Answer
$4$
Work Step by Step
Here in a double-slit experiment, we have to find the largest ratio of d to a so that the fourth bright side fringe is eliminated by a diffraction minimum.
The angular locations of the bright fringes of the double-slit interference pattern are
given by
$d\sin\theta=m\lambda$
For, the fourth bright side fringe $(m=4)$
$d\sin\theta=4\lambda$
or, $\sin\theta=\frac{4\lambda}{d}\;.........(1)$
The angular locations of the diffraction minima are given by
$a\sin\theta=m\lambda \;..........(2)$
To eliminate the fourth bright side fringe by a diffraction minimum, the value of $\theta$ in eq. 2 has to be equal to the value of $\theta$ in eq. 1
Substituting eq. 1 in eq. 2, we obtain
$a\times \frac{4\lambda}{d}=m\lambda$
or, $\frac{d}{a}=\frac{4}{m}$
For the largest ratio of d to a, the value of m should be 1.
Therefore,
$$\boxed{\frac{d}{a}=4}$$
