Question

The Young's experiment is performed with lights of blue ($\lambda =4360\AA$) and green colour ($\lambda =5460\AA$). If the distance of the $4^{th}$ fringe from the centre is $x$, then

• A. $x\left(Blue\right)>x\left(Green\right)$
• B. $x\left(Blue\right)<x\left(Green\right)$
• C. $x\left(Blue\right)=x\left(Green\right)$
• D.
  $\frac{x\left(Blue\right)}{x\left(Green\right)}=\frac{5460}{4360}$

Answer 1

The correct option is B $x\left(Blue\right)<x\left(Green\right)$

We know that
Distance of the $n^{th}$ bright fringe $y_n=\frac{n\lambda D}{d}$
which means
$y_n\propto \lambda \therefore \frac{x\left(Blue\right)}{x\left(Green\right)}=\frac{{\lambda }_1}{{\lambda }_2}\Rightarrow \frac{x\left(Blue\right)}{x\left(Green\right)}=\frac{4360}{5460}\therefore x\left(Green\right)>x\left(Blue\right)$