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 × 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

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

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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