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(A) For a single-slit diffraction pattern,the condition for minima is given by $d \sin 	heta = n \lambda$,where $n = 1, 2, 3, ...$.For the first mini

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$440$

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(A) For a single-slit diffraction pattern,the condition for minima is given by $d \sin 	heta = n \lambda$,where $n = 1, 2, 3, ...$.For the first minimum $(n=1)$,we have $d \sin 	heta_1 = \lambda_1$,which implies $\sin 	heta_1 = \frac{\lambda_1}{d}$.For the first secondary maximum,the condition is $d \sin 	heta_2 = (n + \frac{1}{2}) \lambda_2$. For the first secondary maximum,$n=1$,so $d \sin 	heta_2 = \frac{3}{2} \lambda_2$,which implies $\sin 	heta_2 = \frac{3 \lambda_2}{2d}$.Since the first minimum of $\lambda_1$ coincides with the first maximum of $\lambda_2$,we have $	heta_1 = 	heta_2$,hence $\sin 	heta_1 = \sin 	heta_2$.Equating the two expressions: $\frac{\lambda_1}{d} = \frac{3 \lambda_2}{2d}$.Solving for $\lambda_2$: $\lambda_2 = \frac{2}{3} \lambda_1$.Substituting $\lambda_1 = 660 \ nm$: $\lambda_2 = \frac{2}{3} 	imes 660 \ nm = 440 \ nm$.

In a single-slit diffraction experiment,the first minimum for a wavelength $\lambda_1 = 660 \ nm$ coincides with the first maximum for a wavelength $\lambda_2$. Find the value of $\lambda_2$ in $nm$.

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