Explanation

(b) For possible interference maxima on the screen, the condition is
$d\sin{\theta = n\lambda}$ …(i)
Given : $d =$ slit-width $= 2\lambda$
$\therefore\ \ \ 2\lambda\sin\theta = n\lambda$
$\Rightarrow \ 2\sin\theta = n$
The maximum value of $\sin\theta$ is 1, hence, $n = 2 \times 1 = 2$
Thus, Eq. (i) must be satisfied by 5 integer values ie, -2 -1, 0, 1, 2. Hence, the maximum number of possible interference maxima is 5.