\(5^{\text {th }} \text { maximum }=\frac{5 \lambda D}{d}\)

\(3^{\text {nd }} \text { minimum }=\frac{(2 n-1) \lambda . D}{2 d}=\frac{5 \lambda D}{2 d}\)

\(\therefore \quad \text { Distance }=\frac{5 \lambda D}{d}-\frac{5 \lambda D}{2 d}\)

\(=\frac{5 \lambda D}{2 d}\)

\(\therefore \text { Distance }=2.5 \text { times }\)