Wave Optics Question 8
Question 8 - 2024 (29 Jan Shift 2)
In a single slit diffraction pattern, a light of wavelength $6000 \mathring{A}$ is used. The distance between the first and third minima in the diffraction pattern is found to be $3 mm$ when the screen in placed $50 cm$ away from slits. The width of the slit is $\times 10^{-4} m$.
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Answer: (2)
Solution:
For $n^{\text {th }}$ minima
$b \sin \theta=n \lambda$
( $\lambda$ is small so $\sin \theta$ is small, hence $\sin \theta \simeq \tan \theta$ )
$\operatorname{btan} \theta=n \lambda$
$b \frac{y}{D}=n \lambda$
$\Rightarrow y _n=\frac{n \lambda D}{b}\left(\right.$ Position of $n^{\text {th }}$ minima)
$B \rightarrow 1^{\text {st }}$ minima, $A \rightarrow 3^{\text {rd }}$ minima
$y _3=\frac{3 \lambda D}{b}, y _1=\frac{\lambda D}{b}$
$\Delta y=y _3-y _1=\frac{2 \lambda D}{b}$
$3 \times 10^{-3}=\frac{2 \times 6000 \times 10^{-10} \times 0.5}{b}$
$b=\frac{2 \times 6000 \times 10^{-10} \times 0.5}{3 \times 10^{-3}}$
$b=2 \times 10^{-4} m$
$x=2$
