# (a) Why cannot the phenomenon of interference be observed by illuminating two pinholes with two sodium lamps?(b) Two monochromatic waves have displacements $y_{1}= a\cos \omega t$ and $y_{2}= a\cos \left (\omega t + \phi \right )$ from coherent sources interfere to produce an interference pattern. Derive the expression for the resultant intensity and obtain the conditions for constructive and destructive interference.(c) Two wavelengths of sodium light of 590nm and 596 nm are used in turn to study the diffraction taking place at a single slit of aperture $2\times 10^{-6}m.$ If the distance between the silt and the screen is 1.5m, calculate the separation between the positions of the second maxima of diffraction pattern obtained in two cases.

(a) The phenomenon of interference can't be observed by illuminating two pinholes with sodium lamps because the two independent sources can't be coherent.

(b)Let,

$y_{1}= a\cos \omega t$

$y_{2}= a\cos \left (\omega t + \phi \right )$

where y1 and y2 are displacements.

So, the resultant displacement is given by;

$y = y_{1}+y_{2}$

$y = a \cos \omega t+ a\cos \left (\omega t + \phi \right )$

$y = 2a \cos\left ( \frac{\phi}{2} \right ) \cos \left (\omega t + \frac{\phi}{2} \right )$

The amplitude of the resultant displacement is $y = 2a \cos\left ( \frac{\phi}{2} \right )$ and therefore intensity at that point will be:

$I = 4I_{0}\cos ^{2}\left (\frac{\phi}{2} \right )$

Now,

for constructive interference; $\phi =0, \pm 2\pi ,\pm 4\pi,....$

for destructive interference; $\phi =\pm \pi, \pm 3\pi ,\pm 5\pi,....$

(c) The position of second maxima is given by, $y_{1}= \frac{5}{2}\frac{\lambda D}{a}$

$\Delta y= \frac{D (\lambda_{2}-\lambda_{1})}{2a}$

where D = distance between the slit and screen = 1.5

$\lambda _{1 }and \lambda _{2}$ are wavelengths of sodium lights, $\lambda _{1 }= 590, \lambda _{2}=596$

a = aperture = $2\times 10^{-6}$

Hence,

$\Delta y = \frac{5\times 1.5\times (596 -590)\times 10^{-9}}{2\times 2\times 10^{-6}} = 11.25\times 10^{-3}m$

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