Question

The Young's experiment is performed with the lights of blue $(\lambda =4360\mathring{A})$ and green colour $(\lambda =5460\mathring{A}),$ 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. $x\left(green\right)<X\left(blue\right)$
• D. $\frac{x\left(Blue\right)}{x\left(Green\right)}=\frac{5460}{4360}$

Answer 1

The correct option is A $x\left(Blue\right)=x\left(Green\right)$

Distance of $n$ bright fringe $y_n=\frac{n\lambda D}{d}$ i.e, $y_n\propto \lambda$

$\therefore \frac{x_{n_1}}{x_{n_2}}=\frac{{\lambda }_1}{{\lambda }_2}\Rightarrow \frac{x\left(Blue\right)}{x\left(Green\right)}=\frac{4360}{5460}$

$\therefore x\left(Green\right)>x\left(Blue\right)$