$\beta=\frac{\lambda D}{d} \quad \therefore \quad \lambda_2>\lambda_1 \quad \text { so } \quad \beta_2>\beta_1$
No of fringes in a given width $( m )=\frac{ y }{\beta} \Rightarrow m _2< m _1$
$3^{\text {rd }}$ maximum of $\lambda_2=\frac{3 \lambda_2 D }{d}=\frac{1800 D }{ d }$
$5^{\text {th }}$ minimum of $\lambda_1=\frac{9 \lambda_1 D }{2 d }=\frac{1800 D }{ d }$
So, $3^{\text {rd }}$ maxima of $\lambda_2$ will meet with $5^{\text {th }}$ minimum of $\lambda_1$
Angular sepration $=\frac{\lambda}{ d } \Rightarrow$ Angular seperation for $\lambda_1$ will be lesser