In the two-slit interference experiment, how can we determine where a point P on the screen is in the diffraction pattern by giving the maximum or minimum on which it lies or the maximum and minimum between which it lies. Given that the slit widths are each $a$, their separation is $d$, the wavelength is $\lambda$, the viewing screen is at a distance $D$ and the distance between P and the central maxima is $y$.
I am not sure if we use the formula $\arcsin\theta=m\lambda$ to calculate $m$, since I don't know whether this formula can be used in double slit diffraction. Or we should use another method to find out whether P is the maximum/minimum?
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3$\begingroup$ Look at the answer to this question Young's double split experiment, the slit width. $\endgroup$– FarcherCommented Mar 29, 2019 at 13:02
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2 Answers
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For Young's Double Slit Experiment the $n^{th}$ order maxima is given by $$Y_{n}=\frac{n\lambda D}{d}$$ where D is the distance to the viewing screen and d the spacing between the two slits.
For the $n^{th}$ order minima we use $$Y_{n}=\frac{(2n-1)\lambda D}{2d}$$
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2$\begingroup$ But what I want to know is the position of P in the diffraction pattern, not the interference pattern. $\endgroup$– RLeeCommented Mar 29, 2019 at 10:51
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1$\begingroup$ Y gives us the distance of point P from the centre of the screen. Whether there is constructive or destructive interference can be known from which multiple of (lambda)D/d it is. $\endgroup$– NikitaCommented Mar 30, 2019 at 10:10
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With your formula an interference maximum occurs when, m, is an integer. With a flat screen: y/D = tan(θ). (For small angles sin(θ) is close to tan(θ).) The interference pattern will be modulated by single slit diffraction.
