Question

In a Young's double slit experiment, $\lambda =500nm,d=10mmandD=1.0m$ . Find the minimum distance from the central maximum for which the intensity is half of the maximum intensity.

Answer 1

Given :

$D=1m$

$d=10mm={10}^{-2}m$

$l=500nm=500\times 10⁻⁹m=5\times 10⁻⁷m$

For intensity to be half the maximum intensity :

$y=\frac{\lambda D}{4d}$

$=\frac{5\times {10}^{-7}\times 1}{4\times {10}^{-2}}$

$y=1.25\times {10}^{-5}m$

$\therefore$ The minimum distance from the central maximum for which the intensity is half of the maximum intensity is $y=1.25\times {10}^{-5}m$