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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.

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