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A monochromatic light of wavelenght $\lambda$ is incident normally on a narrow slit of width 'a' to produce a diffraction pattern on the screen placed at a distance D from the slit. With the help of a relevant diagram, deduce the conditions for obtaining maxima and minima on the screen. Use these conditions to show that angular width of central maximum is twice the angular width of secondary maximum.                                                                                              3marks

Explanation:

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The path difference

NP - LP= NQ

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By dividing the slit into an appropriate number of parts, we find that points P for which

(i)                                                              $1}{2}$mark

(ii)                                                    $1}{2}$mark

Angular width of central maxima, $\theta ={\theta }_{1}-{\theta }_{-1}$

$=\frac{\lambda }{a}-\left(-\frac{\lambda }{a}\right)$                                                                      $1}{2}$mark

$\theta =\frac{2\lambda }{a}$

Angular width of secondary maxima =${\theta }_{2}-{\theta }_{1}$

$=\frac{2\lambda }{a}-\frac{\lambda }{a}=\frac{\lambda }{a}$                                                     $1}{2}$mark

$=\frac{1}{2}X$ Angular width of central maxima

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