Q. The Young’ experiment is performed with the lights of blue $(\lambda = 4360\ \mathring{\mathrm{A}})$ and green colour $(\lambda = 5460\mathring{\mathrm{A}})$, if the distance of the 4th fringe from the centre is $x$, then
A) $x$ (Blue) $= x$ (Green)
B) $x(Blue) > x(Green)$
C) $x(Blue) < x(Green)$
D) $\frac{x(Blue)}{x(Green)} = \frac{5460}{4360}$
Explanation
(c) Distance of $n^{th}$ bright fringe $y_{n} = \frac{n\lambda D}{d},\ i.e.,\ y_{n} \propto \lambda$
$\therefore\frac{x_{n_{1}}}{x_{n_{2}}} = \frac{\lambda_{1}}{\lambda_{2}} \Rightarrow \frac{x(Blue)}{x(Green)} = \frac{4360}{5460}$
$\therefore x(Green) > x(Blue)$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.