In a single-slit diffraction pattern, the distance between the fi… - Vidyadip

Question

In a single-slit diffraction pattern, the distance between the first minimum on the right and the first minimum on the left is $5.2 \ mm$. The screen on which the pattern is displayed is $80 \ cm$ from the slit and the wavelength of light is $5460 \mathring A $. Calculate the slit width.

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Answer

$\text { Data }: y _1 \text { (minimum, right) }+ y _1 \text { (minimum, left) }$
$\text { Data }: y_1 \text { (minimum, right) }+y_1 \text { (minimum, left) }$
$=5.2 mm =5.2 \times 10^{-3} m , D=80 \ cm =0.8 m ,$
$\lambda=5460 \mathring A =5.460 \times 10^{-7} m$
$ y_m=\frac{m \lambda D}{a}$
$\text { (for minima) }$
$\therefore y_1 \text { (right) }+y_1(\text { left })=\frac{\lambda D}{a}+\frac{\lambda D}{a}=\frac{2 \lambda D}{a}$
$\therefore 5.2 \times 10^{-3}=\frac{2 \times 5.460 \times 10^{-7} \times 0.8}{a}$
$\therefore \text { The slit width, } a=\frac{2 \times 5.46 \times 0.8 \times 10^{-4}}{5.2}$
$\quad=1.68 \times 10^{-4} m = 0 . 1 6 8 \ m m $

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