Question

In a Young's double slit experiment, the separation between the two slits is d and distance of the screen from the slits is 1000 d. If the first minima falls at a distance d from the central maximum, obtain the relation between d and $\lambda$.

Answer 1

d = 500$\lambda$

Answer 2

The position of the $m$-th minima in Young's double-slit experiment is given by $y_m = \frac{(2m-1)\lambda D}{2d_{slit}}$. Given slit separation $d_{slit} = d$, screen distance $D = 1000d$, and the position of the first minima $y_1 = d$. For the first minima ($m=1$), the formula becomes $y_1 = \frac{\lambda D}{2d_{slit}}$. Substituting the given values: $d = \frac{\lambda (1000d)}{2d}$. Simplifying this equation yields $d = \frac{1000\lambda}{2}$, which results in the relation $d = 500\lambda$.