A single slit of width a is illuminated by a | vedclass

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(B) For the first minimum in a single-slit diffraction pattern, the condition is given by the formula:$a \sin 	heta = n \lambda$For the first minimum

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$1.2$

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(B) For the first minimum in a single-slit diffraction pattern, the condition is given by the formula:$a \sin 	heta = n \lambda$For the first minimum, $n = 1$, so the equation becomes:$a \sin 	heta = \lambda$Given:$\lambda = 600 \, nm = 600 	imes 10^{-9} \, m$$	heta = 30^{\circ}$Substituting the values into the equation:$a \sin 30^{\circ} = 600 	imes 10^{-9} \, m$Since $\sin 30^{\circ} = 0.5$:$a 	imes 0.5 = 600 	imes 10^{-9} \, m$$a = 1200 	imes 10^{-9} \, m$$a = 1.2 	imes 10^{-6} \, m$$a = 1.2 \, \mu m$

$A$ single slit of width $a$ is illuminated by a monochromatic light of wavelength $600 \, nm$. The value of $a$ for which the first minimum appears at $\theta = 30^{\circ}$ on the screen will be ........... $\mu m$.

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