In a single slit diffraction pattern, a light of wavelength \(6000\) A 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 _______ \(\times 10^{-4} \mathrm{~m}\).

\( \text {For } n^{\text {th }} \text { minima } \)
\( b \sin \theta=n \lambda \)
\( (\lambda \text { is } \operatorname{small} \text { so } \sin \theta \text { is small, hence } \sin \theta \simeq \tan \theta) \)
\( b \tan \theta=n \lambda \)
\( b \frac{y}{D}=n \lambda \)
\( \Rightarrow y_n=\frac{n \lambda D}{b} \text { (Position of } n^{\text {th }} \text { minima) }\)
\( \mathrm{B} \rightarrow 1^{\mathrm{st}} \text { minima, } \mathrm{A} \rightarrow 3^{\mathrm{rc}} \text { minima } \)
\( \mathrm{y}_3=\frac{3 \lambda \mathrm{D}}{\mathrm{b}}, \mathrm{y}_1=\frac{\lambda \mathrm{D}}{\mathrm{b}} \)
\( \Delta \mathrm{y}=\mathrm{y}_3-\mathrm{y}_1=\frac{2 \lambda \mathrm{D}}{\mathrm{b}} \)
\( 3 \times 10^{-3}=\frac{2 \times 6000 \times 10^{-10} \times 0.5}{\mathrm{~b}} \)
\( \mathrm{~b}=\frac{2 \times 6000 \times 10^{-10} \times 0.5}{3 \times 10^{-3}} \)
\( \mathrm{~b}=2 \times 10^{-4} \mathrm{~m} \)
\( x=2\)