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(B) For a single slit diffraction pattern of slit width $a$,the condition for minima is given by $a \sin 	heta = n \lambda$,where $n = 1, 2, 3, ...$.

Vedclass · Accepted Answer

$360$

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(B) For a single slit diffraction pattern of slit width $a$,the condition for minima is given by $a \sin 	heta = n \lambda$,where $n = 1, 2, 3, ...$.For the first minima $(n = 1)$,$\sin 	heta_1 = \frac{\lambda}{a}$.The condition for maxima is given by $a \sin 	heta = (n + \frac{1}{2}) \lambda$,where $n = 1, 2, 3, ...$.For the first secondary maxima $(n = 1)$,$\sin 	heta_m = \frac{3 \lambda}{2a}$.Given that the first maxima of light with wavelength $\lambda$ coincides with the first minima of light with wavelength $\lambda_0 = 540 \ nm$:$\frac{3 \lambda}{2a} = \frac{\lambda_0}{a}$.Canceling $a$ from both sides,we get $\frac{3 \lambda}{2} = \lambda_0$.$\lambda = \frac{2}{3} \lambda_0 = \frac{2}{3} 	imes 540 \ nm = 360 \ nm$.

In a single slit diffraction pattern,if the first maxima of light of wavelength $\lambda$ coincides with the first minima of yellow light $(540 \ nm)$,then $\lambda$ is ......... $nm$.

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