The ratio of intensities of consecutive maxim | vedclass

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(C) The intensity of the diffraction pattern is given by $I = I_0 \left[ \frac{\sin \alpha}{\alpha} ight]^2$,where $\alpha = \frac{\pi d \sin 	heta

Vedclass · Accepted Answer

$1 : \frac{4}{9\pi^2} : \frac{4}{25\pi^2}$

Vedclass · Answer

(C) The intensity of the diffraction pattern is given by $I = I_0 \left[ \frac{\sin \alpha}{\alpha} ight]^2$,where $\alpha = \frac{\pi d \sin 	heta}{\lambda}$.For the central maximum,$n=0$,$\alpha = 0$,and $I = I_0$.For secondary maxima,the condition is $\alpha = \left( n + \frac{1}{2} ight) \pi$,where $n = 1, 2, 3, \dots$.Substituting this into the intensity formula:$I_n = I_0 \left[ \frac{\sin((n + 1/2)\pi)}{(n + 1/2)\pi} ight]^2 = I_0 \left[ \frac{\pm 1}{(n + 1/2)\pi} ight]^2 = \frac{I_0}{(n + 1/2)^2 \pi^2} = \frac{4 I_0}{(2n + 1)^2 \pi^2}$.For $n=0$ (central),$I_0 = I_0$.For $n=1$ (first secondary),$I_1 = \frac{4 I_0}{9 \pi^2}$.For $n=2$ (second secondary),$I_2 = \frac{4 I_0}{25 \pi^2}$.Thus,the ratio is $I_0 : I_1 : I_2 = 1 : \frac{4}{9 \pi^2} : \frac{4}{25 \pi^2}$.

The ratio of intensities of consecutive maxima in the diffraction pattern due to a single slit is

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