Question

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

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

Answer 1

The correct option is C x (blue) $<$ x (green)

Distance of $n^{th}$ bright fringe $y_n=\frac{n\lambda D}{d}$
i.e. $y_n=\alpha \lambda$
$\therefore \frac{x_b}{x_g}=\frac{{\lambda }_b}{{\lambda }_g}$
$\Rightarrow \frac{x_b}{x_g}=\frac{4360}{5460}$
$\therefore x_b<x_g$