# Three slit minima and maxima and path differences

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1. Nov 15, 2016

### nso09

1. The problem statement, all variables and given/known data
Find at which angles θ the interference picture that appears on a distant screen
made by three thin slits separated by distance d and enlightened by a source of wavelength λ (see figure)

a) Shows its maxima.
b) Shows its minima.

multiple slit diffraction, d, $\lambda$, $\theta$
2. Relevant equations
$$dsin\theta=m\lambda$$
$$dsin\theta=(m\lambda)/3$$
$$E(t)=E_0cos(k(r_2-r_1)-\omega(t))$$
$$t=top,m=middle,b=bottom$$

3. The attempt at a solution
$$E_t=E_0cos(k(r_t-r_b)-\omega(t))$$
$$E_m=E_0cos(k(r_t-r_m)-\omega(t))$$
$$E_b=E_0cos(-\omega(t))$$

However, the solutions gives me something like this.

My thought process was that the top ray travels the most or has the biggest path difference. I don't get why the solutions give me these equations. Or am I getting my path differences mixed up? Thanks for the help.

2. Nov 15, 2016

I don't know how advanced your textbook is, but if you have the formula for the interference pattern for $N$ slits (It works for any integer N that is 2 or greater): $I(\theta)=I_o \frac{\sin^2(N \phi/2)}{\sin^2(\phi/2)}$ where $\phi= (\frac{2 \pi}{\lambda}) d \sin{\theta}$, you can use it to solve for $I(\theta)$ for the case where $N=3$. $\\$ The zero's of the function $I(\theta)$ occur when the numerator is equal to zero, except when the denominator is also zero. When the denominator is also zero, along with the numerator, you get a maximum and $I(\theta_{max})=I_o N^2$.(You can take the limit where $\theta$ approaches that angle. At that precise angle, the expression is undefined, but assumed to be the same value as the limit.) $\\$ I will let you try to work out this result assuming you are allowed to use this formula. It's a simple matter to determine the angles $\theta$ for which the numerator and/or the denominator is equal to zero. $\\$ Note: If you wanted to work with sinusoidal $E$ fields, instead of simply using the formula for $I(\theta)$, $\phi$ would be the phase of the middle slit, with the top slit at zero (relative phase), and the bottom slit at $2 \phi$. The algebra for the case of just 3 sources isn't terribly complex (using trigonometric identities to compute the sums, etc.), but it is really easier to just use the general formula given above.

Last edited: Nov 15, 2016
3. Nov 16, 2016

One additional comment is the formula for the intensity for N slits is derived by first summing sinusoidal E fields using complex variables ($exp(i \omega t)$ etc.,) and summing the geometric series. For N=2 and maybe for N=3, it is possible to do this calculation using $cos( \omega t)$, but for anything larger, it is much easier to do this calculation with complex variables.