Is the formula, for finding the distance between two slits, my books gives correct? In my book'a section on Young's double-slit experiment, the formula, $d = m \lambda \sin\theta$, is given. In this equation $d$ is the distance between two slits, $\lambda$ is the wavelength of light coming through the slits, and $\theta$ is the angle between the central reference to the brightest maximum on the screen opposite the slits.

I am assuming this formula's derivation involves some degree of approximation, because another formula in the same section assumes the distance between the slit and the screen is similar in length to the hypotenuse the picture above.
Using the same approximation, I got something similar but not the same:
$$d = \frac{m\lambda}{\sin\theta} \, .$$ 
Which result is correct??
 A: If the slits are on top of each other, then the light travelling through each slit goes the same distance and therefore has the same phase.
In this case, the distance between the fringes is infinite.
On the other hand, if the slits are very far apart, then even a small angle incurs a large path difference, so the fringes are very close together.
Thus we have reasoned that the distance between the fringes goes down as $d$ goes up.
Consider the formula written by OP:
$$d = \frac{m \lambda}{\sin \theta} \, .$$
The first intensity maximum occurs when $m=1$ giving
$$ \sin \theta_1 = \frac{\lambda}{d} \, .$$
Expanding the $\sin$ to lowest order we get
$$\theta_1 = \frac{\lambda}{d}$$
which says that increasing $d$ makes the angle of the first maximum smaller, as we predicted above.
From this reasoning, we see that OP's formula is probably correct and that putting the $\sin$ in the numerator would give the wrong behavior.
Note that I didn't actually derive the correct answer, I just showed that moving the $\sin$ function from denominator to numerator would probably be incorrect.
That said, given the definitions in the question, the correct formula for the maxima of the two slit interference is in fact
$$d = \frac{m \lambda}{\sin \theta}$$
as written by OP.
