Answer
Therefore, every fourth bright fringe is eliminated.
Work Step by Step
If a bright interference fringe occurs at a diffraction minimum, then the bright fringe will vanish.
The angular locations of the bright fringes of the double-slit interference pattern are
given by
$d\sin\theta=m_i\lambda$
or, $\sin\theta=\frac{m_i\lambda}{d}............(1)$
The angular locations of the diffraction minima are given by
$a\sin\theta=m\lambda$
or, $\sin\theta=\frac{m_d\lambda}{a}............(2)$
Equating eq. 1 and eq. 2, we obtain
$\frac{m_i\lambda}{d}=\frac{m_d\lambda}{a}$
Again, applying the condition obtained in part (a) of this question: $d=4a$, we get
$\frac{m_i\lambda}{4a}=\frac{m_d\lambda}{a}$
or, $m_i=4m_d$
The result indicates that, for first, second, third, fourth........... diffraction minima, the 4th, 8th, 12th, 16th..........bright fringes will be disappeared respectively.
Therefore, every fourth bright fringe is eliminated.
