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Diffraction pattern from a single slit of width \mathrm{0.25 \mathrm{~mm}} is observed with light of wavelength \mathrm{4890\, \AA}. Angular separation between first order minimum and third order maximum, falling on the same side, is:

Option: 1

\mathrm{5.89 \times 10^{-3} \mathrm{rad}}


Option: 2

\mathrm{5.89 \times 10^{-7} \mathrm{rad}}


Option: 3

\mathrm{5.89 \times 10^{-10} \mathrm{rad}}


Option: 4

\mathrm{5.89 \times 10^{-4} \mathrm{rad}}


Answers (1)

best_answer

In single slit diffraction pattern, positions of secondary maxima and minima are given by,

\mathrm{\theta_{\max }= \pm(2 \mathrm{n}+1) \frac{\lambda}{2 \mathrm{~d}} \, \, and \, \, \theta_{\min }= \pm \mathrm{n} \frac{\lambda}{\mathrm{d}}, \, \, respectively.}
\mathrm{\therefore \quad \theta_{1 \min }=\frac{\lambda}{\mathrm{d}}\, \, and \, \, \theta_{3 \max }=\frac{7 \lambda}{2 \mathrm{~d}}}

Angular separation\mathrm{ =\theta_{3 \max }-\theta_{1 \min }=\frac{7 \lambda}{2 d}-\frac{\lambda}{d}}

\mathrm{=\frac{5 \lambda}{2 \mathrm{~d}}=\frac{5}{2} \times \frac{5890 \times 10^{-10}}{0.25 \times 10^{-3}}=5.89 \times 10^{-3} \mathrm{rad}}

Posted by

Pankaj Sanodiya

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