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(C) For a single slit diffraction pattern,the condition for the $n^{th}$ minimum is given by $a \sin 	heta = n \lambda$,where $a$ is the slit width,$

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$6$

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(C) For a single slit diffraction pattern,the condition for the $n^{th}$ minimum is given by $a \sin 	heta = n \lambda$,where $a$ is the slit width,$\lambda$ is the wavelength,and $	heta$ is the angular position.For the second minimum $(n=2)$,$\sin 	heta_1 = \frac{2 \lambda}{a}$.For the third minimum $(n=3)$,$\sin 	heta_2 = \frac{3 \lambda}{a}$.The total angular separation is $	heta_1 + 	heta_2 = 30^{\circ} = \frac{\pi}{6} 	ext{ radians}$.Using the small angle approximation,$\sin 	heta \approx 	heta$ (in radians),we have:$	heta_1 \approx \frac{2 \lambda}{a}$ and $	heta_2 \approx \frac{3 \lambda}{a}$.Adding these,we get $	heta_1 + 	heta_2 \approx \frac{2 \lambda}{a} + \frac{3 \lambda}{a} = \frac{5 \lambda}{a}$.Given $	heta_1 + 	heta_2 = 30^{\circ} = \frac{\pi}{6} 	ext{ radians}$,we have $\frac{5 \lambda}{a} = \frac{\pi}{6}$.Substituting $\lambda = 628 \ nm = 0.628 \ \mu m$:$a = \frac{5 	imes 0.628 	imes 6}{\pi} \approx \frac{18.84}{3.14} \approx 6 \ \mu m$.

If the measured angular separation between the second minimum to the left of the central maximum and the third minimum to the right of the central maximum is $30^{\circ}$ in a single slit diffraction pattern recorded using $628 \ nm$ light,then the width of the slit is . . . . . . $\mu m$.

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