At any point P on circumference, the path difference, $$\Delta$$x = d sin$$\theta$$

At P₁, $$\theta$$ = 0$$^\circ$$ $$\Rightarrow$$ $$\Delta$$x = 0

At P₂, $$\theta$$ = 90$$^\circ$$ $$\Rightarrow$$ $$\Delta$$x = d

For constructive interference, $$\Delta$$x = n$$\lambda$$

$$\therefore$$ $$n = {d \over \lambda } = {{1.8\,mm} \over {600\,nm}} = {{1.8 \times {{10}^{ - 3}}m} \over {600 \times {{10}^{ - 9}}m}} = 3000$$

Since, n is integer, hence P₂ corresponds to bright spot and also it corresponds to maximum order of fringe.

Now, $$\Delta$$x = d sin$$\theta$$

d$$\Delta$$x = d cos$$\theta$$ d$$\theta$$ or R$$\lambda$$ = d cos$$\theta$$ d$$\theta$$

$$\therefore$$ $$d\theta \propto {1 \over {\cos \theta }}$$

Hence, as $$\theta$$ increases, cos $$\theta$$ decreases and consequently d$$\theta$$ increases.