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  • The ratio of intensities of consecutive maxima in the diffraction pattern due to a single slit is Option: 1 <img alt="1: 4: 9" src="https://entr

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

Option: 1

1: 4: 9


Option: 2

1: 2: 3


Option: 3

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


Option: 4

\mathrm{1: \frac{1}{\pi^2}: \frac{9}{\pi^2}}


Answers (1)

best_answer

\mathrm{\mathrm{I}=\mathrm{I}_0\left[\frac{\sin \alpha}{\alpha}\right]^2}, where \mathrm{ \alpha=\frac{\varphi}{2}}
For \mathrm{ n^{\text {th }}} secondary maxima \mathrm{d \sin \theta=\left(\frac{2 n+1}{2}\right) \lambda}

\mathrm{ \begin{aligned} & \Rightarrow \alpha=\frac{\varphi}{2}=\frac{\pi}{\lambda}[d \sin \theta]=\left(\frac{2 n+1}{2}\right) \pi \\\\ & \therefore I=I_0\left[\frac{\sin \left(\frac{2 n+1}{2}\right) \pi}{\left(\frac{2 n+1}{n}\right) \pi}\right]^2=\frac{I_0}{\left\{\frac{(2 n+1)}{2} \pi\right\}^2} \end{aligned} }

So \mathrm{I_0: I_1: I_2=I_0: \frac{4}{9 \pi^2} I_0: \frac{4}{25 \pi^2} I_0=1: \frac{4}{9 \pi^2}: \frac{4}{25 \pi^2}}

Posted by

Sanket Gandhi

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