Double slit interference question a level physics The question attached asks you to calculate the angle between the central fringe and the second bright fringe. The mark scheme says that you should use tan to work out this angle using the distance between the slits and the plane and the fringe separation. But I'm wondering why you can't use dsinθ  = nλ? Using the second formula, I get 0.19º (to 2 s.f. as is required in the question), but the mark scheme is looking for 0.18º. Why is it wrong to use dsinθ = nλ?


 A: It's actually correct (and, IMHO, better) to use the equation you used.  The problem solution uses a couple of approximations that are basically correct but lead to small enough errors that the final answer differs at 2 significant figures.
Using $s \sin \theta = n \lambda$ as you did, you would obtain $\theta = 0.1865...^\circ$.  This is a direct calculation of the angle, and does not rely on any trigonometric approximations (or at least none that significantly affect the answer.)
The equations used in the solution, though, relies on a couple of approximations that introduce small errors.  Specifically, we have the distance between the first bright fringe and the central spot (call it $w_1$) would satisfy
$$
\frac{w_1}{D} = \tan \theta \approx \theta \qquad \frac{\lambda}{s} = \sin \theta \approx \theta \\
\Rightarrow \frac{w_1}{D} \approx \frac{\lambda}{s}.
$$
This introduces a small error relative to the exact solution, though it's pretty close to being true.
They then assume that the displacements between the consecutive fringes are evenly spaced, and in particular that the displacement of the second fringe (let's call it $w_2$) is exactly twice that of the first fringe.  But this isn't exactly true either.  what is instead true is that
$$
\sin \theta_2 = n \frac{\lambda}{s} = 2 \sin \theta_1.
$$
If $\theta_1$ and $\theta_2$ are both "small", then this reduces to $\theta_2 \approx 2 \theta_1$, which then implies (under the same approximations) that $\tan 
\theta_2 \approx 2 \tan \theta_1$, which then implies that $w_2 \approx 2 w_1$.  So while it is true that $w_2$ is pretty close to twice $w_1$ for small angles, it's not exactly equal to twice $w_1$.
The net effect of these approximations is to change the answer in the solution by about 1% relative to the answer that you calculated directly.  This is enough to change it from 0.19° to 0.18° after rounding to two significant figures.
The moral of this story is that approximations can be useful in simplifying your life, but you should always keep in mind that you have made them, and remember that your answers are only approximately true.  The folks who wrote out that answer key (and anyone who blindly relied on it to mark an assignment without keeping it in mind) apparently forgot that.
