Single Slit Diffraction : Problem with the condition of diffraction The condition for diffraction is that the effects of diffraction become more noticeable when the width of aperture is comparable to the wavelength, and from that we can say the spacing of fringes become more significant when $d \approx \lambda$. 
Now taking the equation :
$$\lambda=\frac{d\sin\theta}{n}$$
where :
$d \to$ size of aperture
$\lambda \to$ wavelength of light used
$n = \{1,2,3,4....\}$
Since $|\sin \theta|$ lies between $0$ and $1$ , $\frac{n}{|\sin\theta|} > 1$ (because $n$'s least value is $1$ in the case of destructive interference) and it keeps increasing as the value of $n$ taken becomes larger, whereas $\frac{d}{\lambda} = 1$. 
Isn't this a contradiction?
Edit : Elaborating on what my doubt exactly is
I took three cases where $d \approx \lambda$ (considering only the dark fringes for simplicity):

*

*$d = \lambda$ :- In this case, for the first dark fringe, $n = 1$, so (as $d$ and $\lambda$ cancel each other) $\sin(t) = 1$, hence $t = 90$. Further increase in $n$ (to $2,3,4.....$) contradicts the relationship as $\sin(t)$ cannot increase further.


*$d < \lambda$ (slightly less than) :- as $n$'s minimum value is $1$ and $|\sin \theta|$ lies between $0$ and $1$, $d$ cannot be less than $\lambda$, so our assumption is wrong.


*$d > \lambda$ (slightly greater than) :- This works for a few fringes at most (if $d \approx \lambda$ still stands). Since $d$ is only slightly greater than $\lambda$, $\frac{n}{\sin(t)}$ should only be slightly more than $1$. But as we put in higher and higher values of $n$ (say multiply $n$ to $5\times$ its initial value), to accommodate the same change in $\sin(t)$, our starting value of $\sin(t)$ keeps going down (highest possible value in the example case being $\sin(t) = 0.2$, because the maximum value of $\sin(t)$ is already fixed). So from this we get (on the other side of the equation), $d = 5\cdot\lambda$ (and this keeps increasing as the $n$ value we want to have increases), which again contradicts the condition saying $d \approx \lambda$.
So even in the third case, limiting to only a certain number of fringes, contradicts the condition.
Are my conclusions correct? If they are could you explain how the condition for diffraction fits in?
 A: Let's think of it this way.
For a given setting of the apparatus, $\frac{d}{\lambda}$ is constant. So the other ratio,  $\frac{n}{\sin{\theta}}$, must be governed by this constraint because if our analytical analysis (the equations) is correct then this constraint property must be respected by the other ratio.
So when n increases $\theta$ increases and consequently, as you know, $\sin{\theta}$ increases. However, the overall effect of this on the ratio is that it stays the same since both the numerator and denominator are increasing.
A: Well after reading more on it, I realized the third condition I mentioned is right, but my interpretation of comparable was wrong. I assumed that comparable meant the values of aperture size and wavelength have to be close to each other. But what the condition was actually stating was that the order of λ and d are comparable (and since λ is of the order $10^{-9}$ , the difference in order is allowed even upto 6)
So the starting assumption "d≈λ" was were I went wrong. 
