Question

In a young's double slit experiment $\lambda =500$ nm , d = 1.0 mm and D = 1.0 m. Find the minimum distance frm the central maximum fro which the intensity is half of the maximum intensity.

Answer 1

Given that, $D=1m,d=1mm={10}^{-3}m,l=500nm=5\times {10}^{-7}m$
For intensity to be half the maximum intensity,
$y=\frac{\lambda D}{4d}\left(\text{As in problem No. -32}\right)\Rightarrow yo=\frac{5\times {10}^{-7}\times 1}{4\times {10}^{-3}}=1.25\times {10}^{-4}m.$