# What frequency range can be tolerated with Young's fringes? (Problem 9.4, Brooker, Modern Optics)

I am having problems with part 2 but provided part 1 for context - I believe I have part 1 correct but could be mistaken.

Let a Young-slits apparatus have slits separated by d. Let the incoming light contain frequencies in range $$\Delta\nu$$ centred on $$\nu_0$$. We wish to be able to see clearly ten fringes, five each side of the pattern's centre.

1) Suppose that we set the following criterion for "seeing clearly": in the vicinity of the fifth fringe, where $$d\sin\theta=p\lambda_0=pc/\nu_0$$ with $$p=5$$, the order of interference $$p$$ shall be permitted to cover a range $$\frac 12$$. Show that this permits $$\Delta \nu/\nu_0 = 1/10$$

Ans: We are told the path difference can vary by half an order $$p$$ or half a wavelength in terms of $$\lambda_0$$. The way I thought about it was: The actual path length from the slits to the fringe for $$p=5$$ is defined by $$\theta$$ ($$PD=5\lambda_0=5c/\nu_0$$) but after a change in frequency/wavelength the wave travels an extra (or less) distance of half wavelength of the original $$\lambda_0$$ relative to the original light. So the change in frequency gives rise to an accumulation of $$\pi$$ radians phase over the original path to the $$p=5$$, $$\lambda_0$$ fringe. In terms of wavenumbers:

$$\Delta k \cdot PD=\pi$$ $$2\pi\frac{\Delta\nu}{c} \cdot \frac{5c}{\nu_0} = \pi$$ $$\frac{\Delta\nu}{\nu_0}=\frac{1}{10}$$

2) The condition set in part (1) is unduly pessimistic. Show that the fringe contrast in the vicinity of the fifth bright fringe is zero if the range permitted to $$p$$ is 1. Show that this permits $$\Delta\nu/\nu_0=1/5$$ -- the hint given here asks one to "remember the elementary way for locating the first zero in the diffraction pattern due to a single slit"

-- I do not understand how to show the contrast is zero near the vicinity of the fifth bright fringe - nor really what this means, I imagine it is something to do with the hint, I have seen this elementary example of rays cancelling across either side of the slit but don't see how to apply it... Please do not show me how to plug in p=1 and obtain the 1/5 ratio.