i. Using Huygens's construction of secondary wavelets explains ho… - Vidyadip

Question

$i$. Using Huygens's construction of secondary wavelets explains how a diffraction pattern is obtained on a screen due to a narrow slit on which a monochromatic beam of light is incident normally.
$ii.$ Show that the angular width of first diffraction fringe is half that of the central fringe.
$iii.$ Explain why the maxima at $\theta=\left(n+\frac{1}{2}\right) \frac{\lambda}{a}$ a become weaker and weaker with increasing $n$.

✓

Answer

$i$. We can regard the total contributions of the wavefront $LN$ at some point $P$ on the screen, as the resultant effect of the superposition of its wavelets like $LM , MM _2, M _2 N$.
These have to be superposed taking into account their proper phase differences.

We, therefore, get maxima and minima, i.e., a diffraction pattern, on the screen.
Maxima and minima are produced when the path difference between waves is a whole number of wavelengths or an odd number of half wavelengths respectively.
Conditions for first minima on the screen
$\operatorname{a sin} \theta=\lambda$
$\Rightarrow \theta=\frac{\lambda}{a}$
$\operatorname{a sin} \theta=\lambda$
$\Rightarrow \theta=\frac{\lambda}{a}$
$\therefore$ Angular width of the central fringe on the screen $($from the figure$)$
$=2 \theta=\frac{2 \lambda}{a}$
Angular width of first diffraction fringe $($From fig$)$
$=\frac{\lambda}{a}$
For the first diffraction, the angular width of the fringe is half that of the central fringe.
$iii$. Maxima becomes weaker and weaker with increasing $n$.
This is because the effective part of the wavefront, contributing to the maxima becomes smaller and smaller, with increasing $n$.

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