Detecting the smoke

For the wave that pass through the smoke to undergo destructive interference with the wave that doesn't, the wave in the smoke has to undergo a path difference of half its wavelength( also known as a phase difference of ${\pi}$ ).

For a mathematical explanation (hope this makes things clearer), this is because $sin(x) + sin(x - \pi) = 0$

Therefore, we can equate the path difference to half its wavelength. $2 \pi$ radians corresponds to $\lambda$ therefore $\pi$ radians corresponds to $\frac{\lambda}{2}$ , so therefore we can get

$( {n}_{smoke} - {n}_{air} ){l} = \frac{\lambda}{2}$

We can immediately solve for the length $l$ by substituting $\lambda$ = 1000mm ${n}_{smoke}$ = 1.53

We should get $l$ = 943mm

Increase in the phase of a plane wave when it passes through a distance of $\Delta x$ in air is $\Delta \phi = \frac{2\pi \Delta x}{\lambda},$ where, here, $\Delta x = \ell$ to be determined and $\lambda = 1000 \rm{mm}$ .

For a passage through a medium of refractive index $n$ , $\lambda' = \frac{c'}{\nu} = \frac{c}{n\nu} = \frac{\lambda}{n}.$

Thus increase in the phase of a plane wave in the smoky air is $\Delta \phi' = \frac{2\pi \Delta x}{\lambda'} = \Delta \phi' = \frac{2\pi n \Delta x}{\lambda}.$

The phase difference must be equal to $\pi$ for the first order minimum: $\frac{2\pi \Delta x}{\lambda} (n-1) = \pi.$ This gives $\Delta x = \ell = \frac{\lambda}{2(n-1)}.$