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