0.8\beta$
$\therefore Percentagecharge=(1-\frac{{\beta }^{\prime }}{\beta })100=$
$=(1-0.8)\times 100=20/$