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  • In a single slit diffraction experiment first minimum for red light <img alt="\mathrm{(660 \mathrm{~nm})}" src="https://entrancecorner.oncodecogs.com/gif.latex?%5Cmathrm%7B%28660%20%5Cmathrm%7B%7En

In a single slit diffraction experiment first minimum for red light \mathrm{(660 \mathrm{~nm})} coincides with first maximum of some other wavelength \mathrm{\lambda^{\prime}}. The value of \mathrm{\lambda^{\prime}} is

Option: 1

\mathrm{4400 \AA}


Option: 2

\mathrm{6600 \AA}


Option: 3

\mathrm{2000 \AA}


Option: 4

\mathrm{3500 \AA}


Answers (1)

best_answer

In a single slit diffraction experiment, position of minima is given by \mathrm{d \sin \theta=\mathrm{n} \lambda}
So for first minima of red \mathrm{\sin \theta=1 \times\left(\frac{\lambda_R}{d}\right)}
and as first maxima is midway between first and second minima, for wavelength \mathrm{\lambda^{\prime}}, its position will be

\mathrm{ \mathrm{d} \sin \theta^{\prime}=\frac{\lambda^{\prime}+2 \lambda^{\prime}}{2} \Rightarrow \sin \theta^{\prime}=\frac{3 \lambda^{\prime}}{2 \mathrm{~d}} }
According to given condition \mathrm{\sin \theta=\sin \theta^{\prime} }

\mathrm{ \Rightarrow \lambda^{\prime}=\frac{2}{3} \lambda_{\mathrm{R}} \text { so } \lambda^{\prime}=\frac{2}{3} \times 6600=440 \mathrm{~nm}=4400 \AA }

Posted by

Kuldeep Maurya

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