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Wave Optics — Physics STD 12 Science — Question

Maharashtra BoardEnglish MediumSTD 12 SciencePhysicsWave Optics3 Marks
Question
In a single$-$slit diffraction pattern, the distance between the first minimum on the right and the first minimum on the left of the central maximum is $4\ mm$. The screen is $2 m$ from the slit and the wavelength of light used is $6000 \mathring A$ . Calculate the width of the slit and the width of the central maximum.
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Answer

Data $: y _1 \ ($minimum, right$) + y _1\ ($minimum, left$)$
$=4 mm =4 \times 10^{-3} m , D =2 m , \lambda=6 \times 10^{-7} m$
$y_m=\frac{m \lambda D}{a}\ ($for minima$)$
$\therefore$ The width of the central maximum,
$W_{ c }=y_1(\min , \text { right })+y_1(\min , \text { left })=4\ mm$
Also, $W_{ c }=\frac{\lambda D}{a}+\frac{\lambda D}{a}=\frac{2 \lambda D}{a}$
$\therefore$ The slit width$, a=\frac{2 \lambda D}{W_c}$
$=\frac{2\left(6 \times 10^{-7}\right)(2)}{4 \times 10^{-3}}=6 \times 10^{-4} m$

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