Question

You may use the figure below as a guide.
Starting from:

You may use the figure below as a guide. Starting from: Psi = 2pi deltar/ lambda + gradient psi 0 A) Derive an equation for theytam the angle distance from the central fringe to both the constructive (bright) and destructive (dark) fringes in a double-slit spectrum. You may use small angle approximation. B)Now also derive equations for ym, the distance from the central maximum to both the constructive (bright) and destructive (dark) fringes in a double-slit spectrum.

A) Derive an equation for theyta_m the angle distance from the central fringe to both the constructive (bright) and destructive (dark) fringes in a double-slit spectrum. You may use small angle approximation.

B)Now also derive equations for y_m, the distance from the central maximum to both the constructive (bright) and destructive (dark) fringes in a double-slit spectrum.

Answer 1

$(a) \phi=\frac{2\pi \Delta r}{\lambda}+\Delta \phi_0 \par \phi-\Delta \phi_0= \frac{2\pi \Delta r}{\lambda} \par \Delta \phi= \frac{2\pi dsin \theta}{\lambda}\par For constructive interference \par \Delta \phi=2n\pi \par So \par d sin\theta =n \lambda \par For small theta\par \theta =\frac{n \lambda}{d}\par$

$\par For destructive interference \par \Delta \phi=2(n+\frac{1}{2})\pi \par So \par d sin\theta =(n+\frac{1}{2}) \lambda \par For small theta\par \theta =(n+\frac{1}{2})\frac{ \lambda}{d}\par$

(b) As can be seen from the figure

$tan \theta=\frac{y}{L}\par For small \theta\par tan\theta=\theta \par So \par \theta=\frac{y}{L}\par For constructive interference$

$\frac{y}{L}= n\frac{ \lambda}{d} \par$

$y=\frac{nL \lambda}{d} \par For destructive interference$

$\frac{y}{L}= (n+\frac{1}{2})\frac{ \lambda}{d} \par$

$y= (n+\frac{1}{2})\frac{ L\lambda}{d}$