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(B) For a single slit diffraction pattern,the condition for the $n^{th}$ minimum is given by $a \sin 	heta = n \lambda_1$,where $a$ is the slit width

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$4400$

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(B) For a single slit diffraction pattern,the condition for the $n^{th}$ minimum is given by $a \sin 	heta = n \lambda_1$,where $a$ is the slit width and $\lambda_1$ is the wavelength.For the first minimum $(n=1)$,$a \sin 	heta = \lambda_R = 6600\,\mathring{A}$.The condition for the $n^{th}$ secondary maximum is $a \sin 	heta = (2n + 1) \frac{\lambda_2}{2}$.For the first secondary maximum $(n=1)$,$a \sin 	heta = (2(1) + 1) \frac{\lambda_2}{2} = \frac{3}{2} \lambda_2$.Since the first minimum of red light coincides with the first secondary maximum of the other wavelength,we equate the two expressions:$6600 = \frac{3}{2} \lambda_2$.Solving for $\lambda_2$: $\lambda_2 = \frac{6600 	imes 2}{3} = 2200 	imes 2 = 4400\,\mathring{A}$.

In an experiment of single slit diffraction pattern,the first minimum for red light coincides with the first maximum of some other wavelength. If the wavelength of red light is $6600\,\mathring{A}$,then the wavelength of the first maximum will be.....$\mathring{A}$

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