1. In what way is diffraction from each slit related to the inter… - Vidyadip

Question

In what way is diffraction from each slit related to the interference pattern in a double slit experiment?
Two wavelengths of sodium light $590\ nm$ and $596\ nm$ are used, in turn, to study the diffraction taking place at a single slit of aperture $2 \times 10^{–4} m$. The distance between the slit and the screen is $1.5 \ m$. Calculate the separation between the positions of the first maxima of the diffraction pattern obtained in the two cases.

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Answer

When a plane wavefront of monochromatic light illuminates, the slit $LN,$ each point in the slit $LN$ becomes the source of secondary wavelets. The secondary wavelets originating from different points superpose on each, while travelling towards the point $C$ and point $P;$ at angle$\theta$. However the superposition of the secondary wavelets produces a diffraction pattern of varying intensity, as shown in fig.

For maxima other than central maxima

$\theta = \bigg(\text{n} + \frac{1}{2}\bigg)\lambda$
and $\theta = \frac{\text{y}}{\text{D}}$
$\therefore\text{a}.\frac{\text{y}}{\text{D}} = \bigg(\text{n} + \frac{1}{2}\bigg)\lambda$

For light of wavelength $\lambda_{1}= 590 \ nm$
$2 \times10^{-14}\times\frac{\text{y}_{1}}{1.5} =\big(1+ \frac{1}{2}\big)\times590$
$\text{y}_{1} = \frac{3}{2}\times\frac{590\times10^{-9}\times1.5}{2\times10^{-4}}$
$= 6.64 \ mm$
For light of wavelength $\lambda_{2}=596 \ mm$
$2 \times10^{-4}\times\frac{\text{y}_{2}}{1.5} =\big(1+ \frac{1}{2}\big)\times596\text{ nm}$
$\Rightarrow\text{y}_{2} =\frac{3}{2}\times\frac{596\times10^{-9}\times1.5}{2\times10^{-4}}$
$= 6.705\ mm$
Separation between two positions of first maxima
$\Delta\text{y} = \text{y}_{2} - \text{y}_{1}$
$= 6.705 – 6.64$
$= 0.065\ mm.$

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